1
\$\begingroup\$

I have a circuit that outputs a voltage of -10V to +10V. Then I needed to get all values as positive output voltages from 0V to 10V i.e. I absolute the negative voltage outputs.So, I searched online and got a an "op-amp absolute value amplifier circuit with one diode" from the following links:

Circuit Details : http://www.cirvirlab.com/index.php/tutorials/77-op-amp-absolute-value-amplifier-with-one-diode.html

Circuit Simulation : http://www.cirvirlab.com/simulation/op_amp_abs_value_ampl_with_one_diode_online.php

So, I rigged up the circuit as shown in the figure with the following components and values:

Op-amps - OP177 (Analog Devices)
R1 = R2 = 1k
R3 = 2k
Vcc = +10V and -10V
Diode = IN4007

How I tested:

First I gave a Vin = 5V, for which I got around approx 2.2V output and as I varied the Vin from 5V to around 7V, the output varied from 2.2V to around 3.6V approx.

Then, second, I gave an input voltage, Vin = -5V, for which I got a very low negative output value of -0.8V.

The problem is, I am not getting the absolute output voltage and getting lower output voltages compared to Vin.

I have attached an image of the circuit digram and IC. Please help me figure out what I am doing wrong, is it with the component values? or anything else.Thanks in advance.

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ I think I would change the 1N4007 (mainly used for power rectification) for something like a 1N4148. \$\endgroup\$ Commented May 23, 2015 at 9:36
  • \$\begingroup\$ OP117 or OP1117? With +10/-10 volt supplies you won't get the input and output to work with those levels as inputs. \$\endgroup\$
    – Andy aka
    Commented May 23, 2015 at 12:03
  • \$\begingroup\$ Yesterday I found another simple absolute voltage circuit. I have implemented and tested it. It works flawlessly with high linearity, gain of 0.999 and with an error of 0.8%. I will post the schematic with all the circuit details as an answer tomorrow. \$\endgroup\$
    – PsychedGuy
    Commented May 24, 2015 at 6:26

3 Answers 3

2
\$\begingroup\$

I put your circuit into LTspice but it didn't work very well even when I upped the supply voltage to +/- 12V, so I Googled "precision full wave rectifier" and then clicked on the "More images..." header.

I found the circuit below, which looked pretty good, and tried it.

Unfortunately, I got oscillations on the rectified negative half-cycles and, suspecting the phase response through part(?) of the circuit to be the reason, I plugged in a couple of LT1007s to see if I could speed the thing up and brute force the oscillation out of there, and it cleaned up everything nicely!

If you want to play with the circuit, here's the LTspice circuit list.

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ Saw a similar circuit like this by TI in "Precision Absolute Value Circuits". This too works fine, simulated it on multisim and checked, however did not try in hardware yet. But since the simulation results were right, I presume the hardware will behave appropriately and on this note, I will give your answer a tick. Between I found another simple absolute value circuit that works flawlessly with precise outputs. I have already built and tested it. Will post the schematic and details as an answer by tomorrow evening. Thanks. \$\endgroup\$
    – PsychedGuy
    Commented May 24, 2015 at 6:40
  • \$\begingroup\$ Regarding the oscillations - you want a small capacitor in parallel with D1. \$\endgroup\$
    – TLW
    Commented Jan 26, 2017 at 3:13
1
\$\begingroup\$

The image below represents the absolute value circuit that I have checked with simulation and hardware. It works perfectly fine. Reference: DavidL. Terrell, "Op Amps:Design, Applications and Troubleshooting",1996.

enter image description here

TESTED OK

\$\endgroup\$
0
\$\begingroup\$

This circuit should work as drawn. I suspect you have a wiring error or possibly a damaged part. You can measure the output voltage of the left-hand amplifier which should be railed near the negative supply for positive inputs and equal to -Vin for negative inputs.

Please correct the part number in your question to agree with the OP177 on the image.

Note that for positive input voltage greater than 2 diode drops significant current will flow between the inputs of the left-hand amplifier.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.