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What kind of op-amp is that?

by Femme_physics
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Femme_physics
#1
Feb27-12, 12:53 AM
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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.
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rude man
#2
Feb27-12, 11:31 AM
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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.).
gneill
#3
Feb27-12, 11:35 AM
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Quote Quote by Femme_physics View Post
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.
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?

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Feb27-12, 11:59 AM
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What kind of op-amp is that?

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
#5
Feb27-12, 01:07 PM
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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
#6
Feb27-12, 01:24 PM
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There are a few opamps with open collector outputs. This could be one of them.
technician
#7
Feb27-12, 02:25 PM
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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
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Feb27-12, 03:07 PM
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Quote Quote by technician View Post
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
Nope on your gain expression ... but you're warm ...
technician
#9
Feb27-12, 04:13 PM
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gain of (R1+Rb)/R1 = 12/5 = 2.4
Femme_physics
#10
Feb29-12, 03:40 AM
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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
#11
Feb29-12, 07:00 AM
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+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
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Feb29-12, 07:05 AM
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I know that there's only 1 mA going through Rb and R1
Proof please!
NascentOxygen
#13
Feb29-12, 07:13 AM
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Quote Quote by Femme_physics View Post
Well, this is the solution my classmate offered
Are you referring to the words he's written to the right of the resistor string?
Femme_physics
#14
Feb29-12, 01:12 PM
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+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).
Shouldn't the power supply ENTER the op-amp as opposed to EXIT from it?


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

3 -2 -I1 = 0
I1 = 1 mA

There you go...

Are you referring to the words he's written to the right of the resistor string?
LOL I didn't c that..sorry..ignore that part.
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#15
Feb29-12, 03:38 PM
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Quote Quote by Femme_physics View Post
Shouldn't the power supply ENTER the op-amp as opposed to EXIT from it?
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.



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

3 -2 -I1 = 0
I1 = 1 mA

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


LOL I didn't c that..sorry..ignore that part.
How can we now that it's out there! ;)
Femme_physics
#16
Feb29-12, 04:36 PM
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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.
OOOOOhhhhhhhhh!!!! AHHHHHHA!!!

OH! OH!

Now I get it :)


What happened to the current coming from the 25V power supply that is coming in through Rx?
It's kinda late now but i'll sit with it tomorrow trying t finalize my results based on this new evidence!


How can we now that it's out there! ;)
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 :)
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Feb29-12, 05:05 PM
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OOOOOhhhhhhhhh!!!! AHHHHHHA!!!
Or take it!

Yes, I think I understand now what he meant. ;)
NascentOxygen
#18
Feb29-12, 05:27 PM
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Quote Quote by Femme_physics View Post
Well, sum of all I entering that crossection I marked in red

3 -2 -I1 = 0
I1 = 1 mA

There you go.../
I believe you realise this is so not right that you need to make a fresh start.

Hint: you know V+ so determine V_.


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