# who can explain this circuits?

Discussion in 'Electronic Design' started by [email protected], Nov 21, 2006.

1. ### Guest

On the above datasheet, page 11th, there is a peak detector circuits. I
will adapt it but I didn't figure out why they put two diodes like this
and I don't know how the 1k ohm work between the two op-amps either.

Could you explain it to me, thanks a lot.

2. ### John PopelishGuest

When the input voltage is positive, the right diode
conducts, and the pair of opamps act, together to produce a
follower (100% feedback to the first opamp) that drives the
up. The resistor in series with the capacitor keeps the
capacitor from causing enough phase shift to destabilize the
pair of opamps in a feedback loop. The 1 k feedback
resistor has no current passing through it, except the bias
current of the first opamp, so it acts essentially like a
wire during this phase.

When the signal is negative, the left diode conducts,
keeping the first opamp from saturating its output all the
way against the negative rail so that it will recover
quickly when the signal goes positive, again. During that
time, the right diode isolates the first opamp form the
voltage that has been stored across the capacitor.

The 1 k resistor conducts current from the second opamp
output back to the first. This current is wasted, but keeps
from having to disconnect the the feedback path between peaks.

3. ### Guest

thanks. then, why not just put a wire instead of a 1 K ohm? and how to
pick this resistor's value?

4. ### Tom BruhnsGuest

Because if it were zero ohms, A1's more negative output would be
fighting A2's more positive output. With 1k in between, the current is
limited to (V(A2out)-V(A1out))*1mA/volt.

Cheers,
Tom

5. ### Guest

to limit current, a larger value > 1K would be better, apparently
there are other considerations like phase compensation?

6. ### John PopelishGuest

You can't tie the two outputs directly together with only a
diode between them, and have them battle for control of the
voltage. The value is not critical, but lower resistance
looks more like a wire when the feedback diode is off, than
a 10k resistor does (as far as driving the input and stray
capacitance with little delay or phase shift. But it is
about as low as you can go to keep the outputs from going
into current limit when it is being used for isolation (of
the two outputs) during the lower than peak parts of the
waveform.