# Homework Help: What kind of op-amp is that?

1. Feb 27, 2012

### Femme_physics

1. The problem statement, all variables and given/known data

http://img38.imageshack.us/img38/8373/circuit950.jpg [Broken]

At first I thought a comparator, then I noticed a line connecting the Vout and the lines that connect to V- and V+, so it can't be a comparator, can it? I don't see any other option....

As far as the question, I need to find out the unknown resistors, but I'll start working on it once I figure that basic fact.

Last edited by a moderator: May 5, 2017
2. Feb 27, 2012

### rude man

As a start, you can either ignore Rx or the circuit is in saturated mode ... this is a consideration since the output current is defined to be 3 mA. The rest is just the usual KVL equations of an active op amp circuit (input voltages equal etc.).

3. Feb 27, 2012

### Staff: Mentor

Looks like an amplifier with feedback, although the feedback mechanism is made a bit trickier due to the presence of Rx. What effect do you suppose the zener diode is going to have?

Rb and R1 form a voltage divider. What voltage are they dividing? What's the op-amp going to try to do with that voltage?

Last edited by a moderator: May 5, 2017
4. Feb 27, 2012

### I like Serena

I'd say it looks a bit like an inverting amplifier, but it's not quite it.
KVL and KCL should tell what it does.

5. Feb 27, 2012

### rude man

I hadn't noticed the output voltage is given. Since R1 is given, all components are defined implicitly.

You can solve this problem without writing any KVL equations. Hints: what must Ra be to satisfy Iz_min? What must Rb be to give a 12V output? And finally, what must Rx be to give 3 mA of current flowing into the op amp?

6. Feb 27, 2012

### skeptic2

There are a few opamps with open collector outputs. This could be one of them.

7. Feb 27, 2012

### technician

I would say it is a non-inverting amplifier with voltage of 5V on the + input and voltage gain of R1/(R1+Rb).
Vout =+12V so Rb can be calculated.... and so on

8. Feb 27, 2012

### rude man

Nope on your gain expression ... but you're warm ...

9. Feb 27, 2012

### technician

gain of (R1+Rb)/R1 = 12/5 = 2.4

10. Feb 29, 2012

### Femme_physics

Well, this is the solution my classmate offered

http://img88.imageshack.us/img88/5638/classmate.jpg [Broken]

I agree on Ra. It's just KVL.

But on Rb-- I'm confused as far as how to use a voltage divider.

I CAN indeed apply KVL

Since I know that the 3 mA split at this point marked in red:

http://img191.imageshack.us/img191/5854/markedl.jpg [Broken]

I know that there's only 1 mA going through Rb and R1, and I know they have a 12V potential difference to the ground. So,

Sum of all V = 0 ; 12 - 1ma x Rb -1ma x R1 = 0

I get that Rb 2000 ohms. Makes sense?

As far as Rx -- well, I don't really understand something fundemental about the circuit. Is Vs some type of another Vout? Or does it just define the limits of the Op-Amp like we see in those Vcc+ Vcc- sort of thing? I really don't know how to approach Vs. How can 25 Volt comes out of an op-amp who only produces a Vout of 12v?!? And how does any of that helps me with Rx?

Sorry-- A lot of questions, I know. Just a confusing circuit!

Last edited by a moderator: May 5, 2017
11. Feb 29, 2012

### Staff: Mentor

+Vs is the power supply (you've previously seen it as +Vcc) to power the OP-AMP. There is no -Vcc here, the negative power supply here is ground (you've previously seen OP-AMP's negative supply as -Vcc, but not with this arrangement here).

12. Feb 29, 2012

### Staff: Mentor

13. Feb 29, 2012

### Staff: Mentor

Are you referring to the words he's written to the right of the resistor string?

14. Feb 29, 2012

### Femme_physics

Shouldn't the power supply ENTER the op-amp as opposed to EXIT from it?

Well, sum of all I entering that crossection I marked in red

3 -2 -I1 = 0
I1 = 1 mA

There you go...

LOL I didn't c that..sorry..ignore that part.

15. Feb 29, 2012

### I like Serena

It does.
The arrow indicates that the wire goes to a power supply.
It does not indicate the direction of the current.
The actual current flows away from the 25V power supply.

What happened to the current coming from the 25V power supply that is coming in through Rx?

How can we now that it's out there! ;)

16. Feb 29, 2012

### Femme_physics

OOOOOhhhhhhhhh!!!! AHHHHHHA!!!

OH! OH!

Now I get it :)

It's kinda late now but i'll sit with it tomorrow trying t finalize my results based on this new evidence!

Well, it just says "Or take it"

Or being my name. Telling me that I should take this exercise and try to solve it. He just put the "e" in front of the "k" by accident :)

17. Feb 29, 2012

### I like Serena

OOOOOhhhhhhhhh!!!! AHHHHHHA!!!
Or take it!

Yes, I think I understand now what he meant. ;)

18. Feb 29, 2012

### Staff: Mentor

:yuck: I believe you realise this is so not right that you need to make a fresh start.

Hint: you know V+ so determine V_.

19. Feb 29, 2012

### Femme_physics

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20. Mar 1, 2012

### I like Serena

Looks good...

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21. Mar 1, 2012

### Femme_physics

Ah..the legendary Klaas enthusiasm persists I see...
Thanks.

How?

If I use the Voltage Divider, it gets nullfied. Should I just use KVL to try and find it? Seems like a longer route, but I guess I can... unless I'm not seeing something?

http://img408.imageshack.us/img408/2059/bolshenemagoo.jpg [Broken]

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22. Mar 1, 2012

### Staff: Mentor

Alternating between 2 threads is probably half the trouble, you are not giving yourself a chance to get to grips with either circuit. Let's stay with this one until you get a few fundamental misconceptions sorted out.

The op-amp inputs draw no current (so we say), so the op-amps can be ignored when it comes to affecting any circuit you connect to their inputs. You have a resistor divider here. The op-amp (-) input has no effect on the divider currents or voltages, so you can forget about it and just concentrate on the resistors.

EDITED: corrected

So V_ is set by the 12v and the resistor ratios.

What is V+ set to?

Last edited: Mar 1, 2012
23. Mar 1, 2012

### Femme_physics

V+ is ground.

How did you figure V- is necessarily 12V? Which formula or principle did you use?

24. Mar 1, 2012

### Staff: Mentor

By V+ I mean the voltage on the op-amp's non-inverting input, often denoted V(+).

V_ is what you worked out here. https://www.physicsforums.com/showpost.php?p=3792342&postcount=21

Except you wrongly equated it to 0. It's value is a fraction of the op-amp's output, and the specifications of the problem tell you the op-amp output voltage is +12v.

25. Mar 1, 2012

### Femme_physics

Now I understand. There's still voltage at V-, but no current. I keep forgetting that,that's the op-amp properties.

Yes, Vout is 12
V- = a fraction of the output
V+ = 5 V (Due to the zener diode)
So V- = 7 V

Wait, are you telling me that for any op-amp no matter what Vout = (V-) + (V+) ?

Anyway, here's the new refined solution :)

http://img823.imageshack.us/img823/3241/rbcalculations.jpg [Broken]

I hope!

Last edited by a moderator: May 5, 2017