I don't like how every VU meter circuit has a resistor draining the capacitor, the decay doesn't look good to me.
So I designed this circuit to discharge the cap linearly, so instead of the VU meter needle dropping; Quickly, slower, even slower, slowlllly setlling at -whatever db.
Instead, it goes; dropping steadily, still dropping at the same speed, still.
I also made it so it works on both waves of the signal, this way in an LED VU meter, you don't have to problem that when the signal is low enough frequency, the LED's jump between one and another constantly. this happens at like 15hz so it's not a big thing, but still. And the linear discharge also helps with this because you don't have to discharge it as quickly to compensate for the slower discharge over time.
The emitter follower before the buffer op amp is because the input bias current of the op amp will discharge the capacitor not linearly. If you have a bigger capacitor like 22uF and decrease the 100k resistor in the current sink circuit, the problem becomes less aparent and you may delete the emitter follower.
Or you can chain 2 emitter followers together to get high enough gain (otherwise the same problem happens, only even more problematic with this), And then you don't need the buffer. And at this point if you increase the capacitor to like 470uF or something high like that, you increase the attack time too because the op amp has finite current drive capability.
The current is controlled by the 100k resistor on the emitter+base connection in the current sink circuit, less resistance = more current, faster decay.
Ofc you can omit the other phase circuitry and save on op amps.
Below the circuit I drew a single phase variety. You want at least ~700mV at the current sink to keep it stable, The chained emitter followers will drop the voltage around 1.2V and since the circuit keeps the output at 0V, the voltage at the current sink will be 1.2V.
If however you're using a single emitter follower into a buffer or whatever, you may run into problems with the current sink dropping in current. If this happens you can put a diode after the buffer or before the buffer to add another 0.6V drop.
Also you don't need a rail to rail op amp, because of all the diode drops the output of the op amp will be higher than 0V.
I did make a version of this circuit with an ancient LM381, adapting an existing VU meter circuit. It was a bit tricky because one input of that op amp is internally biased.
To make it work I shorted the DIFF negative input to ground, and instead connected the feedback loop to the Single Ended negative input. (Since internally this is connected to a 10k to ground, I didn't need an R2, I just made R1 correspondingly higher value)
Then I connected a 250k resistor from the positive input to ground. Internally the positive input base is biased with a 250k resistor to 1.2V, So if I connect a 250k to ground it will be biased to 0.6V instead. And that will account for the 0.6V drop from the positive input base, to the negative S.E input which is the emitter of the same transistor.
Surprisingly the offset voltage was tiny with that ancient op amp. only 50mV with 5x gain. maybe I got lucky with the silicon lottery or maybe they actually have that small of an offset voltage.
But yeah you want an op amp with low offset voltage. The output voltage at 0 signal will be offsetVoltage*Gain
The output resistor you'll choose to limit current. Since the output voltage can swing to whatever the op amp can swing to, minus the 2-3 diode drops. 10V across a small resistance VU meter wont be healthy for it.
Also, you can limit the output voltage by making the reverse diode a zener instead, The one that goes directly from the output of the op amp to the negative input. The output will then be limited to the zener voltage, minus the 1.2 to 1.8V from the Vf drops.
It really does look better imo than the regular resistor dropper.
TimFox can modify the circuit to be real VU Meter ballistic spec compliant if he wants to.
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You can also make it overshoot like an analog VU meter if you want, by adding a ~10uF capacitor from the negative input to ground.
Keep in mind that the capacitor effect will be dependent on the impedance of that point, for the top circuit it will be R1 || R2, or 1/(1/R1 + 1/R2). So if you instead had higher impedance at that point, you'd choose a lower value capacitor.
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