So let's say you have a circuit, which generates a carrier wave at some frequency (let's say 27MHz) and it's connected to a 50 ohm dummy load (which I gather is equivalent to an antenna for circuit analysis purposes). And it is powered by a regulated 12V power supply.

So imagine the carrier wave is 12 volts peak-peak, which is 4.242 volts RMS. As per the formula \$P = (V_{rms})^2/R\$, this gives a power output of about 0.36W. Even disregarding average power, 12V into 50\$\Omega\$ is 2.88W. And the peak of the waveform is actually 6V, and at 50 ohms that's only 0.72W.

How then do circuits like these output of 5W or more with a 12V (give or take a few volts) power supply?

If you wanted an average 5W out of a 50 ohm load, you'd need a peak-peak voltage of almost 45V. For 100W, you'd need a signal that's 200V peak to peak! Somehow I doubt that people are powering their radios with such high voltages.

What I'm not understanding is how one gets more power out of a circuit with a fixed load and a fixed supply voltage. Even if your amplifier can deliver 100A, I=V/R; With a 12V supply, Ohm's law says that even at the peak, it's only going to be delivering 0.12A, with the load dissipating 0.72W.

I think one could somehow use a step-up transformer to increase the voltage to the necessary level, trading current on the primary for voltage on the secondary, but neither of the circuits above do this. Aside from that, all the impedance matching networks in the world aren't going to get you more voltage across that load.

All of what I explained may well be wrong, and that's why I explained it. Please help me sort out my conceptual misunderstandings :)

  • \$\begingroup\$ What do you think impedance matching really does? There is no loss mechanism in a reactive network and Zout > Zin, say... \$\endgroup\$ Commented Nov 4, 2014 at 0:44
  • \$\begingroup\$ @SpehroPefhany There's no gain in such a network either though. Even perfectly lossless matching won't increase the voltage (and therefore the power) across the load. \$\endgroup\$ Commented Nov 4, 2014 at 0:50
  • 3
    \$\begingroup\$ No power gain. If the output Z is higher than the input Z then the output voltage must be higher for the same power, right? Conservation of energy. Think resonance. \$\endgroup\$ Commented Nov 4, 2014 at 0:56
  • \$\begingroup\$ Indeed one of those circuits DOES have a step-up transformer if you look at it right... \$\endgroup\$
    – user16324
    Commented Nov 4, 2014 at 11:43

5 Answers 5


The key to all of this is "impedance matching". You need the amplifier to think it is driving a low impedance (so it can source plenty of current from the 5 V supply, and thus generate a lot of power). Then you "magically" need to transform those currents to drive 50 ohms at a much higher voltage.

This is done with an impedance matching network. When you write down the equations governing the network, it needs to look (at the frequency of interest - these things have to be tuned to work) like a low impedance at the input, and a high (50 ohm) impedance at the output.

There are many ways to achieve impedance matching: if your input impedance is 5 ohm, and you want to match to an output impedance of 50 ohm at 27 MHz, you can use a simple LC circuit


simulate this circuit – Schematic created using CircuitLab

which I "computed" by using http://home.sandiego.edu/~ekim/e194rfs01/jwmatcher/matcher2.html and entering the appropriate parameters.

What happens here is that the alternating voltage on the source (with impedance R1) drives current into the resonant LC circuit. Because these are series switched, they look like a low impedance - but in reality the voltage swings that can be achieved at the output are very high - much higher than the input voltages. Writing the impedance of C1 as Z1 (=1/jwC) and impedance of L1 as Z2 (jwL), you see they can be combined:

R1 and Z1 in series: \$X_1 = R_1 + Z_1\$
R2 and Z2 in parallel: \$X_2 = R_2 * Z_2 / (R_2 + Z_2)\$

Now the input voltage is divided, so the output voltage is

\$V_{out}/V_{in} = X_2 / (X_1 + X_2)\$

$$\begin{align} &= \frac{(R_2 * j \omega L)}{(R_2 + j \omega L) (R_1 + \frac{1}{j \omega C} + \frac{R_2 * j \omega L}{R_2 + j \omega L})}\\ &= \frac{R_2 * j \omega L} {(R_1 + \frac{1}{j \omega C})(R_2 + j \omega L) + R_2 * j \omega L}\\ &= \frac{R_2 * j \omega L}{R_1R_2 + j(R_1 \omega L - \frac{R_2}{\omega C}) - \frac{L}{C}) + R_2 * j \omega L}\\ &= \frac{R_2 * j \omega L}{R_1R_2 - \frac{L}{C} + j(R_1\omega L - \frac{R_2}{\omega C} + R_2 \omega L)}\\ \end{align} $$

Now the imaginary term in the bottom cancels when

\$R_1 \omega L = R_2 (\omega L - \frac{1}{\omega C})\$


\$\frac{R_1}{R_2} = 1 - \frac{1}{\omega ^2 LC}\$

If R1 is zero and \$\omega = \sqrt{\frac{1}{LC}}\$, you can drive almost any voltage into R2 without ever generating a voltage at the input - because your current through C1 is perfectly matched with current flowing into L1. But those variations in current do generate a voltage across L1 and thus across R2. It's all got to do with the fact that a series LC circuit looks like a much lower impedance at resonance - the voltage at the end varies less than the voltage at the point between L and C.

The above link gives you a lot of alternative circuits that will do the same thing - but ultimately for an efficient transmitter you want to have real impedance at the frequency of interest (no reflection) - and the matching circuit achieves that for you, at almost any impedance (with the right values of components, of course).

  • \$\begingroup\$ I would be really thankfull, if you could explain it in more depth, how a LC circuit (or any resonant circuit) can achieve what you are saying. If the voltage across R1 is 12V, wouldnt that mean that the output voltage (across R2) would be the same? I am sorry, I just dont get it. I understand, how a transformer can be used for impedence matching, but this... \$\endgroup\$
    – Golaž
    Commented Nov 4, 2014 at 18:34
  • 1
    \$\begingroup\$ @Golaž I tried to explain it a little bit more. This is a huge subject - just google "impedance matching circuit" and gorge yourself on the wisdom of the internet... \$\endgroup\$
    – Floris
    Commented Nov 4, 2014 at 19:14
  • 1
    \$\begingroup\$ Golaz, it is well known that a resonant circuit "amplifies" a signal. Regardless of what the voltage is across R1, the important question is what is the voltage across R2? This voltage is being generated across L1. Think of this as the inductor being "pumped" at the right time to generate much larger voltage swings. So having voltage swings of 45v p-p (or larger) across the load, should not be surprising! \$\endgroup\$
    – Guill
    Commented Nov 6, 2014 at 23:41
  • \$\begingroup\$ Once you get to the last portion of your simplification (where the imaginary term in the denominator cancels out) can you show how VOUT is calculated from VIN using the remaining expression? Not sure how to handle the j in the numerator. \$\endgroup\$
    – scuba
    Commented Dec 6, 2018 at 15:17
  • \$\begingroup\$ @scuba: A pure imaginary ratio implies a 90 degree phase shift \$\endgroup\$
    – Ben Voigt
    Commented Aug 2, 2021 at 19:31

If you look at either of those schematics there are inductors all over the place. There are many ways of generating higher voltages without using a transformer. Indeed, look at a spark coil used in cars. You generate huge voltages by building up current and then interrupting it, and that device is "transformerless". These circuits operate in different ways, but the core idea of a voltage boost with a change in current applies to both. The "mighty mike" (first link) is resonant with capacitor coupled "Pi" and "T" chain. The Lythal design (second link) is also resonant but with a transformer, it even notes not to use a ferrite slug (which is lossy) and would dampen the resonance.


The output impedance of the driver transistor can be quite low. So the RF amplifier can draw lots of current. Say half an amp, at 12V that would be around 6 watts. That looks like 24 ohms. Then, go through a transformer to match that up to 50 ohms at the antenna. The voltage is higher, the current is lower, but the power is still the same.

  • 1
    \$\begingroup\$ Yes, low impedance and an RF matching transformer is common for portable gear in the solid state era. In the tube era, generating an 800v or so supply rail wasn't uncommon. \$\endgroup\$ Commented Nov 4, 2014 at 2:16
  • \$\begingroup\$ It occurred to me that one could stick a boost power supply ahead of the whole thing, but it seemed like what I came up with would be more direct and fewer parts. And yes, I grew up with vibrators and then germanium switching transistors. \$\endgroup\$
    – gbarry
    Commented Nov 4, 2014 at 19:14

First off, your voltage calculations are wrong. With the 12V supply going through a transformer or inductor, the midpoint voltage is 12VDC and the maximum voltage swing is 24Vpp. So it could actually produce 4 times more power at 50Ω than you calculated.

You are correct that to put a 5W rms sine wave into 50Ω you need almost 45vpp. If the final amp output is only 24Vpp then you need a step-up transformer or other loss-less impedance matching circuit. To increase voltage the output impedance just has to be higher than the input impedance.


The impedance matching explanation is mostly correct, but a bit imaginary (pun intended). To answer the OP's question, we can obtain "gain" through a RF tuned circuit. It resonates, "amplifying" the input power drive by the loaded Q of the tuned circuit. It does this here by converting lots of switched current, courtesy of the output device, into tuned circuit power. The loaded Qs are fairly low (~5) to provide an adequate operational bandwidth and a reduced dynamic impedance to drive the current through. So limited 12V "current pulses" can be become a 60V sinewave tuned circuit output for a theoretical 72W. But we never reach this nirvana due to losses, but 5W is certainly do-able. More power ? Just increase the output device(s) switching current. Automotive transmitters can achieve hundreds of watts of RF output easily.


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