2
\$\begingroup\$

Wikipedia Image

From Wikipedia: "When the input voltage is negative, there is a negative voltage on the diode, so it works like an open circuit, no current flows through the load, and the output voltage is zero."

When the diode is off, shouldn't V- be equal to Vin if we are assuming infinite open loop gain? In that case, isn't Vout = Vin instead of 0 when the diode is off?

*** EDIT:
To address some of the comments, open loop gain is the gain when no feedback is used. So even if the diode is off, there is an open loop gain A on the node directly in front of the diode on the output. This of course Doesn't reach Vout, but it's there.

If there is an infinite open loop gain with no feedback, as is the case in the ideal d̶i̶o̶d̶e̶ op amp, then this equation must hold: Vout = A(V+ - V-)

If the open loop gain is infinity, then Vout/A = 0 = V+ - V- --> V+ = V-. If the input resistance is also infinity, then there's no current going into Vout and it should be zero, but it's also Vin by virtue of there being infinite open loop gain.

Edit 3:

I've answered my own question below and removed my answer from the question.

\$\endgroup\$
  • \$\begingroup\$ "...if we are assuming infinite open loop gain?" it's cancelled out by the diode's (assumed) infinite loss. So the op amp outputs an (assumed) infinite negative voltage, which is blocked by the diode! Assumptions are fun, no? \$\endgroup\$ – Bruce Abbott Apr 18 '18 at 4:22
  • \$\begingroup\$ When the diode is off there is no feedback, so no mechanism to make V- = V+ \$\endgroup\$ – Chu Apr 18 '18 at 11:34
  • \$\begingroup\$ Bruce, Open loop gain is the gain when no feedback is used. Even if the diode is off, there is still open loop gain. Don't confuse it with closed loop gain. \$\endgroup\$ – jacob Apr 18 '18 at 14:03
  • 1
    \$\begingroup\$ I'm not talking about the open-loop gain or the closed-loop gain. I'm talking about the loop gain, which is a third thing, and you shouldn't confuse it with the other two. \$\endgroup\$ – The Photon Apr 18 '18 at 14:34
  • \$\begingroup\$ I wasn't replying to you Photon, I'm aware you're using the term loop gain. I'm not confusing anything by referring only to the open loop gain in my equation. \$\endgroup\$ – jacob Apr 18 '18 at 18:42
4
\$\begingroup\$

If the diode doesn't conduct, then no current flows through it.

The op-amp has high input impedance, so effectively no current flows into the inverting input pin.

So no current is flowing through the resistor.

Therefore, what is the voltage across the resistor RL?

When the diode is off, shouldn't V- be equal to Vin if we are assuming infinite open loop gain?

If the diode is not conducting, then you don't have a closed feedback loop.

If there is an infinite open loop gain with no feedback, as is the case in the ideal diode, then this equation must hold: Vout = A(V+ - V-)

This is simply incorrect.

V- is driven equal to V+ by having high open-loop gain, and a negative feedback connection. High open-loop gain is not a sufficient condition to force V- equal to V+.

The equivalent circuit with the diode not conducting looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

There is no way for the op-amp's output to drive the inverting input in any way, towards the noninverting input or otherwise. There is no negative feedback and there is no forcing of V- equal to V+.

What happens in reality is that the output voltage is limited by the power supplies of the op-amp, putting the op-amp in saturation mode where the open-loop gain is no longer very high.

But that isn't why the inputs aren't forced together. If the input were 1 uV, and the gain 1,000,000, then the output could be 1 V, and it still wouldn't force the inverting input to 1 uV, because there is no closed feedback loop.

\$\endgroup\$
  • \$\begingroup\$ Why do you need closed feedback in order for Vout = A(V+ - V-)? Isn't this model for the ideal op amp true even without closed feedback? i.imgur.com/MFyY0Fp.png - In this case Vout/A = 0 = V+ - V- --> V+ = V- , no? \$\endgroup\$ – jacob Apr 18 '18 at 2:55
  • \$\begingroup\$ @jacob, see my answer to Op Amp operating conditions. The inputs being equal isn't an inherent feature of the op-amp. It's a consequence of the high gain and the negative feedback circuit. When there is no negative feedback, IN- won't be forced equal to IN+. \$\endgroup\$ – The Photon Apr 18 '18 at 3:01
  • \$\begingroup\$ I don't understand your justification for why V- = V+ is not an inherent feature of the op amp. Vout = A(V+ - V-) is fundamentally what an op amp does through its internal circuitry. It follows from that that V+ = V- if A = ∞ in this ideal example (see my comment above). \$\endgroup\$ – jacob Apr 18 '18 at 3:08
  • \$\begingroup\$ When the diode is not conducting, the loop gain goes 0, regardless of how high A is. The voltage at the op-amp output can go as low below 0 as it wants, and it won't affect the voltage at V-. \$\endgroup\$ – The Photon Apr 18 '18 at 3:13
  • \$\begingroup\$ I'm going to open up a new question and accept this one for now. I don't get why the "open loop gain" A would go to zero when the loop is open. \$\endgroup\$ – jacob Apr 18 '18 at 3:16
0
\$\begingroup\$

When dealing with opamps, we generally use the two rules:

  1. Opamp inputs have infinite impedance

and

  1. The opamp will adjust its output so that the V+ input and V- input are at the same potential.

When the diode is off, that means that it's not conducting any current from the opamp output. The only other thing connected to that node is the V- input, which is high impedance and can't source any current, so no current is flowing and Vout is at 0V (pulled down by the resistor).

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

The real problem I had was with the idealized model of the op amp that I was using.

Below the Open Loop Gain A is a property of the op amp. It does not change depending on the circuit configuration.

Sedra and Smith Microelectronic Circuits 7th Edition page 62

The problem with this model is that it does not capture the effect of the output on the input. From the diagram above, you can attain:

Using V+ = V2 and V- = V1,

Equations

Although you can manipulate the equation of the output voltage to attain V+ = V-, there is no feedback to actually have the output have any effect on the input. In this idealized model, it only captures the effect of the input on the output, but not the output on the input, so solving for the inputs given the output doesn't attain a realizable result. There is no causal link from the output to the input. There is only a link from the input to the output in this model.

The equation for the open loop gain cannot be used to determine the inputs. That's really the confusion that I had.

Given this clarification, it should be clear that with infinite input resistance in the ideal op amp, that there is no current going through R_L, and that V- is not equal to V+. Therefore, Vout (or V-) is pulled down to 0.

\$\endgroup\$
  • 1
    \$\begingroup\$ The logic here is backwards. The final equation is telling you what conditions you have to meet at the inputs to avoid saturation. It isn't telling you that the high gain of the op-amp forces the inputs to be equal. So it doesn't help you analyze your rectifier circuit and how it reacts to changing inputs. \$\endgroup\$ – The Photon Apr 18 '18 at 22:35
  • 1
    \$\begingroup\$ Since the question is how to find out what's the output voltage of the rectifier circuit when the input is less than 0, this does not answer the question that was asked. \$\endgroup\$ – The Photon Apr 18 '18 at 22:38
  • 1
    \$\begingroup\$ The equation is valid when the op-amp is in the linear operating mode. But the circuit and conditions you asked about don't put the op-amp in the linear operating mode. The equation is not valid for the situation you asked about in the posted question. \$\endgroup\$ – The Photon Apr 19 '18 at 0:12
  • 1
    \$\begingroup\$ The equation \$V_{out}=A(V_+-V_-)\$ is only valid when the op-amp is in linear operation. It is not valid in this circuit when the diode is not conducting, which is the situation you asked about. Re-read your Sedra & Smith and be sure you check what are the preconditions for the equation. \$\endgroup\$ – The Photon Apr 19 '18 at 0:37
  • 1
    \$\begingroup\$ See slide 22 here, which seems to be based on Sedra & Smith. Alternately, if you want to assume an op-amp so ideal that it doesn't have saturation behavior, then you can't assume \$V_{out}\$ doesn't itself go to infinity, which is a necessary assumption for the way you showed that \$V_+ = V_-\$. In the rectifier circuit, if there is no saturation then the op-amp output (not the circuit output) will go to \$-\infty\$. \$\endgroup\$ – The Photon Apr 19 '18 at 1:54

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.