EMRFD Message Archive 4459
Message Date From Subject 4459 2010-03-15 14:05:09 Chip Owens Divide by 2 in Class C amplifier A friend is having problems with an 80 meter QRP transmitter. It uses a VFO (with buffer) on 3.5 MHz and then 2SC5739 driver and 2SC5739 PA stage.
He is using a 12V DC supply and is getting approximately 6 watts output power on 80 meters.
Symptoms are some RF output on 1.75 MHz and repeatedly blowing of the 2SC5739 PA transistor. He is using a G5RV antenna with an antenna tuner.
It sounds like the PA stage is dividing the 80 meter signal by two to produce the output on 1.75 MHz.
I'm trying to understand two things here:
1. How does a Class C amplifier perform the divide by two ?
2. What would be a good approach to fix the problem?
[Non-text portions of this message have been removed]
4460 2010-03-15 14:35:06 Tayloe Dan-P26412 Re: Divide by 2 in Class C amplifier Here on some notes that I have seen on this subject in the past.
- Dan, N7VE
I don't have a copy, but the article by Ha-jo Brandt, DJ1ZB, Destroy
subharmonic resonances in transistor PA, SPRAT 91, page 6, claims the
problem is the low pass filter does not attenuate subharmonics and
the PA has a parasitic oscillation at 1/2 the desired frequency. His
cure, I believe, is to change the first inductor in the low pass
filter into a series L-C with the same total inductive reactance at
the output frequency, but at the oscillation frequency it becomes a
capacitive reactance and reduces the gain at this lower frequency
enough to stop the oscillation. You can see this design and a
mention of the same problem in the "Miss MOSKITA" manual available
I hope this helps.
73, Kevin, w9cf
The PA was very unstable at reactive loads (bad SWR), and
showed parasitic on half the working frequency, leaving unacceptable
subharmonics. The efficiency was untolerably low. This f/ 2 effect is described
by Hajo (DJ1ZB) in several publications, and showed a solution with
a serially coupled circuit. Peter (DL2FI) optimised this for the Mosquita
with the help of a simulation program and several measurements. Jürgen
(DL1JGS) built several parallel versions to ensure the ability to build copies.
A real team work, Thanks! - to Hajo, Peter, Jürgen. The result of this work
is the present output filter. The ~40 Ohm output impedandce (at 2 Watts)
is transformed from T6 byt L7 to a much higher value, and then through C51
(+C52)/C53 to a 50 Ohm level. Through resonance, you will experience very
high RF voltages, and the voltage limits of C51/ C52 must be considered.
The serial circuit has a high impedance at f/2, and serves to reduce
subharmonic parasitics quite well. Through the low pass filter of C53 /L8/
C54 you will achieve a total dampening of harmonics of at least 45dBc. The
collector efficiency of the stage is more than > 70%; A high value for class
4461 2010-03-15 18:59:04 Graham Re: Divide by 2 in Class C amplifier Chip:
This can be a common problem in transmitters using bipolar PA transistors.
It is a parametric oscillation at a subharmonic frequency of the transmit
carrier, involving the voltage variable capacitor that is inherent in the
collector of the bipolar transistor.
It is usually supported by a resonance in the inductors used to decouple
the PA from the power source. Try lowering the Q of any inductors in
the power feed by swamping with a resistor, or changing from a high Q
inductor to a low Q ferrite choke, try replacing the power feed inductors to
ones with the minimum inductance to provide your desired isolation.
Or switch to a real RF FET for the output device.
4463 2010-03-16 02:00:54 Russell Shaw Re: Divide by 2 in Class C amplifier Graham wrote:
> Chip Owens wrote:Another way is to de-gain the amplifier, but still aim for a reasonable gain
>> A friend is having problems with an 80 meter QRP transmitter. It uses
>> a VFO (with buffer) on 3.5 MHz and then 2SC5739 driver and 2SC5739 PA
>> He is using a 12V DC supply and is getting approximately 6 watts
>> output power on 80 meters.
>> Symptoms are some RF output on 1.75 MHz and repeatedly blowing of the
>> 2SC5739 PA transistor. He is using a G5RV antenna with an antenna tuner.
>> It sounds like the PA stage is dividing the 80 meter signal by two to
>> produce the output on 1.75 MHz.
>> I'm trying to understand two things here:
>> 1. How does a Class C amplifier perform the divide by two ?
>> 2. What would be a good approach to fix the problem?
> This can be a common problem in transmitters using bipolar PA transistors.
> It is a parametric oscillation at a subharmonic frequency of the transmit
> carrier, involving the voltage variable capacitor that is inherent in the
> collector of the bipolar transistor.
> It is usually supported by a resonance in the inductors used to decouple
> the PA from the power source. Try lowering the Q of any inductors in
> the power feed by swamping with a resistor, or changing from a high Q
> inductor to a low Q ferrite choke, try replacing the power feed inductors to
> ones with the minimum inductance to provide your desired isolation.
> Or switch to a real RF FET for the output device.
> --- Graham / KE9H
such as 13dB. Put longer traces for more inductance in the emitter so that the
voltage swing across the total emitter inductance (including bond-wire) is
around 2-3Vpp. Figure out an approximate base current which is Ic/Beta(f) +
I(Cbc) approx, and add a shunting base resistor that will have an approximately
equal resistive current. Add suitable matching networks.
Russell Shaw, B.Eng, M.Eng(Research)
4466 2010-03-17 17:47:02 victor Re: Divide by 2 in Class C amplifier From my experience in distant past the main cause of the frequency divide by 2 in a PA is the large non-linear output capacitance of the power transistor. The usual soluti 4468 2010-03-20 13:18:26 ha5rxz Re: Divide by 2 in Class C amplifier Alter the low-pass filter. This is difficult to draw using ASCII art so I will try to describe it. A conventional low pass filter has a capacitor to chassis, then a series inductor, then a second capacitor to chassis, sometimes this can cause a PA stage to oscillate. By constructing a low pass filter with a series inductor, a single capacitor to chassis, and then a second series inductor the problem will go away.
4469 2010-03-20 20:39:36 ajparent1 Re: Divide by 2 in Class C amplifier Without a picture of the layout and schematic there is
a lot of guessing. My experience suggests a multiplicity
Most common with bipolar amplifiers is the problem of
HFE/Ft, that is to say with decreasing frequency the gain
goes up. So it is common for Class C amplifiers to
"take off" at a lower frequency and often also at audio
where there are sufficient gain peaks.
The other is class C input being inadequately driven
or worse grossly overdriven. The first case leads to
class B operation and the seconds usually ends up with
reverse breakdown of the emitter-base junction.
Both of the cases listed interact with poor or inadequate
decoupling at the collector. The collector DC feed
point should look like very low impedance at audio and
same at RF. If it doesn't then that path can be a alternate
load at some lower frequency. Be wary of bypasses and chokes
becoming tuned circuits out of band.
The other is the input feed circuit for the base, it should
both match the load of the prior stage to the base and insure
the base sees a low impedance (depending on power usually in
the range of 1-10 ohms resistive and reactance). Due to class
C impedances tending to vary with power out and drive levels
attention and care here is required.
The collector load should both swamp and absorb all of the
the device reactances as well as provide a stable RF match.
The correct circuits for this will vary with frequency and
power for a give set of devices.
Lastly mechanical and grounding issues. If RF from the final
has a path back to the VFO, buffer or driver it (the total
circuit) may decide to oscillate at a different frequency
based on circuit design and layout (and out of band).
The fact that devices are failing is significant. If the amp is producing a spur (1/2 as you say) the feed impedances, load
impedances will be wrong and the devices will have to absorb
the power produced. This is always bad. You ahve suggested
a good match at 3.5mhz but at 1.75 it's sure bet the load will
be mis matched and likely be a very low impedance or worse.
I can only suggest that the layout, grounding and likely
bypassing are supporting an oscillati