The AM Forum

hi, this is my first post here so forgive me if i do anything wrong. i'm new to ham radio and don't yet have my license. i've just been really busy lately and haven't gotten around to scheduling a test. i have read the ARRL license study guide and large sections of the ARRL handbook. 

i found the QST article by Rick Campbell KK7B called "Designing and Building Transistor Linear Power Amplifiers" on the ARRL handbook CD-ROM and thought it would be fun to try working through the methodology he presents. i'm not actually broadcasting anything or hooking it up to an antenna so i don't think i am breaking any rules doing this without a license.

i'm using a 2n3866 transistor which i found in a schematic for a low power linear amplifier in the ARRL handbook (page 13.24). i've gotten so far as building the LM317 bias circuit from Figure 4 in Rick Campbell's article and hooking up the transistor as in Figure 3 from the article. I am able to 'turn on' the transistor and i have gotten modest voltage gain on a signal from my function generator.

i am using a 50ohm dummy load as is recommended in the article. In the section titled Load Network (page 3) it says that with a 50ohm load the power output is a product of the supply voltage and that with a 12v supply i can get up to 1.44W of output. i assume that means i should be able to get ~8.5v P-P output across my dummy load (E^2/R=P).

i am sending a 7MHz signal at 100mV from my function generator into the transistor base along with a bias from the LM317 and i am powering the collector with 12V. my collector current and P-P voltage output to the dummy load seem to shift as the transistor heats up. i only have one multimeter so i cant look at bias voltage and collector current at the same time.

in any case the highest P-P voltage output i have achieved is 3.2V which i get when my collector current is ~30mA. the higher the collector current the lower my P-P voltage output. does that sound right?

also, will a heatsink significantly stabilize the collector current from my transistor? i don't have any to-39 heatsinks currently but i have one coming in the mail next week.

i was under the impression that i should be able to get much higher output from one transistor even without a matching network. how do i achieve the 1.44w output mentioned in Rick Campbell's article?

Welcome to the AM Forum.

I'm not familiar with the acticle you're working from, but did want to congratulate you for trying to build something.

Many hams today manage to get a license and then buy everything for a station.  They never build anything.  Most folks you'll find on this forum are builders.

I built my first radio at age 7, that was 58 years ago.  I was given (by a ham) a 1953 ARRL Handbook in 1953.  I still have the book and refer to it often.  I don't think I've mastered everything in the book but I'm going to stay on it until I do.

Keep up the good work.  I'm sure someone on here will come along with more help with your project.

The one thing I will mention,  1.44 watts on 40 meters??  I think you're going to need a bigger set.

And don't worry about doing anything wrong,  there are plenty of experienced hams on here that screw-up all the time (of course, I'm not one of them).

And I will note, that from reading your post, I can see that you have some knowledge of electronics.

Fred, KA2DZT

thanks for the kind words. i'm really excited about getting more into this hobby. i do have a little experience with electronics, mainly audio stuff, but i have always wanted to get into radio. i recently decided to just go for it and have been reading as much as possible. rf electronics is a lot more complex then the kind of stuff i have done in the past!

i admire the the fancy new ham gear that is available for sale, but i really like the idea of building stuff myself. its nice to know exactly what is going on 'under the hood.'  :)

at some point, likely after i get my license, i plan to try building larger gear but for now i want to try building low power stuff to experiment with on the bench. i would like to know i can properly tune a single transistor stage before trying to add additional ones.

Across your 50 ohm load you should see 20 volts Pk-Pk for 1 watt of output. ( with O'scope AFTER the DC blocking cap)

And are you measuring the the output voltage with a multi-meter? or with an oscilloscope?

If you are using a multi-meter, you likely aren't getting an accurate reading.


Are you using figure 3  in the article ?

Also,
The 2n3866 HAS to have a heatsink!
"We" use these devices for 1 watt  driver amplifiers in the SDR transceiver kits I build.

They can get very hot even with the to-5 heatsink on them.
They are good devices, but they are delicate!
\

thanks for the info. i am measuring on an oscilloscope so i am getting Pk-Pk readings.

i have some heatsinks coming in the mail sometime next week and i don't mind burning out some of these transistors testing them before the heatsinks arrive. the main problem i notice with no heatsink is that as the transistor heats up, the collector current changes which causes me to have to constantly adjust the bias voltage to keep a steady power output. at this point its not a big deal because i am just testing out the circuit but hopefully a heatsink will fix this.


yes, i am following figure 3 in the article. the highest voltage i have achieved is 3v Pk-Pk which is what i get with a direct (no resistor) ground shunt from the emitter, and 90-100mA current on the collector.

The only discrepancy i notice between my circuit and that of the articles is the RF Chokes. the article does not say what inductance values to use for the chokes and so i put a 15uH one on the bias input and a homemade one made of 8turns on a FT82-43 (http://toroids.info/FT82-43.php) on the power input (like in the schematic). these are the only inductors i have available, although i have additional FT82-43s and FT37-43s that i could wind if necessary.

when testing a transistor i plan to use for the first stage of amplification, what sort of signal input should i be using? i have a nice HP function generator i got from a swap meet which i have been using. i set it to 100mV sine wave input at whatever frequency. is that the appropriate voltage i should be using to simulate the output from a transmitter's oscillator?

also, if i wanted to drive a transistor i plan to use in a different stage, what would i use? do i just use E=sqrt(P*R) [where P is whatever power i plan to input into the transistor] to determine the voltage of the function generator?

S,

I'd sugest that u go to the homely despot and buy some copper pipe caps. Drill a hole through the cap and shove the transister case through the hole. If u get something of a tite fit, you'll get some enhanced cooling. Or you can place the case upside down in a shallow plate of water and get to chillin this way. Dont spank the silicon.


Whatt kind of biass does this de vice have??

klc

i only have one multimeter so i can't monitor current and bias voltage at the same time. that combined with the fluctuating current due to heating makes it difficult to correlate particular voltages with particular currents but i am running it around 0.6 to 0.7vdc for bias and 12vdc for supply.

i guess when i think about it getting 3v from .1v is a factor of 30. is that as much as i should hope to achieve from one transistor?

The 2N3866 is a beast but if you want 1 watt output you will need to dissipate over 1 watt in a heatsink. Also look at the transformer output ratio. This will effect the peak to peak output voltage. I've never run them over 1/2 watt output myself. Usually these devices run anroud 17 dB of gain with feedback. They will run higher gain but could oscillate without feedback. They will operate at over 200 MHz.

i just got the heatsinks in the mail. this is the type i got: http://www.westfloridacomponents.com/G280APE12/TO5+TO39+Heat+Sink+Seifert+Electronic+KK-502.html

they are describe as 'snap-on' but they fit very loosely over the transistor. am i supposed to crimp them down?



WA1GFZ, maybe this is exactly my problem. my circuit does not have any transformer at all. heres a drawing of my circuit: http://i.imgur.com/j7Pl7.png

i'm not using any feedback resistors or anything. the bias is coming from an lm317 circuit, eventually i will build a more proper bias circuit but i am using the lm317 as directed in the article i am following.  the 50ohm resistor is my dummy load.

what is the purpose of the transformer you mentioned? is it part of a matching network or something

these heatsinks made a huge difference. i now get a very stable 3v Pk-Pk output from the 2n3866. i'm still not anywhere near the 1w output i want to achieve but at least the heating issues are dealt with.

Hi!

As drawn, if the transistor were an ideal transistor, your circuit would have an RF voltage gain of 50.

That is, the 7MHz current flowing through the 1 ohm emitter resistor would be almost the same as (approximately 101% of) the 7MHz current flowing through the 50 ohm RF load.

As a practical matter, the voltage gain will be somewhat less than 50 because:

1. There is some additional emitter resistance associated with the transistor itself... so the effective emitter resistance is somewhat larger than 1 ohm. Likewise, there is a small effect from the base resistance of the transistor that acts to reduce the 7MHz voltage across the effective emitter resistor (i.e. total of 1 ohm and the transistor's added emitter resistance) v. the applied 7MHz base-to-ground voltage. So, in summary, the rf voltage across the actual 1 ohm emitter resistor will be somewhat less than the applied input rf voltage (measured from base to ground).

2. There is some parasitic capacitance from collector-to-emitter associated with the transistor itself, the wiring, the rf choke, etc. that provides an impedance to ground that is in parallel with the 50 Ohm load. So not all of the 7MHz collector current flows through the 50 Ohm load.

Remember: depending upon the type of probe you are using, your oscilloscope probe, by itself, may have an input capacitance of 10-15 pF. If, instead of using a probe, you run a coaxial cable (or a twisted pair) between the 50 ohm load resistor and the scope... with the scope set to high impedance input... the added parallel capacitance across the 50 ohm load will be around 20pF per foot of cable. At 7MHz, an unterminated 6ft cable (not terminated with 50 Ohms at the scope end) would have an impedance of arround -j189 ohms. You would get a more accurate measurement of the peak-to-peak voltage across the 50 ohm load by placing the 50 ohm load at the scope end of the cable (if you haven't already done so)

Therefore, the actual RF voltage gain will be C x A/B... where A is somewhat less than 50 ohms, and B is somewhat larger than 1 ohm, and C is an input attenuation factor that is somewhat less than 1.

If  C x A/B is 30 (instead of the ideal value of 50), then 100 mV of peak-to-peak rf input will produce 3 volts of peak-to-peak rf output.

Good luck! I'm looking forward to your progress updates

Stu

so what you are saying is that a voltage gain of 30 (100mv to 3v) is perfectly normal and should be expected. is it possible to get greater gain without using additional transistor stages?

if not, what is the best way to go about building additional stages. without any success, i tried running the output of one 2n3866 into a second one, both transistors being setup exactly the same. i the output of the second transistor was only a few hundred mV.

i've done the same thing 2n3904 transistors at audio frequency levels and gotten rather large gain.

There are many ways to obtain more gain. Let's start with using two stages of gain... as you mentioned. [We can then talk about using a 1:2 step up transformer at the output of the 2nd stage, in conjunction with some other changes]

You said that when you placed two identical stages in series, you obtained very little output from the 2nd stage.  

I would suggest that you try it again, keeping the following in mind:

Note: there are a thousand different ways to do this... so let's pick one... and make it work (-:

1. Make sure that the capacitor (C1= 0.1 uF) shown between each stage's collector and its corresponding 50 ohm load resistor is in place in both stages. [I.e., use a 50 load for each stage]

2. In addition, make sure to keep the 0.1uF capacitor at the input of the 2nd stage. This will ensure that the dc biasing of the 2nd stage is not disturbed by the 50 ohm load at the output of the first stage.

3. Use a larger value for the emitter resistor in both stages (try 2.5 Ohms). This should result in a gain of around 13 (maybe a little more) per stage... for a total voltage gain of around 169.

4. If you happen to achieve a total voltage gain that is larger than 200, then you will have to reduce the input peak-to -peak voltage, because your power supply can only support a peak output voltage swing that is around 2 volts less than the power supply voltage. I.e. if your power supply voltage is 12 volts... and if you adjust the rf input voltage level so that the collector voltage of the 2nd stage swings between 22 volts and 2 volts, with rf applied to the input... then that will still leave (at the bottom of the swing) 2 volts from collector to ground. Some of that 2 volts will be across the emitter resistor, and the rest will be needed between collector and emitter to keep the transistor working properly.

5. You will have to adjust the base bias voltage on the 2nd stage to be around 1.7 volts... so that the emitter resistor of the 2nd stage has a voltage across it that swings up and down around a dc value of 1 volt. You should also adjust the base bias of the 1st stage to be around 1 volt, so that there will be around 0.3 volts across the emitter resistor of the 1st stage.

As an aside, in your posted circuit diagram, both the DC voltage sources are drawn with the polarity reversed from what it should be.


Good luck

Stu  

AB2EZ,

i'm wondering what the best way to setup my bias would be. i've been using an lm317 but that doesn't seem like a good long term solution. i just tried a voltage divider but with poor results. i was trying to replicate my current 3v Pk-Pk results before moving on to the suggestions you made.

first i tried 150ohm and 10ohm for R1 and R2 of the voltage divider. i got 200mV output and a burnt finger from an over heating resistor. then i bumped the values up to 15000ohm and 1000ohm. that gave me a 825mV output at 7MHz and a deformed sine wave. at 13MHz i get a normal looking sine wave and a 1v output.

i've read about a couple other bias circuits such as these: http://www.qsl.net/va3iul/Bias/Bias_Circuits_for_RF_Devices.pdf
but haven't found any formulas for generating the proper component values.

how would you recommend i setup the bias?

Getting the bias set and stable (with transistor temperature changes) will be harder with the 1 ohm emitter resistor in the circuit than it will be with a 2 ohm or a 2.5 ohm or a 3 ohm emitter resistor in the circuit. As the transistor heats up, the optimum bias voltage will change a bit.

Try this for the biasing:

[Again... there are a 1000 ways to do this. This approach is not as elegant as some alternative approaches. Resist the temptation of jumping from one approach to another, until you get one approach working. Then, as you begin to understand the details of how the circuit works, you can experiment with the other 999 ways... in order to do some optimizations of: the overall energy efficiency of the amplifier, automatic compensation of the bias voltage for transistor temperature changes, the component count, etc.]

1. Change the emitter resistor from 1 ohm to 3 ohms.

Note: if you use a new emitter resistor value of 3 ohms, you will reduce the single-stage gain from its present value of around 30 to a value of around 30/3 = 10 (probably somewhat more than 10... maybe closer to 13, maybe a bit more than 13). Nevertheless, you should increase the value of the emitter resistor in this particular circuit configuration. After you get a simple 2 stage amplifier working properly... you can consider alternative designs.

2. Add a resistor, from base to ground, whose value is approximately 50 x the value of the new emitter resistor. If the new emitter resistor value is 3 ohms, then the resistor added from base to ground should have a value of 150 ohms.

[Note: looking into the base of the transistor, there is an intrinsic resistance from base to ground... whose value is (approximately) beta x the emitter resistance value. Beta is a number around 100, but according to the 2N3866 data sheet, it could be significantly lower. Beta is temperature dependent. To make the biasing reasonably stable with temperature, the resistor that you add, from base-to-ground, should have a value that is substantially less than: beta x the emitter resistance.]  

3. For the 1st stage (initially the only stage), the resistor from the 12 volt supply to the base should have a value of approximately 12 x the value of the resistor you use from base to ground. If the resistor added between the base and ground is 150 Ohms, then the resistor from the base to the 12 volt supply should have a value of 1800 ohms. The wattage rating of the 1800 ohm resistor should be at least 0.25 watts (not 0.1 watts).

Note: the resistor you add from base to ground will act not only as part of the biasing circuitry. It will also act as an additional rf load on whatever signal source is used to drive the base of the transistor. Therefore, depending upon the characteristics of the 100mV rf source... the rf voltage it produces at the input of the transistor will be somewhat less than before.

4. For the 2nd stage, the resistor from the 12 volt supply to the base should have a value that is around 8 x the value of the resistor that you use from base to ground. If the resistor added between the base and ground is 150 Ohms, then the resistor from the base to the 12 volt supply should have a value of around 1200 ohms. The wattage rating of the 1200 ohm resistor should be at least 0.25 watts (not 0.1 watts).

Note: You are biasing the 2nd stage so that the dc voltage across the emitter resistor (i.e. between the emitter and ground) will be around 0.65 volt. This will allow the emitter-to-ground voltage to swing up to at much as 1.3 volt and down to as little as 0 volts (1.3 volts peak-to-peak of rf emitter voltage, without turning the transistor completely off). The actual peak-to-peak voltage that you will want across the emitter resistor of the 2nd stage, to obtain 20V peak-to-peak output from the 2nd stage will be approximately 1.2V.

With this biasing, the average emitter current will be 0.65V/3 ohms = 217 mA. The heat dissipated in the 2nd stage transistor will be approximately 11.35V x 0.217A = 2.5W. Therefore, only run the amplifier for short periods of time until you determine that the 2nd stage transistor is well enough "heat-sinked".

The bias needs to be temperature compensated.
I talked about an output transformer because usually a 2:1 stepdown is used to drive 50 ohms with a TO5 device like the 3866.
Consider you want 1 watt of power which is about 7 volts RMS across the 50 ohm load. That is about 20 volts peak to peak so not possible with a 12 volt power source. I like the 2N3375 when I want to make power above 1/2 watt.
It is a stud mounted device that will directly drive 50 ohms.
Now you could make the 2N3866 make that power if you increased the collector voltage or added a 1:2 step up ratio transformer. Either of these options would
really push the maximum ratings. You might consider a push pull pair.

Frank

In the circuit shown, the collector is fed by an inductor (a "collector choke"). Therefore, the peak-to-peak output swing can be as large as 2 x [B+ - 2volts] = 20V.

I agree that the final design should employ temperature compensated biasing of the 2nd stage... but since SWL is apparently in a steep learning mode, I suggest that temperature compensated biasing, etc. be left as embellishments of a simple initial design. For example, running the amplifier in Class A (as per the initial design) is probably not the best approach for the final design. Therefore, as part of the embellishment process, an output filter will need to be added.

Stu

Note that I have made some edits to the suggested values for the biasing resistors in my prior post.

Stu,
I'm not sure a single ended amplifier would be clean if run outside of class A mode unless the inductor was changed to a tuned tank.

Frank

The inductor will block the fundamental and all of the significant harmonics. The "output filter" that I referred to is equivalent (for the purposes at hand, where we are not looking for a large impedance change between one side of the filter and the other side of the filter) to a "tuned tank" circuit. So I think we are saying the same thing. [Some would say that every LC filter stores energy...and is therefore a collection of tuned "tank" circuits].

I also agree that the filtering will be somewhat easier to accomplish... for a given set of target ratios of: 2nd, 3rd, etc. harmonic power, to the fundamental component's power...  if a balanced design is used to cancel out the even harmonics. But, either way (single ended or balanced), at this output power level, obtaining large amounts of harmonic suppression is less of a concern.

The output filter should be designed to provide a relatively low impedance path to ground for all but the fundamental frequency component of the collector current (just like a tuned tank circuit would). It would probably be a Chebyshev bandpass or lowpass filter. Therefore only the fundamental frequency component of the collector current will flow into the full value of whatever load impedance / resistance is used.

Separately, as you pointed out, a broadband transformer can be used, if needed, to better match the transistor(s) in the output stage to the load.


Stu

sorry i didn't respond sooner. i don't have those R values in 0.25w rating so i am heading over to the electronics shop today to try to find them.  i will report back when i have tested out the circuit.

for the base-ground resistor you mention that it should be significantly lower than Beta*Emitter Resistance. with that in mind, why do you recommend a 150ohm resistor?

You asked:

"for the base-ground resistor you mention that it should be significantly lower than Beta*Emitter Resistance. with that in mind, why do you recommend a 150 ohm resistor?"

If the base-to-ground resistor is too small, then the electrical power consumption in the voltage divider (and the associated heating of the resistors) will be too high. The power consumed in each base-bias voltage divider will be (roughly) 12V x 12V / [the sum of the two resistors that comprise the voltage divider]. The electrical power consumed by the 2nd stage biasing circuit will be 12V x 12V / [9 x 150 ohms] = 0.1 watt. I didn't want to make the power consumption any larger than that, because I don't think it is necessary for this iteration of the design.


I think that 150 ohms will be a good compromise (for the time being) if the emitter resistor value is 3 ohms and if beta is greater than 50 for the specific transistors you have. The 2N3866 specification sheets (from various manufacturers) specify the value of beta as being somewhere between 10 and 200 (depending upon the specific device). That's an enormous range for this parameter. The specified range for a 2N3866A is between 25 and 200. Hopefully the devices you have are on the high side of the range.The earlier results you reported suggest that betas of your devices are on the high side of this range.

After you get the basic design working... your next step will be to transition to a more sophisticated biasing approach that will do a decent job of keeping the bias adjusted properly (as the transistor temperature changes) without wasting a lot of electrical power.

stu,

i just finished testing the setup you recommended and i didn't get the exact results you described.

with the first stage transistor setup i get ~300mv Pk-Pk output from a 100mV signal input. with the second tranasitor setup as well, i got 1.2v Pk-Pk which rapidly declined to ~600mV at which point i turned off the circuit.

i checked the current between the emitter and ground and it was at ~130mA. the current between the collector and the coil was ~120mA.

Okay

Can you produce a complete schematic of what you have now and post it here?

Stu

I suggest you find yousellf a copy of the ARRL Soild State handbook. It has a great section on broadband amplifiers. Sounds to me the second stage is putting a heavy load on the first. also I have never biased a 3866 over about 65 ma. That poor thing is begging for mercey

heres my schematic:
http://i.imgur.com/Nlh8a.png
i forgot to label all the capacitors but they are all 0.1uF. for some reason i i didn't get any signal if i used only one capacitor on the output of Q1 like this: http://i.imgur.com/1cc8T.png
why is that?

should i be using a different transistor for the second stage? i have a couple of these guys: http://www.fairchildsemi.com/ds/KS/KSE44H.pdf

Okay

The circuit you have shown is fine (except the 12V voltage source is drawn upside down). The schematic would look better (but the circuit would not work any differently) if the left side of capacitor C3 was connected to the point where C1 and R2 come together.

The reason you need C3 is to prevent R2 from affecting the biasing of the 2nd stage. Without C3, the combination of R2 (50 ohms) and R6 (150 ohms) in parallel would form the bottom resistor of the voltage divider. The result of that would be to have much too low a base bias voltage on the 2nd stage transistor, and the 2nd stage transistor will not be turned on.

Now to proceed...

Step 1. Temporarily disconnect the top of R5 from the 12 volt supply. This will remove the bias from the 2nd stage... and therefore the 2nd stage's transistor will not be turned on.

Step 2. Do not (yet) apply the 100 mV input signal (i.e. disconnect the input signal from the left side of C2).

Set 3. Turn on the 12V supply (the 100mV rf signal is not connected), and measure the following: a) the DC voltage from the base of the 1st transistor to ground; b) the DC voltage from the emitter of the 1st transistor to ground (i.e. the voltage across the 3 ohm emitter resistor).

Let me know what your measurement results are

Stu

emitter voltage: 0.23V
base voltage: 0.8V

how do those numbers look?

The emitter voltage is okay (and about what I was expecting it to be)

The base voltage is too low. There are several possible reasons for this

a) Recheck your measurment. Make sure you are measuring the voltage from base to ground (not from base to emitter)

b) Make sure your measurement instrument has a high input impedance... i.e., you don't want to use a measurement instrument that has a 50 Ohm load built into its input. If you are using the oscilloscope to measure the dc bias, make sure the oscilloscope's front panel input impedance control is set to the "high impedance" input position.

c) Check the voltage between the collector of the 1st stage transistor and ground. What is your measurement result?

i've been using a Fluke 117 digital multimeter. i checked again and i got roughly the same values. i don't have my oscilloscope (or the second transistor) hooked up at all when doing these tests. the collector-ground voltage is 12.8V



what sort of voltages should i be expecting?

I see you are using LT Spice which is fine. I have used Sandia Labs 2n3866 model in switcher CAD with good results at frequencies to 200 MHz. If you bias the device near class C and allow yourself to adjust the collector impedance, than collector eff. can be quite good, above 50 %. Now the thermal problems assocaited with class A are not to restrictive and getting a watt from this device is possible. At a class A bias point it will be quite difficult! In any case, the nonlinear model for the 3866 will function quite nicely in LT Spice and you can experiment there and try a vast number of impedance matching approaches without picking up a soldering iron! Of course, this is a model, and in the end you will have to build the network and see how results match simulation. If you need more power... I have placed multiple 3866 devices in parallel... a bigger BIP device. Last count, I placed 10 in // and easily achieved several watts with 50+% collector efficiency. Again bias points are critical and all of this was CW, not AM. Linearity in my case was not an issue.

Alan

On this post for bias. To the extent that the device beta is not infinite, you are going to have to support a base bias current. So... your base bias network may not be "low enough in impedance value"... as a rule of thumb, run a base bias network current at least 1/10 x the collector current... so 50 mA Icq, set the base bias network up for 5 mA. Again, if there is a disconnect on the bench measure, your LT Spice and 3866 model should alert you to the disconnect. I have not looked at the details of your sch. ok, your base bias string is a bit lite at 6.2 mA. I would stiffen it up a bit. GL.

I would have expected the dc base voltage (to ground) to be around 0.7 volts + the voltage from emitter to ground... but let's not worry about that for now.

Step 4. Keep the 2nd stage transistor on the "biased off" mode. Disconnect 1 side of C3, so that the 2nd stage is completely disconnected from the 1st stage

Step 5. Reconnect the 100 mV RF source to the input of the 1st stage (via C2), and measure the peak-to peak rf voltage at the following 3 points: base to ground; emitter to ground; collector to ground.

Let me know what the peak-to-peak reading are.

Stu

heres the readings stu:

emitter-ground: 60-70mV
base-ground: 60-65mV
collector-ground: 400-430mV
across the 50ohm load: 360-370mV
*all these values taken with the ground probe plugged into the breadboard directly next to the power supply negative probe.

its hard to confirming any of these readings. they start to fluctuate if i leave the power supply on, if i adjust any of the parts in their seating on the breadboard, and depending where on the breadboard i ground my oscilloscope and function generator. changes in any of those 3 factors can result in up to 100mV (and sometimes more) shifts in the readings on the collector and across the 50ohm resistor.

i'm also getting some weird 'theramin' like behavior when i connect the scope to the collector. i have the scope connected to the collector and ground. when i move my hand close to the transistor the voltage jumps to 680mV but when i am away from the transistor it is 400-430mV.

the emitter-ground and base-ground waves looked kinda noisy and the emitter-ground one looked a little 'squashed' and noisy. is there a website that shows images of common waveform distortions that i could browse to give you better visual descriptions?


w4amv,

i am using ltspice to draw schematics but i haven't gotten around to learning how to actually run and understand the simulations. is there a good book you can recommend on using spice/ltspice?

First off, the base bias on the 3866 per your schematic would be .923 V and assuming 0.7 V drop (nominal VBE) would place the emitter at .225 V. This is just a voltage divider rule calculation. So with your emitter R at 3 ohms, you would expect an emitter current of ~ 75 mA. Of-course this is ideal and you may not see this since device beta is not infinite.

I suggest you open up LT Spice and start by loading the EXAMPLE files and study how these circuits are built in simulation. There is an extensive help menu in LT SPICE and the example files are rich in content. I would put the 3866 to bed. Grab yourself a 2N2222... I nice simple Silicon transistor or better yet a 2N3904 or its complement, a 3906. The models for these devices ARE ALREADY IN LTSPICE. Now go build yourself a nice simple audio amp in LTSpice and get it working in LT Spice.  You already know how to do schematic entry, BUT YOU REALLY ARE missing a very powerful and cool simulator that is FREE. I have built some pretty involved circuits in LT Spice and its capabilities are great. It will provide you with a DC solution so you can check your DC bias points and then you can begin using the AC and transient analysis capabilities to look at far more complex questions. Bottom line... it is a great teaching tool. You can even model tubes and simulate those networks as well. Have FUN. On your direct answer on LT Spice text. LT Spice IS JUST SPICE... Berkley Spice developed back in the ~ 70's been around for quite some time but LT has a excellent GUI. SPICE (Special Program for Integrated Circuit Evaluation) has numerous texts available, you may be able to find them at your local library. Your time invested in learning how to use it will be well invested.

Alan 

Based on your latest measurements and observations... I suggest that the next step is to work on incorporating a ground plane and improving the overall component layout.

At radio frequencies, long leads can act as inductors, and also radiate electromagnetic fields that cause circuits to exhibit (among other things) the kinds of proximity effects you describe.

Here is what I suggest

1. The entire circuit should be (re)built on top of (but, of course, not touching) an electrically conducting surface (usually referred to as a "ground plane"). For this prototyping activity, you can use something ad-hoc. For example, you could take a piece of scrap wood and cover it with a single, continuous piece of aluminum foil or a disposable aluminum cookie sheet.

2. You need to provide some electrical connection points in order to make connectons between some of the components and the ground plane. You could do this (for eample) by attaching a few short pieces of copper wire to the aluminum ground plane with screws and washers. [You can't solder to aluminum; but you can make a good electrical connection by just pressing the copper wire against the aluminum]. Then, when you need to connect a component to "ground", you can just solder it to the nearest copper wire connection point.

3. Lay out your circuit (starting with just the 1st stage, for now) on top of the ground plane... keeping all of the connections as short as possible (preferably every connection between two components, or between a component and the ground plane, should be less than 1 inch in length. If necessary, you can use pieces of electrical tape on top of the ground plane to keep components from touching the ground plane.

4. With respect to the 12 volt power supply:

Use a piece of electrical zip cord, or a twisted pair of wires to connect the + and - sides of the power supply to the circuit. Also (important), at the circuit end end of this connection, close to the circuit components, place one of your 0.1uF capacitors from the +12V lead to the nearest connection point on the ground plane. This will have the effect of making your circuit behave as if the 12 volt power supply was right next to the circuit.

5. With respect to the 100mV rf input signal. Make sure that the connection beteen the 100 mV rf source and the circuit is via a cable with 2 conductors (coaxial cable or a twisted pair). At the circuit board end of this cable, one conductor should be connected to the left side of C2 and the other conductor should be connected to the nearest connection point on the ground plane.

6. When making measurements, use one of the connection points on the ground plane as the place to attach the ground lead of your oscilloscope probe.

Use your ohm meter to verify that all of the connection points attached to the ground plane are electrically connected via the ground plane.

Doing the above (remember, you are still working with just 1 transistor stage) should: greatly reduce the proximity effect you observed, and greatly reduce the noise levels you observed... and will probably increase the ratio of the rf signal between the collector and ground to the rf signal between the base and ground.

Then you can move on to the next steps (e.g., to make some adjustments in the biasing to get rid of the distortion you observed on output waveforms).

Stu

Here are the LTSPICE simulations for the 1st stage:

L= 100uH
V1 = 12.8V
V2= 0.1V peak-to-peak, 7MHz

Blue= base-to-ground: 0.95V average (DC) value and 0.1V peak-to-peak
Green = emitter-to-ground: 0.085V average value (DC) and 0.08V peak-to-peak
Red = across the 50 ohm load resistor: 0V average (DC) and 1.25V peak-to-peak (some distortion)

Voltage gain = 1.25 V / .1V =12.5
Average collector current = 28.4mA

Stu

Here are the LTSPICE simulations for the 2nd stage:

L= 100uH
V1 = 12.8V
V2= 1.25V peak-to-peak, 7MHz
Note R4= 900 Ohms

Blue= base-to-ground: 1.6V average (DC) value and 1.25V peak-to-peak
Green = emitter-to-ground: 0.64V average value (DC) and 1.17V peak-to-peak
Red = across the 50 ohm load resistor: 0V average (DC) and 19.4V peak-to-peak (some distortion)

Voltage gain = 19.4 V / 1.25V = 15.5
Average collector current: 214 mA

Stu

Here are the LTSPICE simulations for both stages:

L= 10uH [As long as L>10uH, the value of L doesn't matter]
V1 = 12.8V
V2= 0.08V peak-to-peak, 7MHz [Note that I have reduced the rf input slightly from 100mV to 80mV peak-to-peak]

Note R8= 900 Ohms; and I have also removed the 1st stage's 50 ohm load. That is: the combination of R2 and R8, and the input impedance of the 2nd stage's transistor, is now the rf load on the 1st stage. I have also added a 50 ohm equivalent series resistor, R5, to the input rf source (V2).

Blue= stage 1 base-to-ground: 0.95V average (DC) value and .057V peak-to-peak
Green = stage 1 emitter-to-ground: .085V average value (DC) and .045V peak-to-peak
Purple = stage 2 base-to-ground: 1.6V average (DC) and 1.21V peak-to-peak
Red = across the 50 ohm stage 2 load resistor: 0V average (DC) and 18.8V peak-to-peak

Voltage gain = 18.8V / .08V = 235
Average stage 2 collector current: 216 mA

Stu

good luck running a 2N3866 at that current

Frank

I believe you!

I think the 2N3866 is the wrong choice for operation at this frequency and this power level:

a) It has too low a value of HFE (beta), in general; and certainly at collector currents above 50mA
b) Its f_T (unity gain frequency) is much higher than is needed for this application... leading to stability problems with simple layouts that would normally be adequate for this application.
c) Its maximum power dissipation rating is marginal for this application.

Once SWL gets the hang of things... he will probably need to switch to a different transistor in the final design.

Stu

Hi Stu. I believe in you simulation that NPN Q1 and Q2 are the default "ideal" LT models. You need to right click on the device, pick a new device, and load in a real model. There are about 30 or so NPN devices in the model library, starting with the 2N2222, 3904 next, etc.... The 3866 may be in the latest version of LT, not sure. If not the Sandia Labs site has a reasonable model. Or, you can build your own model from measured data.

Alan

Here is one model description for the 2N3866, not the one from Sandia Labs, but give it a try. The BF seems to large. It might be E2, not E3 !


******
* Spice Model
* Item: 2N3866
* Date: 8/11/10
* Revision History: A
* ==========================================================
* This model was developed by:
* Central Semiconductor Corp.
* 145 Adams Avenue
* Hauppauge, NY 11788
*
* These models are subject to change without notice.
* Users may not directly or indirectly re-sell or
* re-distribute this model. This model may not
* be modified, or altered without the consent of Central Semiconductor Corp.
*
* For more information on this model contact
* Central Semiconductor Corp. at:
* (631) 435-1110 or Engineering@centralsemi.com
* http://www.centralsemi.com
* ==========================================================
******
*SRC=2N3866;2N3866;BJTs NPN; Si;55.0V  0.40A  500MHz   Central Semi Central Semi
.MODEL 2N3866  NPN (
+ IS=13.232E-12
+ BF=1.2948E3
+ VAF=100
+ IKF=.13116
+ ISE=785.43E-12
+ NE=1.7362
+ BR=1.0408
+ VAR=100
+ IKR=3.3880
+ ISC=13.232E-12
+ NC=1.4860
+ NK=.72187
+ RB=8.9057
+ RC=1.7862
+ CJE=31.375E-12
+ VJE=.87558
+ MJE=.31831
+ CJC=7.3111E-12
+ VJC=.76419
+ MJC=.30725
+ TF=178.96E-12
+ XTF=10.044
+ VTF=10.275
+ ITF=5.8742
+ TR=10.000E-9 )
******

you will need a supply of liquid gas to make a 2222 or 3904 run at that current. The 3866 is a beast and as I said a week ago you are asking way too much power from it. The guy who did that amplifier article is smoking dog poo.
I've been playing with the 3866 for almost 30 years.

Alan

I will try using that SPICE model a little latter this evening...

Frank

Off the top of your head, can you suggest a good choice of transistor for:

1 Watt output, class A (single ended), and <30 MHz applications?

Stu

Hi Stu. Here is the Sandia Labs model. I would use this. Note, the BF is reasonable in this listing.  :)

Let us know how it works out. I agree with the comments on 1 watt levels from the 3866... however, if you optimize the drive level, bias point, and conduction angle, it may be possible to do quite well. I know Rick and he conducts a class on high eff. PA design... you can see his notes and lecture slides on the WEB.

.model 2N3866/27C  NPN     (
+         IS = 9.798605E-15
+         BF = 145.568899
+         NF = 1.007933
+        VAF = 64.3030691
+        IKF = 0.3661244
+        ISE = 1.806705E-14
+         NE = 1.6207001
+         BR = 10.471
+         NR = 1.0003673
+        VAR = 8.322
+        IKR = 0.1449443
+        ISC = 3.326752E-15
+         NC = 1.1076801
+         RB = 15.986
+        IRB = 2.530217E-3
+        RBM = 0.01
+         RE = 0.02604
+         RC = 1.0359
+        CJE = 9.055532E-12
+        VJE = 0.6761546
+        MJE = 0.2754969
+         TF = 1.25476E-10
+        XTF = 13.0616413
+        VTF = 0.4699
+        ITF = 0.2828
+        PTF = 18.9645325
+        CJC = 7.054363E-12
+        VJC = 0.5769848
+        MJC = 0.3139067
+       XCJC = 1
+         TR = 9.098362E-8
+        CJS = 0
+        VJS = .75
+        MJS = 0
+        XTB = 1.831
+         EG = 1.11
+        XTI = 5.0205
+         KF = 0
+         AF = 1
+         FC = 0.9
+ )

A 2N3866 as a linear amplifier is good for maybe .25 watts with heavy bias.

The 2N3375 family of stud mounted devices. The 3375 has a 7.5 watt dissipation rating and can be used as a VHF power amplifier. Check the family that goes to 10 watt dissipation if I remember. I used one as the LO amplifier in my homebrew RX making >1/2 watt. The 2N5109 is a higher frequency version of a 3866. I think ft is above 1500 MHz. A pair in push pull will easily make 1/2 watt.  the 2N4427 another good device.
The ARRL Solid State has some very good information on broadband amplifiers for low power and RX use.

hi everyone. i just got home from an insane day at work. all this information looks fantastic and i will go through it all carefully when i am less exhausted tomorrow. thank you so much for all the help!

i figured out how to test circuits in LTSpice, hurray! this will be much more convenient for testing. how do i check current levels in the simulation?

stu, i copied your spice models and got the same results as you. when using the 2n3866 model the wave form is not quite a perfect sinewave. i'm attaching an image of my output with both stages running and a 2n3866.  http://i.imgur.com/kaljg.png

will this biasing method (voltage divider) be okay if i were to switch to a transistor that can better handle high current or am i going to need to setup a biasing system with temperature compensation?

Hi

Good! LTSPICE is very useful for designing circuits... and the price is right. It has lots of capabilities that I have discovered, over time... and lots more that I haven't yet discovered.

A good tutorial reference is http://denethor.wlu.ca/ltspice/#IIIC

Plotting a branch current:

If you put your mouse pointer over a component that is in a branch (e.g. a resistor or an inductor), an arrow will appear. If you click on the arrow, the current in that branch v. time will be plotted. The reference direction of the plotted current will correspond to the direction of the arrow. There are (of course) ways to make the arrow point the other way. One way is to rotate the component by 180 degrees.

Modeling with a real transistor:

Yes... I tried using a number of different real transistors (but not yet a 2N3866) and obtained the same flattening of the lower half of the output waveform. The flattening appears to be associated with the voltage between the collector and the emitter being a bit too low. To fix the problem (at least in the simulation) I added a separate 18V voltage source for the collector (not the biasing) of the 2nd stage. I.e. the top of the 2nd stage inductor is moved from the 12.8V voltage source to the separate 18V voltage source, but everything else stays the same.

In a future implementation of a real amplifier, this moderate amount of distortion will not matter, because higher harmonics will be removed from the output by an output filter.

Using other transistors/biasing:
 
This biasing approach is okay for your first prototype. Once you get a basic, prototype, amplifier working there are a number of alternative approaches for biasing that will do a good job (or even a very good job) of keeping the bias adjusted properly as the transistor parameters change (with temperature or aging). My choice would be to use an op-amp to sample and compare the average emitter voltage to a stable reference voltage... and then employ negative feedback control to adjust the base bias voltage to force the average emitter voltage to equal the reference voltage.

Stu

wow! theres about 370mA on the emitter of Q2. isnt that going to just burn out the transistor?

this circuit gets 10v Pk-Pk which i believe makes it 2watts. is it always the case that when you are putting out that much power you have really high emitter current or would a different amplifier design allow me to get this kind of power output without such high current?

the 2N3375 isn't available on digikey and NTE16003—which is an equivalent transistor according to mouser—is $45. is there some other transistors i could use to get into this power range 1-2w that are cheap enough that i can buy a bunch to experiment with?

The rf power flowing into the load, at the fundamental frequency of operation = i x i x R/2;  where i= the peak value of the fundamental frequency component of the total current flowing through the load, in amperes, and R=the resistance of the load, in ohms.

Therefore, if you want the output stage to directly drive 1 watt of rf power, at frequency f, into a 50 Ohm load.... then the associated sine wave component, at frequency f, of the total current flowing through the load must have a peak value of 200 mA.  

For class A operation, this means that the average current in the output stage must be at least 200 mA.... and as a practical matter somewhat more than 200 mA.

By using a transformer between the output stage and the 50 ohm load, you can trade off current for voltage. I.e. if you use a 2:1 step down transformer, the minimum required average current in the class A output stage would be half as much (i.e. 100 mA), but the rf voltage swing in the output stage would have to be twice as much (20 volts peak instead of 10 volts peak).

Note: 20V peak-to-peak (i.e. 10V peak) across a 50 ohm resistor is 1 watt.


At this point, it is probably appropriate to point out that... depending upon what you want to use this amplifier for... there are definitely much less expensive, much more power-efficient designs than a single ended class A amplifier. This has been alluded to in some of the earlier posts.

As just one example... take a look at this design. This product's web page allows you to download the manual, along with the schematic and pictures that show the physical design. Some AMer's have built their own linear amplifiers using FET-based designs similar to this one. Of note, the peak output power is around 50 watts if the amplifier is used for SSB or CW. I use one of these as an intermediate power amplifier... mainly for AM... with about 250 mW of carrier going in (from a QRP AM transmitter), and 5 watts of carrier going out. I am very happy with its performance.

http://www.eham.net/reviews/detail/9827


Stu

funny that you mention the hfprojects kits. i actually have one of their SDR 5w preamps sitting on my desk. its fully assembled but as yet untested. my plan was to use it as a final output stage off a QRP transmitter (which i have yet to build).

my actual goal is to try out low power 1-10w QRP broadcasting before eventually moving up to a larger station. i have that kit assembled essentially ready to go (i need a power supply and transformer for it) but i really want to try successfully build an amp from scratch.

this may seem dumb, but why is it such a challenge to get 1 watt out of a 2n3866 if its 'total device dissipation' is 5w? does that spec not mean maximum power output?

"....why is it such a challenge to get 1 watt out of a 2n3866 if its 'total device dissipation' is 5w? does that spec not mean maximum power output?"

No... it means the amount of total electrical input power that is converted to heat within the device.

If you use a step down transformer to trade off less average output current for more voltage swing, and you properly heat sink/cool the device... then I suspect that you can get 1 watt out.

Remember, the device is specified to produce 1 watt of rf output with a power supply voltage of 28VDC. [Note, that the circuit shown in the device's specification sheet, that corresponds to the specified 1W of rf output power, appears to have the device biased to operate in Class C, not class A. Class C operation will produce a higher efficiency... more r.f. output and less dissipation... than class A]

http://www.datasheetcatalog.org/datasheets/400/74256_DS.pdf

The device is also specified to have a maximum allowable device dissipation (at a heat sink temperature of 25 degrees C) of 5 watts. The specification sheet also shows an LC circuit (which acts both as an impedance transformer and a filter) at the output of the test circuit that corresponds to the 1W rf output power specification. Thus, when producing 1 watt of rf output, the amplifier would likely be operating with some distortion of the rf output waveform on the left side of the LC output circuit. [Actually, considerable distortion if the circuit is operating in Class C... as the biasing suggests.]

For an amplifier operating in class A mode, the maximum device dissipation will occur when there is no rf input. When there is no rf input, the electrical input power will all be converted to heat within the device. Also, in a class A amplifier, the electrial input power is the same whether rf is applied or not. Thus the electrical input power (if the device is operated as a class A amplifier) must be less than 5 watts.

With no rf input applied: 5 watts of electrical input power, at 28 volts of collector voltage, corresponds to 178 mA of collector current. With r.f. input applied, and for class A operation, the collector current will increase and decrease in each rf cycle... but its average value will remain at 178mA.  

An ideal class A amplifier, operating at full rf output power, has an efficiency of conversion of electrical input power to rf output power of 50%. I.e. 50% of the electrical input power is converted to heat, and 50% of the electrical input power is converted to rf output.

A practical class A amplifier (designed to have reasonably low distortion) would typically have an efficiency (at peak rf output power) of around 12.5-32% (corresponding to a peak rf output voltage that is around 0.5 - 0.8 x the average collector voltage); and it would have an efficiency at 0 rf output power of 0%. Thus with 5 watts of electrical input power... corresponding to 5 watts of dissipation in the device when there is no rf input applied, the maximum rf output power would (as a practical matter) be around 12.5-32% of 5 watts = 0.625 watts -1.6 watts.

In the test circuit shown in the device's specification sheet, the circuit appears to be biased to be operating in class C. That is, the transistor will only be conducting for less than half of each rf cycle.

Stu

why do they give the heat dissipation spec for when there is no input applied? wouldn't it make more sense to give the spec as 1.67w at peak rf input rather than 5w at 0 input?


right now i am a little confused about measuring power output. i had been using (E^2/R)=P  where E is the peak to voltage measured at the collector load. is that wrong? you said "20V peak-to-peak (i.e. 10V peak) across a 50 ohm resistor is 1 watt." however 20^2/50=8.

where does the formula "p = i x i X r/2" come from? according to ohms law "p  = i x e"

what is the proper way to calculate power?

These are all very good questions. Note that I have made some edits/changes to my previous post to clarify, somewhat, what I was trying to say... and to provide more realistic numbers for the efficiency of a real class A amplifier (operating with low distortion).

The manufacturer doesn't know how the device is going to be used... so it specifies the maximum amount of dissipation that the device can handle without causing too high a temperature rise of the device within its package. The dissipation is, by definition, the short time average of the heat being produced within the device, per unit time. This heat is removed from the device, by conduction, via the heat sink.

The heat being produced within the device, per unit time, at any given time, t, is a result of: the currents, at that instant of time,  flowing from collector to emitter and from base to emitter... multiplied by the corresponding voltages at that instant of time, from collector to emitter and from base to emitter. This corresponds to the electrical energy per unit time flowing into the device from the electrical circuit that the device is connected to.

It is up to the user to determine how much electrical energy per unit time the device will have to accept from the circuit (again, this energy is converted to heat within the device, and then delivered to the heat sink), under the full range of operating conditions.

In the particular case of a class A amplifier, the maximum heat per unit time being produced within the device (and then removed from the device, via the heat sink) occurs when there is no rf input.

Heat flow has the units of joules per second = watts. The temperature of the device, inside the package, will rise until the amount of electrical energy per unit time flowing in from the circuit, and being converted to heat, is equal to the amount of heat per unit time flowing out via the package/heatsink/ambient environment. This is analogous to the case of a vacuum tube, in which the termperature of the filament will rise until the electrical power flowing into the filament equals the energy per unit time, radiated (ref: black body radiation), as infrared and visible radiation, by the filament into the surrounding vacuum. The temperature of the tube's glass envelope will rise until the energy per unit time that the glass is absorbing from the radiating filament and other hot objects inside the vacuum, is equal to the amount of heat per unit time being carried away by the surrounding air. [Note: if the tube's glass envelope didn't absorb any of the infrared and visible radiated power produced by the hot filament and the hot plate, etc. inside the vacuum, (energy absorbed and converted to heat per unit time), then the glass envelope wouldn't get hot! Any infrared and visible radiation, being produced by the filament and the other hot objects inside the vacuum, that is not absorbed and converted to heat by the tube's glass envelope, will be absorbed and converted to heat by the surrounding air... and by objects surrounding the tube]

The other questions you are asking relate to topics that are covered in a typical 1st course in electrical engineering (usually sophomore year) in an undergraduate electircal engineering program. There are many good textbooks that are used in conjunction with those courses; and, recently, some universities are making the lectures from their versions of those courses available for free downloading.

Rather than answering your other questions here... I suggest that you refer to the existing textbooks and the materials that are available via the web. Textbooks that I have used in conjunction with courses I have taught include:

"Fundamentals of Electrical Engineering" by Giorgio Rizzoni
"Electric Circuits" by Nilsson and Riedel

Good luck!

Stu

thanks for all the help stu. you have been incredibly helpful and really gotten me on my way. i'll keep reporting in this thread as my make progress.

The fine print says a 3866 will dissipate 5 watts if you can hold the case temperature to 25 degrees C. Not an easy task. This is why they added a 10-32 stud on larger devices to help sink heat away from the device. In the real world with a top hat dissipator heat sink you want to hold the input to 1 watt and get 1/4 watt out to get a clean linear waveform. Many devices have a derating curve as the case temperature heats up power goes down.
Welcome to the school of hard knocks.

what do you mean by "hold the input to 1 watt"?


haha, thanks for the welcome. i hope i can pass muster. this thread has been a wild ride.

I believe Frank is referring to input power (P=IxE).  With the amplifiers inefficiency, will yield an output power of around .25W. 

Joe, W3GMS   

I have used 3866s and 5109s in many receiver projects and have carefully measured IMD. I found running them at around 65 ma at 15 volts is about the top of the range for clean operation. I usually use a 3 fin top hat heat sink at that power level. I run push pull Norton configuration for the next step in performance with two devices running 65 ma each. Cubic uses a pair of 2N5109s push pull as the LO amp to the first mixer in many of their HF receiver designs set to 1/2 watt output.

been a while since i posted here. i ended up buying the HF Projects 5watt amp. I had some trouble assembling it but Virgil (the vendor) is a really amazing person and helped me trouble shoot things and get it working. my little transmitter wasn't actually able to drive it but i put together a one stage 2n3866 circuit to use as a pre-amp which enabled me to drive the hf projects amp decently. i only need a 100mw or so to drive the hf projects amp which was pretty easy to achieve with the information all of you provided.

thanks everyone for helping get me this far.

i am currently getting 12.7V RMS which i believe works out to 3.2watts. :)

i would like to continue exploring simple low power amplifiers. next i want to try building a push-pull type but before i jump into that i think i need to learn more about proper biasing techniques. right now i'm just using a basic voltage divider. i think its okay for now because i am only using the 2n3866 to get my signal up to ~100mW. however i would like to try getting more power input into my hfprojects amp and that will require a temperature stabilizing bias (i think?).

can any of you point me in the right direction for learning about voltage biasing?

i've been messing around some more little one and two stage amps and biasing circuits. i'm using germanium diodes in series with the base->ground resistor and placed in physical contact with the transistors. this seems to work well at preventing thermal runaway. i've seen some more complex methods of temperature stabilized biasing but diodes seem okay to me.

is this simply because i am still operating at fairly low power levels (less than one watt)?

if i was building a bigger amplifier, would i want to use a different method?

down load the following:

www.avagotech.com/docs/5988-6173EN

Also APPCAD from the old HP is an excellent resource. In their routine, they automate the bias calculations for you. For bipolar, an emitter R is a solution to the thermal run away. However, by passing the R is required to achieve gain etc... and that is an issue especially for large power devices. So the R is not used. However, there are other solutions, the matched diode-transistor pair you refer to is one and another is active bias. This requires a second device. For power devices, proper 3-R method will work if chosen correctly. If a emitter R is required, it usually is kept small. A little emitter R goes a long way to linearizing the device without killing gain. You will see that in many of the 3866 designs, a combo of two R-s in series, one for reduction of thermal (it is bypassed for AC) and one smaller one to tailor the frequency response and improve linearity. Hope this helps.

thanks for the article. it looks really useful!

i am using a small emitter R to limit the gain on the amplifier. i am using it to drive a larger amp that i dont want to overdrive. is there any particular reason to use active bias instead of diode bias?

no, the diode bias is fine, although for that technique to work well, the diode-transistor should be of similar characteristics. The active bias does not require any sort of matching requirement. As long as you follow the usual rule of thumbs for the classic 4-R or H - bias for bipolar, you should be fine and the emitter R for gain control i.e. un bypassed and thermal control should work fine.