Hi!

I partly understand how inverting op amp circuits work. There are some details I'm unsure of though. Take a look at this circuit picture and maybe you can explain.

This op amp is powered by a single sided power supply of 3.3V. I know that since the non-inverting op amp pin is at ground then the inverting pin also stays at 0V. I understand what happens when the input is 0 to 10V, the output should be 0 to -3.3V, maybe a little off of that depending on the rail-to-rail specs. But my main concern is what happens during the rest of my large input range? The negative input goes all the way down to -30V? The highest input voltage of +30V? Will the op amp get damaged? What will the output be? The output will eventually go into an ADC pin of a microcontroller. The pin can only accept 0 to 3.3V.

I know you can voltage divide the input down and convert it to positive only, but these questions above are more theoretical.

Thanks for the help!

 

Hi!

I partly understand how inverting op amp circuits work. There are some details I'm unsure of though. Take a look at this circuit picture and maybe you can explain.

This op amp is powered by a single sided power supply of 3.3V. I know that since the non-inverting op amp pin is at ground then the inverting pin also stays at 0V. I understand what happens when the input is 0 to 10V, the output should be 0 to -3.3V, maybe a little off of that depending on the rail-to-rail specs. But my main concern is what happens during the rest of my large input range? The negative input goes all the way down to -30V? The highest input voltage of +30V? Will the op amp get damaged? What will the output be? The output will eventually go into an ADC pin of a microcontroller. The pin can only accept 0 to 3.3V.

I know you can voltage divide the input down and convert it to positive only, but these questions above are more theoretical.

Thanks for the help!

View attachment 113270

With a single 3.3 V rail the output cannot go negative.

 

With a single 3.3 V rail the output cannot go negative.

ok. so the output will be 0V? Will the op amp pin get damaged?

 

ok. so the output will be 0V? Will the op amp pin get damaged?

I will have to go look at a data sheet for the part.
(edited to add ...)
The data sheet says the applied voltage should not exceed the power rail plus 300 mV. It doesn't say at what point damage will occur. It would probably depend on the current level. With enough resistance it might survive 30 V in.

 

I will have to go look at a data sheet for the part.
(edited to add ...)
The data sheet says the applied voltage should not exceed the power rail plus 300 mV. It doesn't say at what point damage will occur. It would probably depend on the current level. With enough resistance it might survive 30 V in.

are u sure the op amp pin sees the 30V? isn't it held at virtual ground by the inverting setup? or does this not matter after it goes above the supply?

 

are u sure the op amp pin sees the 30V? isn't it held at virtual ground by the inverting setup? or does this not matter after it goes above the supply?

Correct. The input of the op amp is virtual ground ... if the op amp is working correctly. The 30 V input causes current to be drawn out of the pin. The force is still felt at the pin. There is a sucking force apparent. If the op amp is working correctly the input should be at virtual ground. The question is can the op amp operate correctly under these conditions. Build it and see what you get.

 

Hello there,

There could be many ways we can loose that "virtual ground" in the operation of this circuit.
Virtual ground is a concept that is usually only valid when the op amp is operating in it's normal range of input and output. If the input goes to ANY level that will cause the output to saturate in either direction, it is goodbye virtual ground. After that we depend only on the external parts and the internal ESD diodes that act as clamps.

An op amp with equal input and feedback resistors is easiest to visualize, so say we have 100k in and 100k feedback, and a 3v positive supply and no negative supply. When the input is 0v the output is around 0v, when the input is -1v the output is around +1v, but when the input is +1v the output is zero, so to make a little table:

-5v, +3v
-4v, +3v
-3v, +3v
-2v, +2v
-1v, +1v
0v, 0v
+1v 0v
+2v, 0v
+3v, 0v
+4v, 0v
+5v, 0v

Looking at this table, we see that the op amp is out of it's normal linear operation range when the voltage is either positive or is less than -3v. We only assume that the op amp has rail to rail output, but if that is not the case then the above has to be modified to reflect that.

So the question then is what happens to the input of the op amp when the input to the circuit goes out of range.
With +5v in and no clamping diode, we essentially remove the op amp from the circuit except for the output and analyze with just the two resistors. This gives us +2.5v at the input of the op amp. If we increase the input to +10v however, then the input would be +5v which is too high unless the clamp diodes kick in which would clamp the input to somewhere around +3.2 to 3.5v. The op amp may survive if the clamp diodes can handle the current. If they cant, the op amp input blows out.
With -5v in, the input of the op amp itself would be -2.5v without a clamp diode, but if the op amp clamp diode kicks in then the input will be about -0.2 to -0.5v and the op amp will survive if the current is low enough as before.
So the questions when the op amp goes out of it's normal range is:
1. Is the op amp at any time driven out of it's normal operating range?
2. If #1 is true, then using the external circuit is the voltage above the max supply voltage or below the min supply voltage?
3. If #2 is true, can the ESD clamp diodes handle the current?

If #1 is false then all is well, else if #2 is false then all is well, else if #3 is true then all is well.
Of course when the op amp is taken out of it's normal linear operation range it wont work the same way because the output saturates.

 

All you need is a voltage divider