I built a Pierce oscillator that generates 13.5MHz @ 7Vpp signal from a V_cc=12V. But I need to drive a capacitive load of 5 ohms @ 6Vpp. I'm not sure how to do this.

First I need a buffer to prevent loading of the signal source. FETs have a large input impedance so I think it's appropriate, so I used a source-follower with a large \$R_S\$. However 7Vpp swing seems too large and leads to distortions, is there any way around this? Using a resistor ladder to scale down the voltage first, then connecting it to the gate seems stupid.

I also tried to use a 2n3904 emitter follower, but I noticed that the emitter voltage rises fast but falls slowly and cannot keep up with the signal, therefore I got reduced and distorted output. The falling waveform looks like the RC voltage decay, to make it fall faster I have to substitute a smaller emitter resistor, therefore increasing the loading of the source.

So the first question is how to buffer the signal without significant distortions.

The next problem is how to match the impedance. If I got 7Vpp and 50 ohm of output impedance, using a transformer to match 50 to 5 ohm would require a 3:1 turns ratio. That is, my 7Vpp would become 2.3Vpp, which is too low. Even if I somehow managed to get 12Vpp (Vcc=12V), output would only become 4 Vpp, still not enough.

Is there any way around this?

  • \$\begingroup\$ For the buffer, look at video opamps, they should be able to faithfully drive something like 50 ohms. \$\endgroup\$
    – user16324
    Dec 17, 2014 at 16:49
  • \$\begingroup\$ I don't want to use opamps or ICs, because I think I would learn nothing that way. I'm asking because I want to learn the thinking behind the design of these circuits. \$\endgroup\$
    – seilgu
    Dec 17, 2014 at 16:52
  • 1
    \$\begingroup\$ You should try to minimize source impedance, not raise load impedance to match. Since you ruled out IC's, you might consider using a different amplifier topology that preserves more of the inherent bandwidth of the transistors. For example, the cascode (look it up). Maybe you can try a cascode with a current source load? Outside of my experience, but I have read about such things in school. \$\endgroup\$
    – user57037
    Dec 17, 2014 at 20:43

2 Answers 2


As in your description, the FET source follower works to certain extent but with distortion. The source of the distortion is probably not exactly because the signal amplitude is too large, but a secondary effect to that. You mentioned a 12V power supply and 7V signal (assuming the signal is centered), so at the low end, the signal would be at around 2.5V. Most likely, that is too low to turn on the FET adequately.

So maybe give this a try -- bias the base of the source follower FET to around 8V with couple of resistors. Capacitively coupled the input signal to that.

As to matching the impedance, typically, the requirement to match the impedance is because the amplifier is designed to only work at a certain load impedance. That does not seem to apply in what you are doing, so there is no need to match impedance. The amplifier just needs to have enough drive (low enough impedance) to get the signal level that you want.


You may want to look at the push pull driver circuit like that:

enter image description here

I hope that helps.

  • \$\begingroup\$ That is what seiglu initially tried (in a recent post). It did not work well with 3904/3906 because of limited beta at 13.5 MHz, and possibly other reasons. Do you know of some transistors that can maintain high beta at 13.5 MHz that are worth considering? \$\endgroup\$
    – user57037
    Dec 17, 2014 at 20:28
  • 3
    \$\begingroup\$ That's actually a bad idea for high frequencies. The problem is called slew-rate-limited crossover distortion. If the emitter-base voltage is assumed to be 0.7 volts, then consider a positive ramp which passes through zero. At an output of -.0001 volts, the common base has to be -.7 volts, but at +.0001 volts the base has to be +.7 volts, so the base voltage has to jump by 1.4 volts in a very short time. For low frequencies this is not a problem - for 13 MHz it's a killer. Incidentally, look up "Class AB amplifier" for a discussion. \$\endgroup\$ Dec 17, 2014 at 23:39
  • \$\begingroup\$ Converting to class AB is not going to salvage this. The problem is not distortion. The problems is that the beta of the transistors is so low at 13 MHz that it loads down the source. The OP already said no IC's (discrete only). \$\endgroup\$
    – user57037
    Dec 18, 2014 at 3:55

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