# Precision op amp voltage

Thread Starter

#### Zaryab Saeed

Joined Feb 10, 2016
3
Hi, I have a precision op amp and a sinusoidal input voltage at the non-inverting input (as in the figure) and am required to find the voltage waveforms at X and Y. According to my knowledge, when Vin is positive, Vout is equal to Vin and hence the graph for Y should be the same as for Vin and since X is being taken from that voltage, its waveform should be the same too, but with a lower amplitude. When Vin is negative, Vout (and hence X and Y) is zero. So what I essentially get is sinusoidal wave for parts where sine(x) is positive and zero elsewhere. Is this approach right or am I missing something?

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
Your logic is correct for Vin and Vout (X). Be careful with the voltages the op amp is producing (Y) when the input it negative.

#### crutschow

Joined Mar 14, 2008
27,237
.................., Vout is equal to Vin and hence the graph for Y should be the same as for Vin and since X is being taken from that voltage, its waveform should be the same too, but with a lower amplitude. ..................
Looks like X is the same as Vout with the same amplitude.
Why do you think X has a lower amplitude? #### MrAl

Joined Jun 17, 2014
8,256
Hi,

For positive portions the output can either be slightly lower or slightly higher but only by a small amount due to input offset. Typical offsets range from 10uv to 10mv depending on the op amp part.

#### WBahn

Joined Mar 31, 2012
26,398
But the 'X' signal IS the 'Vout' signal! How can they be any different at all. Even if you argue that there is a wire between them and, therefore, an IR voltage drop, we would be talking about milliohms (or microohms) times nanoamps (or picoamp). Input offsets don't change that at all.

#### WBahn

Joined Mar 31, 2012
26,398
Hi, I have a precision op amp and a sinusoidal input voltage at the non-inverting input (as in the figure) and am required to find the voltage waveforms at X and Y. According to my knowledge, when Vin is positive, Vout is equal to Vin and hence the graph for Y should be the same as for Vin and since X is being taken from that voltage, its waveform should be the same too, but with a lower amplitude. When Vin is negative, Vout (and hence X and Y) is zero. So what I essentially get is sinusoidal wave for parts where sine(x) is positive and zero elsewhere. Is this approach right or am I missing something?
If X is the voltage at the inverting input of the op-amp, then how can it ever, under any condition, be any different than Vout since Vout is hard connected to X by a piece of wire (which is almost certainly assumed to be resistanceless)?

When Vin is negative, even if Vout is zero, why does this mean that Y is zero. If Vin is -1 V and Vout is 0 V, what is the differential input voltage to the op-amp? What is the corresponding output voltage Y?

#### MrAl

Joined Jun 17, 2014
8,256
But the 'X' signal IS the 'Vout' signal! How can they be any different at all. Even if you argue that there is a wire between them and, therefore, an IR voltage drop, we would be talking about milliohms (or microohms) times nanoamps (or picoamp). Input offsets don't change that at all.
Hi,

I am referring to the nodes labeled "Vin" and "Vout", not X or Y, but let me restate this anyway and include Y (X is connected directly to Vout as noted).

When Vin goes positive, the output at Vout goes positive, but may be slightly lower or higher than the Vin because of the input offset of the op amp. The reason Vout can go positive is because the output Y goes positive enough to forward bias the diode up to the level of Vin plus the diode drop.
In order for the diode to conduct Y has to go about 0.7v above the node Vout.

Now when Vin goes negative, the output at Y goes negative, and the diode is reverse biased so does not conduct much, so the resistor pulls down the output at Vout to nearly zero. The output at Y however has no limit as to how negative it can go except for the minus power supply rail voltage plus any internal saturation voltage or whatever limits it by the op amp internally. With a -15v negative rail the output might get down to -13.5v for example, or maybe all the way down to -15v if it was a rail to rail output op amp. It will remain in this state until the input again goes above zero.

There are some finer points i am not sure if we want to get into right now, but for example, when the input STARTS to go positive the output may not go positive immediately because the internal gain may not be enough with extremely low positive inputs, or it may START to go positive but only by a very small amount, less than the diode drop, until the input goes positive enough to force Y up to at least around 0.5 to 0.7 volts (the low current forward drop of the diode). We're talking very low input levels here though, like 1uv, which could only take the output up to 0.1v with some op amps.
The same thing will happen when the input goes negative. With only -1uv on the input the output may not get very far negative, at least until the input starts declining more and more down to maybe -10mv or so. We probably dont want to get into this level of detail though. We probably dont want to think about the slew rate either, which means when the input does go positive there will be a delay before the output can get up to 0.7v and force the diode into conduction.

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