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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

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    \$\begingroup\$ could you be probing with a 1M scope probe? \$\endgroup\$
    – glen_geek
    Commented Oct 18, 2018 at 22:52
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    \$\begingroup\$ Just confirmed, I am. Was I loading the circuit? \$\endgroup\$
    – Leoc
    Commented Oct 18, 2018 at 22:53

3 Answers 3

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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.

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  • \$\begingroup\$ Sorry, didn't realize it was important factor to take in. I am using a 1MEG probe. \$\endgroup\$
    – Leoc
    Commented Oct 18, 2018 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
    Commented Oct 18, 2018 at 23:33
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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.

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  • \$\begingroup\$ Appreciate thank you, Ill shall study this and give it a try. \$\endgroup\$
    – Leoc
    Commented Oct 18, 2018 at 23:14
  • \$\begingroup\$ Just checked your topology. It acts more so of a half bride than a full bridge \$\endgroup\$
    – Leoc
    Commented Oct 18, 2018 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\$ Commented Oct 19, 2018 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
    Commented Oct 19, 2018 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
    Commented Oct 19, 2018 at 2:33
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Looking for the idea

This circuit solution is much more clever than it seems at first glance. It gives the illusion of being a very imperfect half-wave rectifier... and has led some to improve it (including a second diode to maintain a virtual ground on the inverting input) to make it a perfect half-wave rectifier. But I think the idea here is different:

This circuit is intended as a "slightly imperfect full-wave rectifier", and the "improvement" only destroys this idea.

Operation

Here is how the circuit works shown in a slightly different way than generally accepted.

Negative input voltage

The diode is forward biased (on) and the circuit is simply an inverting amplifier. In this configuration, we can see (with a little more imagination) two "voltage sources" - negative and positive, connected in parallel to the load.

The first of them consists of the input voltage source and the two 2k resistors in series; so it behaves as a real (imperfect) voltage source with 4k internal resistance.

The second "voltage source" consists of the op-amp output and the low-resistance diode in series. Since the negative feedback voltage is taken after the diode, the op-amp compensates for the forward voltage VF by increasing its output voltage by VF. In addition, through the mechanism of negative feedback, the op-amp compensates for the influence of the load. As a result, the circuit is perfect and behaves as an "ideal" (perfect) voltage source with (almost) zero internal resistance.

In the unequal "struggle" between the two voltage sources, the second one (the op-amp) prevails and its voltage is applied to the load.

More precisely, the voltages of the two sources are summed (subtracted) through their respective resistances and their sum is applied to the load. And because the op-amp output resistance is much less than the input source resistance, the load voltage is almost entirely determined by the op-amp output voltage.

Positive input voltage

Now the diode is backward biased (off) and the second voltage source (the op-amp) is disconnected. Only the input source remains to operate and it supplies its voltage through the network of two resistors to the load. Figuratively speaking, the feedback has become a "feedforward":-)

And because the total resistance of the resistors is relatively low (a few kohms) and the load resistance is assumed to be high (hundreds of kohms), almost all of the input voltage is transferred to the load. A straightforward solution to improve the circuit is to insert a voltage follower before the load.

Summary

Let's finally summarize the operation of this full-wave rectifier.

During the negative half-wave, the op-amp makes an inverted copy of the input voltage and passes this positive voltage through the diode to the load.

During the positive half-wave, the input voltage is applied through the feedback network directly to the load.

Simply put, the rule for making a full-wave rectifier is: the negative half-wave is inverted and the positive half-wave is not.

Generalization

Extracting general rules (principles) from specific circuit solutions is no less important than giving precise answers to specific questions.

Following this approach, we can see this configuration (only at negative input voltage here) in any negative feedback inverting circuit where the input source tries to feed its voltage directly to the load but the op-amp does not allow it and forces its output voltage to the load.

So, if the feedback resistors in op-amp inverting circuits are with too low resistance and the input source is too powerful, it will dominate and its voltage will be transferred directly to the output (undesired).

In contrast, in non-inverting negative feedback circuits this is not observed because there the input voltage source has no connection to the op-amp output.

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