I haven't done a bad job for this amplifier circuit since I got it to work with 470K for RV and 10K for RG but what I'd like to do is create a better quality sound output. Maybe my 1/2 watt 8-ohm speaker is to blame but who knows.

So anyway, I tried a stiffer ratio and used 10K for RV and 10K for RG and I get no sound output. I notice I don't get any sound output for any values of RG until I make RV substantially high (like 470K)

Given the components I chose in this circuit and given each audio device is 8-ohms, how do I calculate the optimal values for RG and RV? I'm trying to avoid the need of tuning multiple potentiometers in the future as I will make multiple sound units later on.

For my tests, AMP_PWR is VCC and VCC is 5V. later on I may make AMP_PWR a higher voltage.


  • \$\begingroup\$ A lot of tolerance errors with Vgs(th) vs Ron vs Vcc , which is far too low and not the best topology with risk of shootthru \$\endgroup\$ – Tony Stewart EE75 Mar 11 '18 at 20:24
  • \$\begingroup\$ Should the polarity on the 22uF connected to the earphone be reversed? \$\endgroup\$ – AlmostDone Mar 11 '18 at 21:04
  • \$\begingroup\$ Crap design from the start, 5V is hardly sufficient to turn on the fets, missing decoupling and I Think pin 6 on the 393 should probably be connected to the top of the 220p cap, not sure how the thing is supposed to work as is. Use a real mosfet driver, add some deadtime and make it self oscillating with a phase lead network in the manner of that Phillips patent, you will be much happier. \$\endgroup\$ – Dan Mills Mar 11 '18 at 21:09
  • \$\begingroup\$ As for polarity I might change the cap type. I did a direct connection from 555 output to 393 input because the output from the 555 is digital. If I added coupling at that point then I'd need extra parts and more math to deal with. The 555 is wired as a typical astable oscillator. \$\endgroup\$ – Mike -- No longer here Mar 11 '18 at 21:51

This circuit will NOT work for four reasons.

  1. Your using an earphone as a microphone. Spend a buck and buy a condenser mike, which needs a 1.5 volt bias through a 2 K to 10 K resistor.

  2. Your signal from the microphone needs to be amplified at least 100 times to 1,000 times. Use a quiet TL072 op-amp

  3. The source waveform for the comparator needs to be a perfect triangle waveform of about 300 KHZ to 500 KHZ.

  4. Tying in with that issue is that you need a very fast comparator, such as the LT1016 (5 nS)

  5. After those changes then you can fine-tune the output stage, which is very reactive to load changes, etc.


Vcc should be +15 volts, just to make the op-amp work and drive the MOSFET's to saturation. The (-) input of the op-amp should go to the 200 pF cap on the 555, where you have a crude triangle waveform. See if the LM339 will work first, as it can be powered by 15 volts on Vcc. The LT1016 needs +/- 5 volts rails, not enough to drive the MOSFETS.

Rg and Rv should be 100 K each, then add 1.5 volts to the microphone side by using a 22 K to +15 volts and a 2.2 K to ground, tied in the middle to the mic input. This will give the mic the 1.5 volts it needs.

  • \$\begingroup\$ Ok I can definitely make changes to 1 and 2. I know it says earphone as input but at the moment the input is audio from the computers earphone port in which the volume can be adjusted to very loud. So I guess I need to change my resistors in the 555 so its output is between 300 and 500Khz? and what about the resistors I mentioned in my question? how do I calculate them? \$\endgroup\$ – Mike -- No longer here Mar 12 '18 at 0:11
  • \$\begingroup\$ See my EDIT: section in the answer. \$\endgroup\$ – user105652 Mar 12 '18 at 0:40
  • \$\begingroup\$ Ok but you did give values but what math was used to calculate those values? \$\endgroup\$ – Mike -- No longer here Mar 12 '18 at 1:09
  • \$\begingroup\$ Because the comparator runs at 15 volts, it needs to be biased at 1/2 Vcc to work properly, so Rg and Rv must be equal, but not load down the mic sound. The CMOS 555 will work at 15 volts as well. \$\endgroup\$ – user105652 Mar 12 '18 at 1:14
  • \$\begingroup\$ Ok, so what math determines why a pair of 100k resistors are better than 10K or even 470K? "load down the mic" isn't math to me \$\endgroup\$ – Mike -- No longer here Mar 12 '18 at 1:59

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.