# Circuit understanding

Discussion in 'The Projects Forum' started by blah2222, Jul 19, 2012.

1. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
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36
Hi, I just have a quick question about the attached circuit. For an input sinusoid with zero DC offset I see how the positive portions of the cycle have a gain of +1, but for the negative portions I am getting a gain of -2.

The intended design was to rectify the signal, but I don't know why anyone would want the rectified negative portions to be of a higher magnitude than the positives.

Any insight is much appreciated.
Thanks

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Is your question related to the theory of operation or a result you are observing in practice / simulation?

3. ### WBahn Moderator

Mar 31, 2012
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When you say you are getting a gain of -2, is that by analysis, simulation, or actual measurement?

By analysis, I get that the gain should be as expected, namely that Vout = |Vin|.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
785
There must be an echo somewhere ...

5. ### WBahn Moderator

Mar 31, 2012
20,416
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Yep, I thought I was the first responder, but it took long enough for me to mentally walk through the negative input case that you snuck in there ahead of me, you *\$&#*%!

Last edited: Jul 19, 2012
6. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
574
36
This is by my understanding of how the circuit works on paper.

On positive portion of input signal D3 is shorted and D4 is opened:

Since all the resistors are all 10k I will just use R.
Vi = input voltage
Vo1 = output of first op amp
Vo2 = output of second op amp

$\frac{Vi}{R} = \frac{-Vo1}{R}$

No current flows to the positive terminal of the second op amp so it is tied to ground.

$\frac{Vo1}{R} = \frac{-Vo2}{R}$

$Vo1 = -Vo2 = -Vi ---> Vo2 = Vi$ Gain+ = +1 V/V

On negative portion of input signal D4 is shorted and D3 is opened:

$\frac{Vi}{R} = \frac{-Vo1}{R} + \frac{-Vo1}{2R}$

$Vo1 = \frac{-2Vi}{3}$

$\frac{-Vo1}{2R} = \frac{Vo1 - Vo2}{R}$

$Vo2 = 3Vo1 ---> Vo2 = -2Vi$ Gain- = -2 V/V

Unless there is something fundamentally wrong with my reasoning, I am confused. I am also assuming ideal diodes.

**EDIT**

NVM I can't do math apparently. I forgot to carry the factor of 2 for the Vo2. Blarghhhhh!

Should be:
$2*Vo2 = 3Vo1 ---> Vo2 = -Vi$ Gain- = -1 V/V

Regardless, is this a common circuit for this rectifying low power signals?

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
785
I'll withdraw and let WBahn handle the rest.

8. ### WBahn Moderator

Mar 31, 2012
20,416
5,824
You don't have to assume ideal diodes.

When there is a negative input, assume D4 is conducting (as a normal, real diode) and that D3 is shut off (and be sure to check that the final result is consistent with these assumptions).

The current drawn from the input pin is Vin/R.

On the upper branch, the only place for current to come from is Vout and the only place for it to go is the summing junction (-ve input of the first amp). So that current is Vout/(3R).

The voltage at the -ve input of the second amp is, by voltage division, (2/3)Vout.

Assuming a virtual short across the inputs, the +ve input will also be a (2/3)Vout. Thus the current in R9 will be (2/3)Vout/R.

Applying KCL at the summing junction, we have:

Vin/R + (1/3)Vout/R + (2/3)Vout/R = 0

Vout / Vin = -1 V/V

The opamp will put out whatever voltage is needed to servo the voltage at the resistor to the ideal value, so it will compensate automatically for the forward voltage drop.

I don't see how you are getting your last line.

The next to last line leads to:

Vo2 = (3/2)Vo1

Using the line above that, you have

Vo2 = (3/2)(-2/3)Vi = -Vi

9. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
574
36
Yeah, I corrected myself in the EDIT. Clumsy math.