Basic RF HF Amps

Nev,

Years ago I built this 7MHz low power AMTX. Schematic
https://ibb.co/f36YvA

Ferrite binocular core has been used at the output. Impedance ratio is
50ohm load * 16 = 800ohm

The final pushpull class C PA was being driven by a LM386. It was supplying half the Vcc or 6V.

Output power assuming 6V RF peak swing=( 6*6 / 800 ) * 2 = 90mW
I confirmed it with the diode power meter. So in this case it is swinging from zero or ground to 6V+ which is supplied by LM386 chip.

But then it’s a class C stage so linearity is not important here. We just want maximum output power. Voltage swing is always lower or equal to V+ minus emitter voltage but never higher so it’s safe to assume a figure of 10V peak to peak in a 100mW class A BC547 or 2n2222 HF amplifier to preserve its linearity.

Hi Nev,

Your power meter is accurate. I use it all the time to measure power of low power transmitters. I use two 100ohm resistors in parallel as load, and an IN4148 diode with its cathode connected to ground with 100nF cap.

If you check Harry’s 40mW VHF TX, the final stage has one air core coil with 7T. It’s tapped at 1 1/2T from V+. Turn radio = 7/1.5 = 4.66, impedance ratio = 21.74 : 1

The transistor sees 50ohm * 21.74 = 1090 at its collector. It’s emitter is around 1V. Total voltage swing at collector would be supply voltage minus emitter voltage or 8V peak.

Air core coils have high Q so the voltage swing is maximum here. For low Q inductors wound on ferrite cores in HF bands, the voltage swing might be a little less hence it’s safe to assume 5V peak, 10Vpp or 3.5V RMS (refer to Harry’s 100mW class A amplifier in Basic HF amps) to preserve the linearity.

If you’re getting more power with a different turn ratio then it’s fine. You are getting more voltage swing hence the higher power output.

If you remove one transistor from class AB amplifier, it would still work. The maximum output power depends on Q of the inductor, voltage swing, and impedance ratio.

Thank you Joy & Harry for your responses.

I have read everything carefully and made reference to Google to further understand. I am happy with my understanding of biases etc.

I'm still bothered by the turns ratio - here's the problem I can't get out of my head (related to the number of turns on the primary of T2 in the class AB amplifier). If one of the transistors was removed from circuit then there would still be a waveform on the output, half a sine wave. My logic is each half of the class AB works in isolation (ignoring the crossover class A bit) so while each half is conducting in class B the load is between the centre tap and collector. I can't get my head around why the load, and therefore impedance, is calculated from collector to collector rather than collector to centre tap. Can you think of a way to help me through this mental block?

To make matters worse I have built up the amplifier and changed the turns winding from 2+2:8 to 4+4:8. I seem to get more power out with 4+4:8 (my power meter is a home built 50ohm dummy load with a diode tacked on the end to measure volts, so maybe the extra power is an illusion caused by my basic equipment). If my power measurements are correct that leads me to think the impedance matching is better giving better power transfer? I'm confused.

Thanks in advance to any contributions.

Nev wrote:...
I follow the maths but not some of the “whys”. I understand why the base current (which I assume to be current through the base emitter junction) needs to be 11mA. Your maths then calculates the R1+R2 bias using the same 11mA. Why do these need to be the same?
...
My logic says the load is between the centre tap and collector of the conducting transistor. I would have thought the turns ratio would be 4+4:8 (not 2+2:. Can you please tell me why the ratio is not based on what I assume is the load.

Many hanks to dare4444 for his explanations.

Hi Nev,
As always, we can look at this from many different angles. But I will try to simplify differently.

Class-A
With a 12V supply, the peak-peak voltage is 12V/13.8V (+6 and -6). But the RMS (average continuous power) is only about 4V RMS. To get 5-Watts you need 15V RMS. So you need a transformer that will step up by about 4x to give you 50 Ohms.

This class needs about 600mA or so as a standing bias current, so that the current can vary between 0 and 1.2 Amperes.

Class-AB
This uses two transistors, and each transistor will give you 12V peaks, or 8V RMS in each half-cycle. So your average output is 8V (from TR1) and another 8V (from TR2) = 16V RMS. So you need a transformer that will step up by about 2x (1 plus 1) to give you 50 Ohms. In other words you need a 1:1 transformer, but with a split primary because the input is coming from two different devices.

This class only needs a small standing current of about 10mA so that the transistors have only-just started conducting, and are at the bottom of the linear portion of the conduction curve (they need a little tickle to get them started).

Has this explained things in a different manner? Have I answered your question.
If yes then read dare4444's reply again and he gives a lot more info on biasing. But biasing can vary a lot between transistors.

Crossover distortion
Crossover distortion occurs when the amplified waveform is around zero volts and neither transistors are conduction. This gives that flat bit in the article "Waveform #4".



Increasing the base voltage causes the two transistors to conduct only a little and the output waveform then looks more like Waveform #3".

With too much standing (no signal current) in the transistors, the output will still be linear, but take more current than necessary.

Best regards to you and dare4444, from Harry - SM0VPO

The primary is full length of the coil or collector to collector when it comes to calculating the output impedance because in a class AB amplifier each of the push-pull transistors is conducting during half cycle of the output signal therefore collector to collector impedance is the correct output impedance in a push-pull configuration for a given peak or maximum output power.

I'll answer the first part of your question here. The current flowing through R1+R2 is 11ma. This is the maximum base current that the transistors will be needing as they conduct on RF peaks and deliver maximum power output into the load. DC current gain or HFe of each transistor = 80. Harry has mentioned that each transistor will be passing 825ma on RF peaks. 825ma/80 = 11ma approx. Therefore the biasing resistors must deliver 11ma on RF peaks for the amplifier to be linear. If they can't deliver 11ma base current then the output peaks of RF waveform will look flattened out on an oscilloscope. The value of 100 ohm resistor has to be adjusted to set the standing DC current on the collector (not base) of each transistor with no RF input. In this case it is 5ma for each transistor or 10ma for both. DC standing current of each transistor in Ampere with no RF input = emitter voltage in Volt divided by emitter resistor in Ohm. So 11ma is the maximum base current with RF drive and 5ma is the collector standing current with no RF drive for each transistor. I hope this answers your question.

none of those … on the home page -> Projects -> Transmitters circuits -> Basic RF AF Amps (then class AB)

Which HF-amplifier is you refering to?:

http://213.114.131.21/left_amps.htm