Non-inverting precision rectifier

"Bitrex" <

** The circuit operates from a single +9V rail with a divider providing a +4.9V input bias to each op-amp.

So it cannot swing negative.

.... Phil

On the G OP-amp, you'll see the (-) input, pin 2 is getting the same signal as the H op-amp how ever, you'll notice the 1 uf cap there just after the 1K R.., This is causing a time delay, so when the signal is in the + region and lets say it's going up, it'll be more on the (-) input of the G op-amp (Pin 2) than will be on the (+) of the G op-amp (pin 3)

So, it looks like its generating a - output only when the reference is ramping in the + direction and when the ramping stops, the output will then drop back to some fix low point.

That whole circuit over there looks like a + & - response output to to the reference moving from a DC component point. In other words, the reference does not have to be setting at 0 for a fixed point, it could be at any point at an idle, much like using a capacitor to block the DC component of the signal...

If you look at IC's F and E, they are performing the same job with the exception that IC F is allow the other polarity...

After looking at the use of the Flip Flop, it is placing one of the FF in it's Set or Reset state depending on which way the reference is going and IC's G and E are nothing more than a comparator switch..

I guess if you were looking for a way to monitor an analog signal for a sudden change in direction, this would be a good way to do it..

Jamie

Indeed you're correct. In the circuit it seems they're using the 4.9V bias as a virtual ground, in my simulation to simplify things I'm using a 4.5 volt positive supply and a 4.5 volt negative supply, with ground as the midpoint. Sorry for not clarifying that. In my alteration I would think that the output should not go below 0 volts, but it does.

Nevermind, I see the problem. It's not going to work properly the way I have it set up, not without that 4.9V input bias. Thanks.

Thansk for your reply - I guess I thought I understood, but I'm confused again now. Isn't the point of having the precision rectifier in the circuit to get only the positive or negative half cycles?

In my simulation now, I have a 1 volt sine wave going into the precision rectifier with a 4.9 volt DC bias applied, as per the original schematic. The op amp positive supply is tied to Vcc, the negative supply is also tied to ground, as per the schematic. I would think that the output of the second diode (before the 1k resistior) would be the signal going up from 4.9 volts to 5.9 volts, and then the part of the signal that goes below 4.9 volts being cut off.

That's not what I'm seeing, in the simulation the signal goes up from

4.9 to 5.9 as expected, but then dips down again to 4.5 volts.

Well it would in that case. If you are doing a split rail supply, you would in fact, drop below 0 Volts on the outputs.. That isn't what that circuit was design to do but most likely will work, I guess.

That part of the circuit you pointed out is noting more than a slope direction detector, that simply generates a pulse to place the FF in a latching state that will indicate the direction..

Jamie

"Bitrex"

** The circuit is NOT a precision rectifier.

It is a form of peak detector.

.... Phil

Well, I'm not sure what I was doing incorrectly, but the circuit is working OK in simulation now. I believe the function of the peak detector (precision rectifier + RC circuit) is to dynamically adjust the threshold of the comparator (G and E on the schematic) so that the output pulses going into the 4013 have a consistent pulse width, regardless of the amplitude of the guitar signal. If there were just a comparator with a fixed reference going into the 4013 low amplitude signals would have a narrower pulse width, or might not trigger the comparator at all.