# Homework Help: Name of an op-amp circuit?

1. Apr 12, 2012

### Femme_physics

I know this is a comparator, but I am asked for the name of this circuit.

Can I just say a comparator circuit, or a comparator op-amp? Or is there a specific name for this circuit arrangement?

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2. Apr 12, 2012

### Staff: Mentor

I'd call it an analog comparator. Because it is powered from a single supply, and not both +Vcc and -Vcc you could mention that in the description.

3. Apr 12, 2012

### Femme_physics

Makes sense.. I will mention that...thank you NascentOxygen!

:)

4. Apr 12, 2012

### Staff: Mentor

By that I mean the output levels are approx. 0V and 8.2V. Perhaps "with single-polarity output" might cover it; or "with positive polarity output".

Could also state whether it's an inverting or a non-inverting comparator, too.

Last edited: Apr 12, 2012
5. Apr 12, 2012

### Femme_physics

You mean to tell me that comparators can be tagged as inverter or non-inverter? I didn't think it would possible!

So,

If V+ > V-

It's a non-inverter comparator

If V- > V+

It's an inventer comparator?

6. Apr 12, 2012

### Staff: Mentor

Noo!

V+ and V- are related to the inner working of the circuit. The description should be concerned with what the user sees. To wit, if S1 represents the input, then the highest of the input voltages forces the output to its lowest voltage.

Sure sounds like an inverter to me!

7. Apr 12, 2012

### Femme_physics

Isn't this exactly what I said? It depends the condition of the switch then.

8. Apr 12, 2012

### Staff: Mentor

It's always an inverting comparator. If S is high, output is low. If S is low, output is high.

It doesn't change from inverting to non-inverting.

9. Apr 12, 2012

### Femme_physics

Last edited by a moderator: May 5, 2017
10. Apr 12, 2012

### Staff: Mentor

The circuit described in the schematic is built around an analog comparator (probably using an op-amp). The op-amp, like all op-amps, has inverting and non-inverting inputs. We are not talking about the op-amp itself here.

We are talking about the complete circuit. The power supply and arrangement of resistors and switching has created a circuit that in my view operates as an inverting comparator. https://www.physicsforums.com/images/icons/icon6.gif [Broken]

Last edited by a moderator: May 5, 2017
11. Apr 12, 2012

### Femme_physics

OK, let's consider the whole circuit,

If we do the calculations in S1 condition,

V+ = 4.6 V

V- = 4.5 V

Since the the higher voltage value enter the POSITIVE end of the op-amp, it comes out as Vs which is POSITIVE. Where in all that do you see an inversion?

Last edited by a moderator: May 5, 2017
12. Apr 12, 2012

### Staff: Mentor

Correct. That's for the 4.5V at switch in position 1. But we are not there yet ....

Now, for the switch position 2, the switch connects a different voltage to the op-amp's (—) input. What is that new voltage, is it greater or less than the 4.5V in position 1, and what does the output of the op-amp do? (Here's an opportunity to bring your potential divider expertise into play.)
You'll see.

13. Apr 12, 2012

### Femme_physics

:)

Well I did it, turns out that in case of S2, V- = 4.75 V

So in the case of S2 V- is bigger than V+ therefor the output from the comparator is the lowest output.

Is this where we call it an inverter?

So in S1, non-inverter

S2 - inverter

I hope I got it

14. Apr 12, 2012

### Staff: Mentor

Summarizing:
✫ when the input is LOW (4.5V) the output is HIGH (∼8V)
✫ when the input is HIGH (4.75V) the output is LOW (∼0V)

Isn't that similar to how a digital NOT gate (i.e., inverter) works? i.e., OUT = ¬(IN)

So this comparator is an inverting comparator. (The fact that the switch delivers the input to the op-amp's inverting input is probably sufficient to guarantee this will be an inverting comparator.)

No, it's all the time an inverter. This circuit is very similar to your digital logic, so think how an inverter operates there.

15. Apr 13, 2012

### Femme_physics

Can't be, in S1 the input to V+ is 4.66, and the output is indeed HIGH (8.2 V). But the input is High as well (4.66 V), since it enters V+

16. Apr 13, 2012

### Staff: Mentor

It all comes down to what we view as our "input" to the circuit. Because the only thing that the user can change is the switch position, I have been viewing that switch as our "input" to the circuit; the remainder of the voltages and components I picture as being fixed in position and not accessible to the user.

So I'm regarding the voltage selected by S as the circuit's input. I admit that sometimes circuits are concocted as an student exercise, and exactly what should be viewed as the input is debatable. :uhh:

But here, in the absence of instructions to the contrary, I'm staying with S.

17. Apr 13, 2012

### Femme_physics

Is that an official thing to do or is that from a veteran's electronics exerciser solver logic?

18. Apr 13, 2012

### Staff: Mentor

I'm basing it on three pieces of supporting evidence. The fact that to make any other changes would require a soldering iron, that often a switch is installed for user convenience, and that I can see no other variable parameter or selectable option offered in the schematic.

19. Apr 14, 2012

### Femme_physics

Makes perfect sense.... I like the way your logic works! I doubt I'll get asked for that but always good to know :) thanks.