What kind of op-amp is that?

Femme_physics

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



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.

rude man

As a start, you can either ignore R_x 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.).

gneill

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?

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.

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?

skeptic2

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

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

rude man

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

technician

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

Femme_physics

Well, this is the solution my classmate offered



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:



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!

NascentOxygen

+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).

NascentOxygen

Proof please!

NascentOxygen

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

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.

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! ;)

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 :)

I like Serena

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

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