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This is sort of the "circuit side" of a question I asked on a different subdomain regarding an RF amplifier that takes in 50 W and outputs 160 W (https://ham.stackexchange.com/questions/21411/how-much-does-a-rf-amplifier-use-of-its-input-power). There, I was wondering if an amplifier like that would simply "dummy load" its input and drive its output from a lower-power sample of that signal. But someone replied, no…:

That amplifier requires a 50-watt drive to push it to its full power output: that is how hard you have to hit the output devices in order to fully turn them on. So what the amp is doing is adding 110 watts of power to a 50 watt signal.

I'm wondering how this works in practice? Most amplifier circuits I find either "waste" a small current through the base of a BJT while controlling the output current, or "sample" a voltage on the gate of a FET. In both cases the output is a multiple of the input signal, and the input signal itself isn't really used to "power" the output — only to control a separate source of power, right? What sort of a circuit is used to add more power on to an input signal that's relatively large already?

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  • \$\begingroup\$ "What sort of a circuit is used to add more power on to an input signal that's relatively large already?" Perhaps consider whether you can think of any non-electrical amplification where this occurs. Can you? \$\endgroup\$
    – DKNguyen
    Nov 1, 2022 at 4:34
  • \$\begingroup\$ @DKNguyen Well to me that sounds like a robot exoskeleton or maybe more simply an ebike. [edit: you said non-electrical so I guess I would have to mean steam-powered versions of these!] But I couldn't really explain how those work either, like they somehow need to get ahead of the user's own movement but without actually ever getting ahead… :-/ \$\endgroup\$
    – natevw
    Nov 1, 2022 at 4:47
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    \$\begingroup\$ @natevw Perhaps to answer DK with something quite common, one could consider power steering or power-assisted steering for a car. \$\endgroup\$
    – jonk
    Nov 1, 2022 at 4:52
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    \$\begingroup\$ @natevw But for the RF case, you often already have a transmitter that is designed to deliver a certain power into a certain load and you want to upgrade. So the add-on RF amplifier must, by definition, present that load correctly to its source. There will be a certain voltage produced. That is then amplified appropriately (at watts you are talking about you are moving towards vacuum tube finals) and delivered into the same kind of load used earlier. \$\endgroup\$
    – jonk
    Nov 1, 2022 at 4:56
  • \$\begingroup\$ @jonk Yes, in terms of load and source this is getting back to my original question which was basically: can the amplifier "pass along" most of the incoming power (its input) to its own output? Or would it just dump nearly all the input energy into a "dummy load" and use just a voltage/small-current sample of the signal to control its own separate source of full output power? It sounds like both options are implemented in practice but it wasn't obvious to me how the former ("pass along" plus additional power) option would be accomplished. \$\endgroup\$
    – natevw
    Nov 1, 2022 at 5:06

6 Answers 6

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Simon Finch's answer gives the parallel case. The series case is exemplified by the common-base amplifier:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that Vin = Vbb - Vbe, and Vbe varies with Ic. Meanwhile, output is relative to input, that is, Vce. And we can reasonably assume (i.e. for Vce > Vce(sat)) the currents are approximately equal. So the output voltage contains a fraction of the input voltage, and thus power as well.

Granted, it's a tenuous argument, especially for small signals. But if we measure all the way up to clipping, we'll clearly see that the output voltage peak can be slightly higher than Vcc.

For the BJT, it's a very small fraction added, anyway; which is to say, the transconductance and amplification factor are both very high. For devices having less, the fraction can be much higher. Vacuum tube triodes especially can have an input (cathode) AC voltage as much as, say, a third of the output.

These two methods (common-base, and follower, bootstrap or parallel-adding) are not at all exclusive; a real circuit can have a linear combination of both, thus giving a mix of both characters, as well as neither (a common-emitter or other topology). They could also be inverted, which seems rather counterproductive, but often input or output power is simply burned wholesale as a means of improving bandwidth, stability or distortion.

To really know what's going on, you'd have to measure the power inputs and output quite accurately. Only additive methods can have seemingly anomalously high efficiency (where a fraction of the input really is adding to the output). Such an amplifier could have apparent efficiency (i.e. taking DC input versus AC output) higher than the theoretical class B maximum, of course on closer inspection it's merely the case that the added power obeys that limit.

Of course, any type can be made to have arbitrarily poor efficiency, so having some figure is not exactly proof of anything!

If I had to guess, my guess would be your amplifier in question uses a combination of attenuation and feedback to ensure stability for ease-of-use, at expense to efficiency. In that case, the topology really doesn't matter. Common-emitter (or source) would be most likely, as most devices are available packaged that way (i.e. with the source/emitter hard grounded to the heatsink tab). But an inspection of the internals, or a review of the service manual, would say for sure.

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The term you want is "power Added Efficiency (PAE)", and it is a thing in some RF power amplifiers, particularly in the upper microwave region where per stage gains are low. It is clearly far less important at lower frequencies where you can get 20dB out of a single stage (Maximum 1% difference from drain efficiency at that point).

While it is a figure of merit, actual designs that couple a meaningful amount of input power to the output are not that common as it requires a combination of low gain and a suitable topology (Common Base/Gate/Grid being the obvious approach), you see it sometimes in microwave applications (Low gain devices) and some old school ham radio stuff does the grounded grid thing, but most modern devices are poorly suited to the approach.

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A simple example of an amplifier that sends some of the input power to the load would be an DC coupled emitter follower BJT amplifier.

An emitter follower BJT amplifier sends whatever current comes into the base of the BJT out to the load. Typically, this is going to be like 1% or less of the total current (and probably power) going to the load.

In many cases its desirable for the amplifier to have high input impedance so it doesn't load down whatever is driving it. In that case, having power go into the amplifier (even for the purpose of powering a load) is not really desirable.

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  • \$\begingroup\$ Ahhh, that's not so complicated then maybe! So assuming the circuit at homemade-circuits.com/… is correct, the base current and the collector current can simply be used in combination at the emitter, albeit some of the power going into a resistor there which I'm assuming is chosen so that the source sees the impedance it expects? \$\endgroup\$
    – natevw
    Nov 1, 2022 at 4:40
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Sure you can. It's not easy, but an approach like this might work:

schematic

simulate this circuit – Schematic created using CircuitLab

This is grossly over-simplified, and distorts the signal somewhat, but it illustrates the idea, which is to reproduce the input signal, and buffer it with a high current stage. I do that here with a push-pull pair Q1 and Q2 inside the feedback loop of a voltage follower.

Then I mix the original signal with this copy using R1 and R2, whose purpose is to force any current required by the load to be shared equally between those two identical voltage sources.

Obviously R1 and R2 should be very small compared to the load, to avoid wasting too much power in those resistors.

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There are different kinds of amplifiers, and a "power amplifier" is one that is fairly low impedance on the input side, so it requires another amplifier in front of it.

Typically, power amplifiers also waste quite a bit of energy because the transistors need to be in the linear region for the entire amplitude swing, or they will behave nonlinearly, which introduces harmonics.

The power transferred with the input signal is small compared to both the output signal and the power that is lost, and the electrons are mixed together, so there is no good answer to whether the input power is forwarded or sunk -- it's "both".

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  • \$\begingroup\$ In radar applications, the final amplifier (PA) is usually operated compressed or as a class C amplifier, not linear. \$\endgroup\$
    – SteveSh
    Nov 2, 2022 at 12:02
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Can an RF power amplifier "reuse" its input power?

No. It does not. It uses it and it needs it to work.

So what the amp is doing is adding 110 watts of power to a 50 watt signal.

This is a little simplistic...
What must be understood is that RF PA needs the input power to work.
Because there are "matched" 50 Ohm for output ... AND for input.
And because individual RF PA (of that kind) can be "chained" for a bigger power output.
Note also that PA RF must be protected against loss of "load".
They have a special circuit for measuring "reflected" wave and shut down immediately power because RF PA work always quasi at maximum power available.

This was often done by "radio amateurs" because of the "price" of such pieces of equipment to use "bigger" power for their transmitters.
Note also the use of a "power" relay activated by the input power to be in OUT TX to the antenna. When power is released, we are in RX IN from the antenna.

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