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Author Topic: Rick Campbell's One Transistor Amp Article  (Read 34233 times)
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ssbothwell SWL
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« Reply #50 on: June 24, 2011, 08:01:26 PM »

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?
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AB2EZ
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"Season's Greetings" looks okay to me...


« Reply #51 on: June 24, 2011, 09:19:09 PM »

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

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ssbothwell SWL
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« Reply #52 on: June 24, 2011, 09:49:30 PM »

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?
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AB2EZ
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« Reply #53 on: June 24, 2011, 10:16:31 PM »

"....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
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ssbothwell SWL
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« Reply #54 on: June 24, 2011, 11:45:29 PM »

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?


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AB2EZ
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« Reply #55 on: June 25, 2011, 06:29:55 AM »

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
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ssbothwell SWL
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« Reply #56 on: June 27, 2011, 03:10:36 PM »

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.
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WA1GFZ
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« Reply #57 on: June 27, 2011, 04:01:33 PM »

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.
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ssbothwell SWL
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« Reply #58 on: June 28, 2011, 03:14:40 AM »

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.
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W3GMS
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« Reply #59 on: June 28, 2011, 08:14:01 AM »






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

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

Joe, W3GMS   
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Simplicity is the Elegance of Design---W3GMS
WA1GFZ
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« Reply #60 on: June 28, 2011, 08:24:43 AM »

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.
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ssbothwell SWL
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« Reply #61 on: July 16, 2011, 12:38:30 AM »

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. Smiley

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?
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ssbothwell SWL
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« Reply #62 on: July 18, 2011, 02:00:50 AM »

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?
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W4AMV
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« Reply #63 on: July 18, 2011, 01:28:48 PM »

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.
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ssbothwell SWL
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« Reply #64 on: July 18, 2011, 06:34:55 PM »

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?
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W4AMV
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« Reply #65 on: July 18, 2011, 10:24:13 PM »

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.
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