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  1. I found this VGA from texas instruments. On page 34 you can see a AGC (automatic gain control) loop, does this set the output voltage (Vout) to the Vref voltage?

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  1. The VCA821 needs a bipolar supply voltage, I wanted to use this switched capacitor inverter to generate the negative supply voltage of -5V. My question is will the switched capacitor inverter be able to deliver enough current for a stable negative voltage on the VCA821s negative supply pin?
electrococuk
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    It says in the datasheet: *When the output level exceeds the reference voltage (VREF), the integrator ramps down reducing the gain of the AGC loop. Conversely, if the output is too small, the integrator ramps up increasing the net gain and the output voltage.* so the point where the gain is **not corrected** is "in the middle" meaning output level = Vref. Don't expect this circuit to **exactly** adjust the gain so that the amplitude at the output is **exactly** Vref. There might be some offset. This will also depend on what type of signal is used. – Bimpelrekkie Jun 01 '21 at 14:16
  • @Bimpelrekkie will this circuit be fast enough to set a small impulse with a width of 25ns to Vref? – electrococuk Jun 01 '21 at 14:18
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    *will the switched capacitor inverter be able to deliver enough current...* That LM2682 can deliver up to 10 mA. Is that enough for the VCA821? Look up "quiescent current" in the VCA8221's datasheet. (hint: no). – Bimpelrekkie Jun 01 '21 at 14:20
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    *will this circuit be fast enough to set a small impulse with a width of 25ns to Vref?* No, this is a circuit that needs continous signals, not a small pulse. and 25 ns is waaaaaaay too fast for the time constant of the gain control loop (0.1 uF cap and 1 kohm resistor, surrounding the OPA820). For short 25 ns pulses, you really need to use a different circuit like a fast comparator. – Bimpelrekkie Jun 01 '21 at 14:22

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