Hi Guys,

I encountered this circuit, a op-amp with a diode feedback. The input of the op-amp is 5V. Can someone tell me what will be the output of the op-amp and why?

See attached file for the circuit.

 

Have you tried to simulate it in LTSpice?

 

We have tried simulating using the Tina-TI software and we are getting a 5V on the output. But I can't explain why it is 5V.

 

I'd expect the output to be anywhere between 5V less a diode drop (about 4.4V) to near the positive supply.

If the the output drifts below 4.4V, the diode tries to pull the inverting input below the non-inverting, forcing the output higher. If the output is higher than that, theortically the inverting input is floating so the voltages depend on the input leakage and offset.

If yout built it, possibly the output would be random or possibly it would be about 5V, depending if the diode leakage is less or greater than the opamp input current.

(If the diode is less that perfect, you can think of it as a resistor so the circuit becomes a voltage follower; the inverting input is only connected to the output.)

 

With such a high impedance the diode will have some problems conducting at all. I don't believe this is a stable circuit, a resistor is needed to bias the diode.

 

I wouldn't trust a simulation to get this one right, necessarily. Also, I think Bill is on the right track. Even if this circuit works, there is a potential problem with the diode current and the input bias current needed on the negative input of the opamp.

It would be more reliable to place a resistor between the positive supply and negative opamp terminal. This will ensure that the opamp has the bias current it needs (albeit, very small), and the diode is forward biased (again little current is needed). A very large resistor is best since it will not draw much current from the output of the opamp.

When this circuit is working, the negative terminal will equal the positive terminal voltage, and the output voltage will be a few tenths of a volt less than that.

This diode technique is sometimes used to provide a clipping voltage limit.

 

Has anyone noticed that the diode is backwards?

Here's how it works, or doesn't...The op-amp wants both its inputs to be at the same voltage, give or take tiny little millivolts of imperfection in the input transistors. When +5 is applied to the + input, the op-amp will output positive volts, ever increasing until the - input arrives at +5 volts. If the diode is "perfect", the output voltage will never get 5 volts to arrive at the - input. The negative input pin of the op-amp will not be happy and the output of the op-amp will be stuck to the positive supply voltage (minus the saturation voltage of the output stage of the op-amp).

If the diode is not perfect, some tiny little microamps of current will flow and the negative input pin will be satisfied because it has 5 volts on it. So, the op-amp will have to output 5 volts at the absolute minimum, and more if the imperfect diode has resistance that is significant compared to the input bias current of the op-amp. 4 instance, If the op-amp needs 100 nanoamps of bias current, and the diode leaks that much, the output will be 5 volts. If the diode has an apparent resistance of 1 million ohms in this backwards condition, the output will have to go to 5.1 volts to get the - input to 5 volts. Then there is the fact that various op-amps use the other polarity of input bias current. In that case, the .1 volt caused by 100 nanoamps flowing through a million ohms will subtract and the output of the op-amp will be 4.9 volts.

Got it?

 

The diode cannot be backward?

A generic opamp is assumed to be symmetrical around 0V.
The direction of the diode sets the possible output range as either 5V -0.6V to V+ (as it is now), or 5V +0.6V to V- (if it were reversed).

A 'perfect' opamp would have no input current so would not need current to 'force' it to 5V.

A real-world one could have offset from zero to significant levels in either direction so that can't be predicted.

The only thing you can know is that the output will not be in a range where the diode would conduct, as that would force the output back to the range where the diode does not conduct.

 

I think it depends on what kind of op amp it is. If it's an LM324 or similar, the input pins source a small amount of current, so the diode will conduct and the output will be one diode drop less than 5 volts; I'd guess that for this tiny current, the diode drop isn't likely to be 0.6V. But for other op amp types, especially FET input ones, I wouldn't try to predict the result. The best you can say is "Output will be at least 4V".

 

this circuit used as a voltage buffer for peak detection circuit , the diode used to charge capacitor as example , and the discharge path is locked by the diode.

 

This reminds me of when I told the board of trustees at a local church that they must recharge their fire extinguishers before the expiration date. They sent a public opion poll to the congregation, asking if the congregation thought they should recharge the fire extinguishers.

I guess I need to understand that this IS a public opinion poll and the congregation will still speculate after I explain things down to the nanoamp level.

 

I discuss this with my co-league. Per his explanation, he said that the output will still be 5V. This is because the diode will still be turned on but because there is no current flowing from the inverting input to the output of the op-amp, there will be no voltage drop accross the diode. He calls this the "3rd state" of the diode. The diode is turned on so there is still attraction of the holes and electrons in the depletion region, but since there is no current flowing there will be no voltage drop accross the diode. This explanation is somewhat correct for me but I am still looking for more ideas regarding this since I can't find any books or references where the "3rd state" of the diode is mentioned.

 

This reminds me of when I told the board of trustees at a local church that they must recharge their fire extinguishers before the expiration date. They sent a public opion poll to the congregation, asking if the congregation thought they should recharge the fire extinguishers.

I guess I need to understand that this IS a public opinion poll and the congregation will still speculate after I explain things down to the nanoamp level.

I sense your frustration. That's often the way of things and it's inevitable that some explanations don't always produce the "Aha! - Yes I get it!" response we might have anticipated. You get used to this and it's not such a bad thing that there are many viewpoints worthy of consideration. I suspect very few things I submit to the forum are considered particularly helpful. But it's fun having a go!

 

I discuss this with my co-league. Per his explanation, he said that the output will still be 5V. This is because the diode will still be turned on but because there is no current flowing from the inverting input to the output of the op-amp, there will be no voltage drop accross the diode. He calls this the "3rd state" of the diode. The diode is turned on so there is still attraction of the holes and electrons in the depletion region, but since there is no current flowing there will be no voltage drop accross the diode. This explanation is somewhat correct for me but I am still looking for more ideas regarding this since I can't find any books or references where the "3rd state" of the diode is mentioned.

I haven't got any answers for this yet...

 

If the diode is perfect, then the output will hit the positive rail as the diode is reverse biased with respect to the output voltage polarity. A real-world diode will have a little reverse leakage current, and if the input bias current to the op-amp is low enough then you've got a unity gain buffer with an (effectively) very large value feedback resistor, in the steady state, so the output voltage will be 5V plus a lot of noise plus a big offset. The extra output noise is a function of the intrinsic noise of the PN junction being applied to the op-amp input, and the big offset is a function of the diode's effective, and very large at small currents, slope resistance.

This is not a partial peak detector circuit. That would have the anode of the diode connected to the op-amp output, and the feedback taken from the diode cathode. The peak detector output (plus cap to gnd) is the diode cathode.

 

Have you read the Wikipedia article on precision rectifiers? That is the most basic schematic, there are better out there.

http://en.wikipedia.org/wiki/Precision_rectifier

 

I get your point guys. I have also taken into consideration that there is a leakage causing the op-amp to have a feedback with a very large feedback resistor. I have also read an article about the diode, that there are 3 biases for the diode. forward bias, reverse bias and the no bias. for the no bias, the diode is not forward nor reverse, the voltage applied across the diode is 0V and there is no current flowing through the diode. is it possible that this is the explanation why the op-amp acts like there is a feedback even though a diode is placed on its feedback?

 

Yes, the diode is in reverse and no it doesn't need a resistor due to the op-amp's high-input impedance. I can't see what the purpose of this ckt is, but when the 5V is applied (assuming it's +5V), the output will slew to the positive supply rail, less a volt or so. The diode, cathode to the output, will block the op-amp output from getting back to the inverting input and the op-amp will remain saturated at near the positive supply rail unless something else is changed. That's all there is to it.
Regards,
Kamran Kazem

 

Has anybody looked at this stuff?
http://en.wikipedia.org/wiki/Operational_amplifier_applications#Logarithmic_output

 

Oh yes, technically, for it to be a precision rectifier, the input would have to be -5V, but the picture only said "5V", which I take to be +5V.

Regards,
Kamran Kazem