In many RF applications I have seen transistor circuit that looks something like this:


simulate this circuit – Schematic created using CircuitLab

Now, I understand the role of all the components, but one thing bothers me. The drain voltage is superposition of DC and AC voltages: \$V_{D} = V_{DC} + v_{AC}\$, where \$V_{DC}\$ is obviously 12 V and \$v_{AC}\$ is \$ G*v_{In}\$. Given the transistor gain G >> 1, the drain voltage swings about 12V, so its amplitude is definitely higher than 12V. But how can it be if the power source dictates the 12V limit??

With normal transistors and op amps, I would assume that this iduces a saturation. But it's not the case from what I have seen - RF transistors I used are highly linear and harmonic level is reasonably low so no effects of saturation have been observed.

So, how do I get drain voltage higher than power supply voltage?

  • 1
    \$\begingroup\$ Where did you get the idea that there is some voltage on top of the supply voltage? The more voltage there is at the fet, the more it pulls down the voltage at the drain, all below 12V \$\endgroup\$ – PlasmaHH May 10 '16 at 9:26
  • 2
    \$\begingroup\$ @PlasmaHH : no the mean voltage (the DC voltage) will be 12V, so positive peaks will exceed 12V and total swing can be twice the supply voltage. Pull V_D low and you are storing energy in the inductor; releasing that energy is where V_D > 12V comes from (half a cycle later). L1 is often tuned to the carrier frequency with a capacitor (not shown here) forming a parallel tuned circuit. \$\endgroup\$ – Brian Drummond May 10 '16 at 9:34
  • \$\begingroup\$ Well the idea I got from V_D=V_DC+v_AC, or V_D = 12V + v_AC. \$\endgroup\$ – Roker Pivic May 10 '16 at 9:35
  • \$\begingroup\$ Indeed the AC voltage is superimposed on the DC voltage which is 12 V so you can for example get a sinewave at the output between 13V and 11V (that is 1 Vpeak on top of the 12 V DC). \$\endgroup\$ – Bimpelrekkie May 10 '16 at 9:37
  • 1
    \$\begingroup\$ More precisely: that would work for any signal (including the DC bias !) for which the inductor behaves as an inductor meaning it does not saturate (because then it has stored the maximum amount of energy it can store, nothing more). \$\endgroup\$ – Bimpelrekkie May 10 '16 at 9:40

The drain voltage will have a DC (average) value of 12 V because the inductor (at DC) is a dead short. There can be no other result here. Therefore it follows that the AC content of drain voltage must rise and fall about the 12 V average level. Maximum swing is nearly 24 Vp-p. This happens because of the formula for an inductor; as the transistor starts to switch off (on part of the AC cycle) the inductor tries to maintain current and will cause the drain to rise above the 12 V level.

This IS NOT restricted to RF transistors - all transistors will behave like this with an inductor in the collector/drain.

The worst case scenario is when the transistor is turned on then rapidly turns off - the resulting back-emf can destroy the transistor BUT, this is a class A amplifier and class A amplifiers behave with a little more dignity (despite the inductor being present).

  • \$\begingroup\$ all right, so the energy is stored and cycled in the inductor. Can you please elaborate more on "class A amplifiers behave with a little more dignity"? \$\endgroup\$ – Roker Pivic May 10 '16 at 9:48
  • \$\begingroup\$ A class A amplifier is always consuming some collector/drain current at all points on the output waveform. This prevents the "back-emf" scenario I mentioned. electronics-tutorials.ws/amplifier/amp_5.html \$\endgroup\$ – Andy aka May 10 '16 at 9:52
  • \$\begingroup\$ but when I turn it off suddently, the inductor generates huge voltage spikes, which are the cause of EMI. constant supply current has nothing to do with preventing it... \$\endgroup\$ – Roker Pivic May 10 '16 at 10:07
  • \$\begingroup\$ Then you are not operating it as a class A amplifier. If you want something other than a class A amplifier you need to describe your application and input / output expectations. \$\endgroup\$ – Andy aka May 10 '16 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.