I am trying to amplify a very high frequency square wave. The waveform is 3.3Vpp (biased so it's going from 0 to 3.3v), around 145MHz. I'd like to amplify it to around 10Vpp (doesn't matter if it's DC biased or not, but I'd prefer to stick with a single-ended power supply).

The simplest approach seems to be a simple FET switch, but I've had trouble finding the right one. Most have rise times on the order 8-20ns, way too slow.

Ideally, it'd be capable of this sort of speed and be enhancement mode, with a threshold voltage below 3.3V. The only things I've found this fast are some Cree Gan FETs which would be perfect, but they seem to all be depletion mode, with negative threshold voltages.

Doesn't need to be able to handle much power at all, 500mW at most, so something small and SMD would be perfect.

Does what I want exist? Or am I barking up the wrong tree entirely?

ETA: The output of this FET would go into a impedance matching network to match a 50ohm transmission line. I don't know the exact parameters of the network as, as far as I understand, it would depend on the FET.

The source is, as stated, a square wave betweeon 0 and 3.3v, coming into the FET on a 50ohm impedance transmission line.

  • \$\begingroup\$ What is the load pF ? This determines the slew rate from driver ESR. \$\endgroup\$ Commented May 3, 2018 at 3:12
  • \$\begingroup\$ Load pF of the FET? Or of whatever the FET is driving? The output of the FET is going into an LC low pass filter. Don't recall the first C but I can go look it up if it's relevant. \$\endgroup\$
    – jgalak
    Commented May 3, 2018 at 3:20
  • \$\begingroup\$ @jgalak At about \$10\:\frac{\text{mA}}{\text{pF}}\$, it is probably relevant. \$\endgroup\$
    – jonk
    Commented May 3, 2018 at 3:28
  • 1
    \$\begingroup\$ Sounds fishy, a X-Y problem likely. With 50 mH inductor, the input impedance will be about 50 MOhms, which is something way outside of usual realms for 145MHz signals. \$\endgroup\$ Commented May 3, 2018 at 5:32
  • 1
    \$\begingroup\$ Load (and source) aspects added to question. Sorry, I'm very new to this, very much a self-taught novice, and thus I literally don't know what info is required and what isn't. \$\endgroup\$
    – jgalak
    Commented May 3, 2018 at 15:20

1 Answer 1


So you want to drive a resonant circuit at 145MHz? Just do that, perhaps with a unity-gain buffer to provide power gain, at 0.5 watts. Using power = Freq * C * Vrms^2, you have C = power/(Freq * Vrms^2). With Vpeak of 1.6v, Vrms of 1.1v, the C = 0.5 / (145MHz * 1.1 * 1.1) = 3,000 picoFarad. Current will be (using P = I * V) = 0.5/1.1 = 0.4 amps at 3.3 volts peakpeak Power dissipation in such cases has the local power equaling the delivered power, or 0.4 watts.

Just use a class-C circuit, over groundplane, to convert the digital swing into a 0.5 watt output. Motorola/ONNN certainly sells such beasts, and has lots of application notes to guide you to success.


This may help. Or not.


  • \$\begingroup\$ Just using an off the shelf MMIC amp was my other plan, I was hoping there was an easier option since I'm starting from a square wave rather than a sine so it would be a simple on/off situation. \$\endgroup\$
    – jgalak
    Commented May 3, 2018 at 11:49
  • \$\begingroup\$ The class C is a current source so the filter does not use mH on the collector unless you want to match that with 50 Ohm R across it or change the filter \$\endgroup\$ Commented May 3, 2018 at 18:33
  • 2
    \$\begingroup\$ Class C is the trivially simple case, and quite sufficient for this sort of thing, L network in, L network out, bias the thing for a few tens of mA of standing bias, job done. There are any number of small 2M band driver amplifiers in the ham literature that will get this done. \$\endgroup\$
    – Dan Mills
    Commented May 4, 2018 at 20:08
  • \$\begingroup\$ @ Tony If the C is a current-source, then the output power is very vulnerable to drive level. That means a cascade of several "C" stages will have enormous output power fluctuations. \$\endgroup\$ Commented May 5, 2018 at 3:27
  • \$\begingroup\$ @analogsystemsrf, can you point me to one of those app notes you mention in our answer? \$\endgroup\$
    – jrive
    Commented Dec 1, 2020 at 15:48

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