RF Feedback

[Gito]

Hi

After looking at the schematic diagram R4 a 33 ohm 5 watt resistor .it is a shunt Resistor(fixed load) for the input transmitter ,to make the Driver transmitter a constand load.

So I  convinced that a 100kohm, value for R1(droping resistor) is wrong,since I think it is a droping resistor to make a stif bias for the base off the MRF 454 .
a Diode(1n4997) has a voltage of maybe 0.6 t0 ...0.8 volt (if I'm not wrong)between it's cathoda and anoda,so we can used it as Zener diode with high current
capability.
So R1 is the droping resistor to Diode (1n4997) conecting from the B+ ....R1 ---to diode ..to ground,and there we got a fixed + bias.

Gito

[Gito]

hi

I found a schematic diagram after googleling /searching in the Internet,it looks similliar
circuit .look at the min bias circuit and the clamping diode ,to make the transistor stable (to prevent run away current ) , the Diodes are in thermal contact with the transistor.

Gito.n

[AB2EZ]

Gito
Bob [see the paragraph at the end of this long post]

Hi!

Keep in mind that these transistors are bipolar devices, not FETs. They are controlled by base current, not base voltage. Also, in this design, there is no emitter resistance to ground. The base current v. base-to-emitter voltage relationship is essentially exponential... so controlling the base-to-emitter voltage doesn't do a good job of controlling the base current.

As a result, it is a challenge to bias the transistors properly... and that is creating a lot of problems.

As you point out, there is a forward biased junction (actually another transistor, but shown as a diode in some versions of the schematic) that is acting in a role that is analogous to the way a Zener diode works (when it is beyond its breakdown voltage). This device is drawing current through R4 (82 Ohms in the present amplifiers... regardless of what some earlier versions of the schematic show as 33 Ohms)... and acting as a (roughly) 0.7 Volt voltage regulator. The thinking is that if this device is in thermal contract with the amplifier's RF power transistors, it will provide a bias vltage (again, around 0.7 Volts) that roughly tracks (with temperature changes) the base-to-emitter voltage of the RF power transistors.

This approach may work, to some extent... but it is still a crude method of achieving two goals:

a. Biasing the RF power transistors so that their resting (no RF applied) base current (and therefore their resting collector current) is some target value.

b. Allowing the total average base current (in the RF power transistors) to increase when RF input is applied to the amplifier. [Remember: the average collector current = H_fe x the average base current... where H_fe is approximately 100 for these RF power transistors]

Doing both a) and b) is hard. Doing just a) (using a current source) or just b) (providing a low DC impedance to ground) would be easy... but doing both is a challenge.

In this particular amplifier, the design was modified from its previous version [R4 is now 82 Ohms instead of its previous value of 33 Ohms] to fix sme shortcomings in that previous design. Unfortunately this change makes the amplifier unsuitable for AM operation.

In the new design, the large electrolytic capacitor (270uF) that is across the bias supply acts as a charge reservoir... to supply the required additional base current on modulation peaks (which is a separate issue from RF peaks). Thus this capacitor performs the role described above as b) during SSB modulation peaks (but it cannot perofrm that role on a continuous basis).

Between SSB modulation peaks, this reservoir of charge (contained in the electrolytic capacitor) refills by drawing current from the 13.8 volt supply ... via R4 (82 Ohms). R4 cannot deliver enough constant current to keep this capacitor charged up in AM operation (i.e. it cannot deliver the average base current needed to hold the amplifier at carrier output)... but it can deliver enough average current to support the positive peaks in base current associated with SSB operation.

For the above reason, one of the changes needed to operate this amplifer in AM mode is to decrease the vaue of R4 (not necessarily all the way back to the original 33 Ohm value).

Right now, Bob is dealing with a different problem. It is necessary that [in addition to the above issues that are related to AM v. SSB operation] the center tap of the input transformer (that provides the base current) have a low RF impedance to ground... i.e. less than 1 Ohm... at 3.885MHz. For some reason, the 270uF electrolytic charge resevoir capacitor is not providing a low enough impedance to ground at that point. It could be that this electrolytic capacitor does not normally have such a low impedance at RF. It could also be that the capacitor is damaged. In addition... I suspect that the biasing transistor (that acts like a forward biased diode) may be damaged. For AM operation, I would suggest that Bob replace that biasing transistor with a Schottky diode that is capable of handling 1 amp of average forward current. A Schottky diode will bias the RF transistors at a point which keeps their resting current essentially at zero... but which is close enough to the point where they would draw current. The application of RF at the input will cause the transistors to turn on... and if the applied RF is AM, the carrier level will keep the transistors (one or the other) turned on for most of each RF cycle. Thus, the amplifier will be operating in (approximately) push-pull Class BC (biased just below Class B)... but this is okay for AM operation.

Stu

[Gito]

Hi

Yes it's a bipolar transistor,the purpose of a fix/stif power suplly is used to set the operating Class of the transmitter,
 a class B transmitter needs a + bias suplly ,to make it a Class B it needs a standing current that is provided by biasing the MRf 454,

Yes  a  bipolar needs a current not voltage, that's why it must be a stiff power suplly /bias suplly,That does'n change it's voltage when current is drawn from it.(base current)
Say it used a Diode as a Zener with A 100 Ohm droping from the !2volt,than about 120ma is flowing in the R and trought the "Zener" , (0.8 V?)
when the grid is taking current,the voltage stay the same, because the grid is in parallel with the Zener, the current trought it is divided between the base current and the Zener.

Stu ,maybe I'm wrong but R4 is not the droping resistor it's a constant load for the the driver transmitter.

R1 is the droping resistor and it's a 100kohm resistor.
in the schematic that I atached before it used a 47 ufd  as a reservor and a 0.01 uf as a bypass capacitor in parallel,and I think the base current is in miliampere,maybe 10ma,So I think the 47ufd reservoar ,has enought ability ,to suplly it

Here I also attached the schematic diagram of EB 63.
unfortunatly it's small. picture


Gito.N

[AB2EZ]

Gito

I think that you are not looking at an accurate schematic of the amplifier that Bob is using. You can see the schematic of Bob's actual amplifier by looking at his feedback reply post #47 in this thread that he submitted on June 27.

Separately... the Harris schematic that you posted above, is a good example of how to do the biasing. The adjustable voltage regulator in the Harris schematic [LM317] is delivering its current through an adustable resistance of between 10 Ohms and 110 Ohms... and the feedback from the output of this series resistor to the regulator's bottom (adj) input makes it an adjustable, regulated current source. In this case, it can deliver up to 1.25 Volts/10 ohms = 125mA of regulated current.

As we agree, the 1N4001 diodes in the Harris schematic act as a low voltage regulator to provide bias voltage (and therefore bias current) to the RF power transistors. The excess current that is flowing out of the regulated current source, but not flowing into the center tap of the input transformer, flows into these diodes.

About 50 mA flows from base to ground through each of the 12 Ohm base-to-ground resistors. So that means (in the Harris design) that, at most (if the 100 Ohm pot is set to 0 Ohms), 25 mA (total) is available for biasing the bases of the RF transistors.  25 mA of average base current will support 2.5A of average collector current (if H_FE is 100)... and that is sufficient to support around 10 watts of carrier in AM operation. [I suspect that the Harris amplifier is designed for SSB operation; and therefore, the average base current will be much lower... because there is no carrier]. I.e. 25mA of average base current => around 2.5A of average collector current. 2.5A x 13.8 volts (max) = 34 watts of DC power to the RF transistors. Assuming 33% efficiency (at carrier for a Class AB amplifier) => 11.4 Watts of RF output (at carrier).

Note that the 1N4001 diodes are serving a role that is similar to the role that I suggest could be served by a Schottky diode in Bob's design. The reason that I suggest a Schottky diode is to bias the RF transistors a little below the voltage at which base (and collector) current will begin to flow.

In the Harris design, there are two capacitors across the biasing bus (the center tap of the input transformer): 47uF and .01uF.

In Bob's amplifier: there was (before he added a capacitor) only a 470uF electrolytic (I said 270uF in some of my earlier posts... but it is shown as 470uF in the schematic that Bob posted).

A smaller capacitor in parallel with Bob's amplifier's 470uF capacitor... having a sufficiently low RF impedance (i.e. less than 1 Ohm at 3.885MHz) is needed. I am suggesting 0.1 uF (or something between 0.1uF and 1uF, that is rated for RF applications).

Stu

[Gito]

Hi

Stu of course a bigger reservoar capacitor and a higher RF bypass value is better,One thing that you and me are in common,is do we look at the same schematic diagram,the schematic I look at is R4 is not a dropping resistor,it's a load for the driver transmitter,
The dropping resistor is R1 that is a 100 kohm resistor,that's why I wrote in my reply it has a wrong value,it's more logical ,it's a 100 Ohm resistor

Gito.n

[AB2EZ]

Gito

The schematic you are looking at is not the correct schematic (and you are also mis-reading the labels on the resistors because the schematic you are looking at has very poor resolution).

Please refer to post number 47 in this thread.

Once you and I are both looking at the schematic shown in post number 47 of this thread... I think we will be pretty much in agreement on most aspects of this.

Stu

[WA1GFZ]

The bias circuit as designed should work. Temperature tracking isn't great but it should work. I wonder if the output transformer has the wrong core material. I suggest you do some testing at 10 mhz. I have seen CCI amplifiers have problems below 3 mHz due to transformer reactance. Also you could try dropping the voltage to 6 or 8 volts on the collector supply. 

[W1RKW]

Frank,
that's a good idea.  the whole time I've been focusing on 75m.  I'll have to try a higher freq.  STand by.

[AB2EZ]

Bob
Frank

Going to a higher frequency will also decrease the impedance of the (presently installed) 0.01uF capacitor that is in parallel with the 470uF electrolytic.

Before changing the frequency, I suggest that you increase the .01uF capcitor to 0.1uF.

That way, you can put aside the issue of whether or not the impedance from the center tap to ground on the input transformer is low enough (i.e. less than 1 Ohm).

Changing the frequency will change many things that are going on in the circuit... all at once.

Changing the frequency will also affect your measurments because (I believe) your scope is only good to 20MHz (therefore you will lose the harmonics of the waveforms that you observe in the measurment process... not in the waveforms themselves)

Changing the capacitor will only change one thing.

Stu

[Gito]

HI

Stu You are right,I'm looking at the wrong schematic.But after all we are talking about the min bias of the transmitter,so I'm not totally wrong .

It does not cross my mind that changing the 33 ohm to 82 ohm can improve the pateren  of the output of the transmitter,since the grid resistor is 12 ohm so in parallel it is 6 ohm (neglecting the internal resistance of the transistors) so the plus bias with 33 ohm is 6: (6 +33) x 12 volt is 1.8 volt with a "zener" in parallel  the voltage  is 0.8 volt
With a 82 Ohm  the plus bias is 6 : (6+82) x 12 volt is also 0.8 v0lt ,So the bias is the same.
the difference is,it lighten the burden of the power supply feeding the plus bias from around 400 ma to around 150ma

Speaking of power supply,something come into my mind,A 80 watt linear (class B} transmitter needs a 160 watt power input (50% efficiency) .

 with a 12 volt power supply it needs around 160 : 12 = 14 A.
Now a transformer has internal resistance in the primary and secondary winding, as the internal resistance(email cable resistance) of the primary winding reflected to the secondary winding + the resistance of the secondary winding it self  is .5 ohm for instance ,than there will be a 14 x 0.5 =  pullsating 8 volt drop in the power supply at the frequency of the transmitter.
With this drop it affect also the min bias of the transmitter ,that make the plus bias close to zero.
so at this point it' s not a class B transmitter ,but  close to  Class C operation.

I think You need a regulated power supply than can supply  maybe 20 A at 12 volt or a  12 volt car battery .That doesn't drop when taken high ampere currents.

again I may be wrong

Gito.n

[AB2EZ]

Gito

I assume that the power supply that Bob is using can deliver the necessary current without a significant voltage drop. I.e., I assume that it can deliver something like 20 Amps without a significant voltage drop from its nominal output voltage.

Separately:

If the transmitter was operating "key down", and the amplifier was drawing a constant 14A of average collector current (using your example, for illustration only)... then the average base current would be 140mA (assuming the current gain, h_FE, =100)

As you pointed out, the 10 Ohm base to ground resistors (not 12 Ohms) will be drawing a total of at least 0.6 Volts/5 Ohms = 120mA.

So I ask you: how can the bias supply (that supplies both of these currents) deliver 260 mA (or more, if the base voltage is higher than 0.6 volts)? If the bias supply can't deliver this average current... where would it come from?

This is why the 82 Ohm resistor is a problem for any type of high duty cycle operation (e.g. AM).

Stu

[W1RKW]

I haven't read Stu's or Gito's recent responses yet because I was messing around with the amp since last post.  

Changed the .01 to a .1uf.  No change in linearity.

Upped the frequency to 14Mcps and the amp produces a nearly perfect sign wave.  It gets better as freq goes up.  Vout seems to increase as well.  

[AB2EZ]

Bob

That's good ... but can your scope see any frequencies beyond 20 MHz? The output waveform has symmetry around the horizontal axis... so it only contans odd harmonics.

If you are operating at 14MHz... the next higher harmonic is at 42 MHz.

So, regardless of what the output waveform really is, it would look like a sine wave on your scope.

Also note that the fundamental of a square wave has a higher peak amplitude than the square wave itself.

Thus the peak output would look bigger at 14MHz as a result of this effect.

[If you start with a square wave of peak amplitude A, and if you filter out all but the harmonics... leaving just the fundamental (i.e. just a sine wave)... the amplitude of the resulting sine wave is A x (4/pi) ~ 1.27 x A]

Stu

[AB2EZ]

Bob

If you go back to 3.885 MHz... can you post a dual trace scope picture of the base voltage (DC-coupled, either transistor) and the output voltage (across the dummy load)

Stu

[W1RKW]

unfortunately the scope I'm using has a bandwidth of 20MHz.  I do have a scope that is capable of 100MHz but the HV supply is toast. Another project. So I'm stuck for the time being.

Stand by for images.

[WA1GFZ]

Bob,
I wonder if you ever tried a low pass filter at the output? The harmonic energy could be the source of distortion.

[W1RKW]

Frank, Yes on the LPF. Not much of an improvement.

Chan2 is Base or bottom trace.

Chan1 is DL or upper trace.

F= 3880KHz
Chan1= .1V/div  probe set at x10 plus a x10 reduction at DL
Chan2= .1v/div probe set at x10

Both channels DC coupled.

[AB2EZ]

Bob

The base voltage, and therefore the base current, is saturating, as I suspected. If you were to add the base voltage trace you are displaying to its inverted image, shifted by half a time slot (in order to account for both transistors)... the resulting waveform would look like the output waveform.

Note that on the trace that's shown... the base voltage drops to zero when the other transistor turns on... because the other transistor is drawing all of its base bias current away from it. Neither transistor has anywhere to draw base current from... except from the complementary transistor. This implies that the impedance from the center tap to ground is still too high.

Even 0.1 uF may not be enough across the 470uF electrolytic (even assuming that the electrolytic is still working*). Try 1uF or more**.

*I am also wondering if the electrolytic is damaged. I believe it is a low voltage electrolytic... and it could have been damaged by excess voltage at some time or other. [Excess voltage => shorted capacitor => excess current => toasted/open capacitor]

**If we want to draw (for example) 50mA of current from this capacitor for one half cycle (0.5 / 3885000 seconds)... then we would draw 6.4 nanoCoulombs during that one half cycle. If we wish to do this, without changing the voltage across the capacitor by more than 0.1 Volts (for example)... then the capacitance has to be 64 nanoFarads = .064uF. So... I would think that 0.1 uF is enough... but perhaps a 0.1 Volt change in the voltage across the capacitor in each half cycle isn't really good enough. Maybe we need to have enough stored charge to keep the change in voltage across the capacitor during each half cycle below 0.01 volts... in which case we need 0.64uF.

Stu

[WA1GFZ]

You should be getting about 300 volts peak to peak on the output. I see about 100 volts or so. What does each collector look like and what is the total input current?
You should be running 45 to 50% efficiency. I wonder if the output transformer is in saturation?  Are you sure the transformer is installed properly? How about the cap across the collectors, does it run hot?

[Gito]

Hi

Stu
,I"m confused when You asked me where the current that supply the bias of the transmitter ,since I see the B+ running trough R1  trough the secondary winding of T1 is in series with R5 and R6 ( R5 in parallel with R6) ,
they form a voltage divider  with R1 33 Ohm in series with  5 Ohm ,so there's  about 300 ma flowing in this circuit ,so actually its a PLUS bias with 1.5 volt, but it was paralleled ,With A "zener" so the  bias voltage is 0.8.v Current is divided between R 5 Ohm and the "Zener"
So in static condition the PLUS bias supply  comes from the B+ trough R! ..through the secondary winding of T1 ... to the R5 and R6 and also to the base of the Transistor.
So of course the current is supplied from The B+ (12 volt)power supply) .
We are dealing with A plus bias not a min bias.

When the transmitter is in operation the more power needed that supply the base current comes from the power of the driver,which trough T1 (secondary winding) is flowing in the base of the Transistor.
So i still suspect the power supply of the the transmitter,with a 14 ampere needed.
it can drop when there's a resistance in Transformer itself ( email cable resistance)

It' just like supplying a min bias for Triode tubes audio amplifier/transmitter

The difference is A transistor if  the bias is is zero ,there's no current flow in the collector,so we must give a plus Bias to make the Transistor take current and depending the bias voltage to make the class  supposed to work.(class A,AB1,AB2,B,C)

In class B tube amplifier  not all class B is Zero bias,if we don't supply a min bias to it,the plate current will soars up,so we must give it a MIN bias,to control the standing current of the Tubes,

So I think there's no "difference" between this Transistor transmitter and tube transmitter ,when coming to Biasing them,what it needs is a Stiff power supply.

So actually this Class B Mrf454 is " The same" as a class B tube transmitter. Of course with different input and output coupling,but as a whole it is the"same"


Gito.N

[AB2EZ]

Gito

Why are you referring to 33 Ohms. If you look at post 47, you will see that R4 is 82 Ohms in the verson that Bob is using. The bias source resistor is R4, not R1; it is 82 Ohms, not 33 Ohms. Lets talk about the same thing if we are going to talk about this.



Stu

[Gito]

Hi
when I read ,that the transmitter get a sine wave output,when the frequency goes higher,
than maybe the problem is in the material used in T1 and T2 is not the material that's needed for 3.7 Mc (the AL factor} or the output network is resonant at the higher frequency and not 'resonant at 3,7 mhz,

I believed even in broadband amplifier that use a C and a L network in parallel ,have a resonant frequency some where

Gito.N

[WA1GFZ]

Bob,
Download the EB63 app note from CCI. One of the scope pictures shows the output voltage swing at 100 watts am modulated. Input power is 2.2 watts.

[Gito]

Hi

Even if it used a 82 Ohm as R4 the current flowing in R5 and R6 is 12 volt divided by 87 ohm is around 150ma so it is enough to supply the base in static conditions.
As I wrote before the extra power that is needed when the transmitter is on is Supplied  by the driver transmitter.
Stu why  I used 33 0hm ,it' s from the original circuit  that you wrote in Your reply.
Or from Motorola schematic diagram.
And when I looked at the article ,the sine wave output is in 30 Mhz(the example)
Who knows the wave looks at 3.7 Mhz.

Gito.n

Sorry for attaching the picture twice