Re: Using Op-Amp to drive Amp Gauge with Audio Signal

On Feb 26, 11:47 am, John Popelish <jpopel...@xxxxxxxx> wrote:
Tom wrote:
Duh, a rectifier. That makes sense. A am a little confused as to
where it goes, based on your description. From what I understand, the
op-amp output goes through a voltage divider to ground. The tap on
this divider feeds the inverting input. I gather that the audio input
goes to the non-inverting op-amp input. (or does it go to the
inverting input with an inverting input resistor to set the gain?)

Either will work. connecting the signal to the non
inverting input will load the signal source the least, so
that might be best if you just want to borrow the signal
from some point in the amplifier with minimum chance of
changing anything. The signal will have to be ground
referenced (zero volts when no signal) not biased at some
DC, or you will have to add a capacitor to block the DC and
add a resistor to ground to bleed the cap voltage to zero on
the opamp side.

I
am getting confused with -> "But replace the output to inverting input
resistor with the AC terminals of a bridge rectifier". Do you mean
the first resistor in the voltage divider, right before the tap?

Yes. the one that normally connects between the output and
the inverting input. The current that would pass through
that resistor, instead passes through the bridge rectifier
and the meter movement. The bridge rectifier makes sure
that for both positive and negative currents, they all go
through the meter in the direction that will deflect it
above zero.

So
if I understand correctly op-amp-out -> rectifier(AC) -> tap ->
resistor -> ground. The tap feeds back to the inverting input?

Right. The opamp produces whatever output voltage it takes
to force the two inputs to have the same voltage. When the
input on the non-inverting input swings, say, positive, the
output must force current through the bridge and meter
movement and then through the grounded divider resistor,
till that resistor's voltage drop matches the input voltage.
The amount of current that takes for a given input voltage
depends on the value of that grounded resistor. For
instance, a 1k resistor will need a milliamp through it to
produce a 1 volt drop. And that milliamp will be delivered
by the output and passing through the meter before it gets
to the resistor.

--
Regards,

John Popelish

Thanks again John,

Its seems to make sense. I will give it a try!

Tom
.

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