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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.

circuit

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  • \$\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 Sunnyskyguy 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 Mar 11 '18 at 21:51
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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.

EDIT:

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.

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  • \$\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 Mar 12 '18 at 0:11
  • \$\begingroup\$ See my EDIT: section in the answer. \$\endgroup\$ – Sparky256 Mar 12 '18 at 0:40
  • \$\begingroup\$ Ok but you did give values but what math was used to calculate those values? \$\endgroup\$ – Mike 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\$ – Sparky256 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 Mar 12 '18 at 1:59

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