1
\$\begingroup\$

Its been a while I took a long hiatus due to not having a scope any longer, but I finally saved up and got myself one.

I was playing around with a precision rectifier (OPA350PA) and I came across something I don't understand.

Link to datasheet: OPA350 DATASHEET

What's going on with the positive cycle? It's somehow losing voltage somewhere. To make things more confusing for me I did it on paper and the diode is open (R.B) during the positive cycle so there's no current flowing thru the resistive network thus making all the voltage at each node the same as the input. Meaning Output should be equal to input.

Two case scenarios:

Case 1:

schematic

(Orange input / Green output)

enter image description here

Case 2:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

Improve this question
\$\endgroup\$
2
  • 1
    \$\begingroup\$ could you be probing with a 1M scope probe? \$\endgroup\$ – glen_geek Oct 18 '18 at 22:52
  • 1
    \$\begingroup\$ Just confirmed, I am. Was I loading the circuit? \$\endgroup\$ – Leoc Oct 18 '18 at 22:53
2
\$\begingroup\$

You haven't provided details about how you are measuring the signals, but I'm going to guess that you are using an oscilloscope with 10 Meg probes.

The drawback of this circuit topology is that the output impedance is different between positive-going half-cycles and negative (on the input). The output impedance is essentially zero for negative half-cycles but is the sum of the two resistors for positive half-cycles.

In your second example, the output impedance of the rectifier is 200k. Work out what the voltage drop is with your 10M scope probe and you will most likely find the value that you calculate matches what you are measuring.

Improve this answer
\$\endgroup\$
2
  • \$\begingroup\$ Sorry, didn't realize it was important factor to take in. I am using a 1MEG probe. \$\endgroup\$ – Leoc Oct 18 '18 at 22:54
  • \$\begingroup\$ So I think I did the math right, however I don't see how the 2k topology is conserving the input voltage (I understand) but on paper isn't working out for me. Vo/Vin = (1/1+(R1/R2))) Voltage divider essentially. Positive Cycle Output impedance is R1 = 2k + 2k + 1M = 1,004,000 Ohms R2 = 1M (The scope is acting like a load). Vo/Vin = 0.499? Why is the scope outputing the exact input then. \$\endgroup\$ – Leoc Oct 18 '18 at 23:33
1
\$\begingroup\$

Your circuit, as shown, is behaving exactly as it should. Dwayne Reid has correctly identified what is going on.

In order to get what you want (which is a different matter), try adding a 10k resistor from the cathode of the diode to ground.

Now, during positive cycles, the diode leakage and the R1/R3 path current will produce a small voltage.

Even better, add another diode D2 like so

schematic

simulate this circuit – Schematic created using CircuitLab

Now, during positive inputs the Vin/R3 current will be supplied by the forward-biased D3, and the output will go to zero. With the -input of the op amp always held to (approximately, but very closely) zero, you'll find the circuit works much better.

Improve this answer
\$\endgroup\$
7
  • \$\begingroup\$ Appreciate thank you, Ill shall study this and give it a try. \$\endgroup\$ – Leoc Oct 18 '18 at 23:14
  • \$\begingroup\$ Just checked your topology. It acts more so of a half bride than a full bridge \$\endgroup\$ – Leoc Oct 18 '18 at 23:23
  • \$\begingroup\$ @Pllsz - Yeah, it is. I apologize for the confusion, but I thought you realized that a "precision" full wave rectifier is not possible with your circuit. You can get fairly close by using small R1/R3 and a large load, but this is not what is usually meant by a precision rectifier. \$\endgroup\$ – WhatRoughBeast Oct 19 '18 at 0:58
  • \$\begingroup\$ Oh Sorry, didn't mean to confuse, however I found out. You guys were right it was due to the loading affect of my probes. \$\endgroup\$ – Leoc Oct 19 '18 at 1:13
  • \$\begingroup\$ @WhatRoughBeast. That circuit is wrong. On positive inputs D2 will short out any bypass ability by conducting it into the op-amps output. In other words positive inputs will have worse than zero gain. The op-amp will have no output on positive inputs. \$\endgroup\$ – user105652 Oct 19 '18 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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