In case it matters, let's discuss solid-state, linear, class AB amplifiers operating below 50 MHz and on the order of 100 W.

Here's my confusion: I'd think ideally, the output impedance of the amplifier would be 0Ω, or at least as low as possible. This would minimize loss and the load impedance wouldn't matter much.

Yet, datasheets for these amplifiers almost always specify the output impedance as 50Ω. Sure, this means any reflections from the load will be absorbed in the source, but for HF and the kinds of things usually transmitted on HF (AM, SSB), reflections will not appreciably distort the signal.

I would think any kind of resistive 50Ω source would mean the amplifier efficiency can't be more than 50%. The 50Ω source could also be realized with reactive components which would improve efficiency, but preclude usage on multiple frequencies.

So what is the output impedance of a typical amplifier of this sort, and why?

  • \$\begingroup\$ Phil, are you saying/hinting that you can have a zero ohm source providing that the cable impedance matches the antenna/load impedance? If so I agree with you. \$\endgroup\$
    – Andy aka
    May 10, 2016 at 7:55
  • \$\begingroup\$ @Andyaka I don't see why it wouldn't be theoretically possible, but I'm wondering how it's actually done in practice. \$\endgroup\$
    – Phil Frost
    May 10, 2016 at 15:56
  • \$\begingroup\$ "In practice" I've seen some RF amps with a big ~50 ohm resistor in series at the output. When I've asked similar questions, the answer is that 50 ohm out can be used by everyone. If you want some more power for a specific load, then maybe it gets modified. \$\endgroup\$ May 10, 2016 at 18:51

3 Answers 3


What happens when a very stiff source drives a transmission line? You get reflections, and those reflections occur from both ends. If the source is not terminated any mild imperfections in the cable or termination at the RX end will cause reflection that then reflect with 100% efficiency back down the cable and cause problems.

The issue then arises, how do I terminate the Tx and the Rx without killing the amplitude by 1/2?, because after all you are forming a voltage divider.

The easy approach is to put a termination (series resistor) inside the amplifier right after the ideal very stiff driver and before it hits the cable. To compensate for the loss you feed-back the voltage signal from the driven end of the cable and cause the driver to drive harder to compensate for the voltage divider termination.

More sophisticated drivers actually design the differential resistance of the amplifier to be matched accordingly and are thus more efficient. Of course you can't drive to DC with that approach but often that is not the issue.

Of course this means that any given driver must then be designed for an explicit termination impedance.

Upon edit: I forgot to mention the simplest case, where you design it to drive at 2X the voltage and have a builtin terminator (series) resistor. This is less power efficient and has the side effect that an non loaded amplifier put out 2X the voltage.

  • \$\begingroup\$ Sure there are reflections, but is that a problem per se? I'd probably match the antenna to the feedline to minimize the reflections first. Then given the choice between dissipating that power at the source, and reflecting it back to the load, wouldn't the latter have a higher efficiency and a less hot transmitter? \$\endgroup\$
    – Phil Frost
    May 10, 2016 at 0:52
  • \$\begingroup\$ Also, wouldn't any kind of resistive termination get really hot? \$\endgroup\$
    – Phil Frost
    May 10, 2016 at 0:54
  • \$\begingroup\$ @PhilFrost Not caring about reflections only counts in a world where you are transmitting power and no information. You're going to have non-idealities in any case so you have to deal with them. Does the resistor get hot in the simple cases? Yes, that is why it's less efficient. I'd suggest looking at analog devices cable driver topologies. \$\endgroup\$ May 10, 2016 at 1:16
  • \$\begingroup\$ I'd think there are many applications where reflections won't cause any appreciable distortion to the information, especially if the antenna is reasonably matched and has return loss >15 dB. Take AM radio broadcast for example, with a bandwidth of 30 kHz. The highest baseband frequency components are so relatively low that the signal might as well be a sine wave. How are reflections going to mess that up? \$\endgroup\$
    – Phil Frost
    May 10, 2016 at 15:54
  • \$\begingroup\$ @PhilFrost ... and how many AM broadcast applications are there vs. surveillance cameras? Crappy cables, high sensitivity to ghost images etc. Your rebuttal is equivalent to saying "just make everything ideal". \$\endgroup\$ May 10, 2016 at 16:17

Generally linear amplifiers have their loads on the drain or on the collector and when frequency is high, feedback is local. This way output impedance is very high. So really you should think of a current source driving a line with distributed constants. This from an impedance viewpoint causes voltage rise at resonance of the system if the load is removed. Circulators or other voltage and current limiters are employed to limit maximum stress on active components.

  • \$\begingroup\$ Is that really true for class AB amplifiers? Most designs I've seen have a final stage of a pair of emitter/source followers. \$\endgroup\$
    – Phil Frost
    May 10, 2016 at 15:43
  • 1
    \$\begingroup\$ @PhilFrost and if you bias the upper and lower stages appropriately you can make the small signal output impedance any value you want (within reason) \$\endgroup\$ May 10, 2016 at 16:18

Most broadband HF linearish amplifiers (remember we are talking UNDER 30MHz here!) actually model as more or less voltage sources, designed to remain within their safe operating areas into a specified LOAD impedance. You can design one as a current source, and this is in fact often what a single ended stage at VHF and up looks like (The drain choke looks like a current source to RF, and the device is a voltage controlled transconductance).

Conjugate matching (and harmonic tuning) are appropriate to narrow band amplifiers but are effectively impossible to apply over the decade of bandwidth typical of a broadband HF amplifier.

In fact for most HF linear amplifiers, the drain impedance is largely determined by the load impedance reflected thru the output matching network, and is only ever optimum at full power, as soon as you back off the drain impedance becomes too low given a constant supply voltage and the efficiency goes completely to pot.

Depending on design they either appear as voltage sources (very low Z) or current sources (Very high Z), but in no case is the OUTPUT impedance of a HF band amplifier typically 50 ohms unless it is part of some test gear (Where the extra losses are acceptable), now the design LOAD impedance, that is another matter!

Comms line drivers are a slightly different animal to things intended to drive aerials as they are often trying to drive into electrically long lines that are severely mismatched at the far end, here source termination and loosing 6dB of drive level is a sane choice.


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