What type of op-amp is being used in this circuit?

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In summary, the conversation was about a circuit problem involving an op-amp and unknown resistors. The conversation discussed various approaches to solving the problem, such as using Kirchoff's Voltage Law and knowing the input and output voltages. The conversation also clarified the function of the zener diode and the role of the power supply in the circuit. The final conclusion was to take the exercise and try to solve it.
  • #1
Femme_physics
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Homework Statement



http://img38.imageshack.us/img38/8373/circuit950.jpg

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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  • #2
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
Femme_physics said:

Homework Statement



http://img38.imageshack.us/img38/8373/circuit950.jpg

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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  • #4
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
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
There are a few opamps with open collector outputs. This could be one of them.
 
  • #7
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
technician said:
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 ...
 
  • #9
gain of (R1+Rb)/R1 = 12/5 = 2.4
 
  • #10
Well, this is the solution my classmate offered

http://img88.imageshack.us/img88/5638/classmate.jpg

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

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 fundamental 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!
 
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  • #11
+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
I know that there's only 1 mA going through Rb and R1
Proof please! :-p
 
  • #13
Femme_physics said:
Well, this is the solution my classmate offered
Are you referring to the words he's written to the right of the resistor string? :frown:
 
  • #14
+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.
 
  • #15
Femme_physics said:
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! ;)
 
  • #16
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 :)
 
  • #17
OOOOOhhhhhhhhh! AHHHHHHA!
Or take it!

Yes, I think I understand now what he meant. ;)
 
  • #18
Femme_physics said:
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 realize this is so not right that you need to make a fresh start.

Hint: you know V+ so determine V_.
 
  • #19
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  • #20
Femme_physics said:
Taking your criticisms into consideration, here is my new idea...

http://img37.imageshack.us/img37/5191/eevso.jpg

Looks good...
 
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  • #21
Looks good...
Ah..the legendary Klaas enthusiasm persists I see... :biggrin:
Thanks.

NascentOxygen said:
I believe you realize this is so not right that you need to make a fresh start.

Hint: you know V+ so determine V_.
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
 
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  • #22
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?
 
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  • #23
V+ is ground.

How did you figure V- is necessarily 12V? Which formula or principle did you use?
 
  • #24
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
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/showpos...2&postcount=21

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

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

I hope!
 
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  • #26
Femme_physics said:
Yes, Vout is 12
V- = a fraction of the output
V+ = 5 V (Due to the zener diode)
So V- = 7 V
Correct for V+. But you are forgetting the important thing about op-amps (on which I expounded at length in your other thread) that V_ = V+ when operating as an amplifier. So V_ is not 7v.

Wait, are you telling me that for any op-amp no matter what Vout = (V-) + (V+) ?
In light of what I've just written, perish the thought!

You are not using your time to best advantage, Femme_physics. You have gone on and done further work, futilely, without waiting until you have had the answers here confirmed. So now you have to recalculate.

And don't forget that you already have worked out the value for the resistor feeding the Zener. I think it is Ra. So you are slowly getting there. :wink:

But as you will note, we haven't mentioned Rx yet.
Cja3i.gif
 
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  • #27
Actually I did use some time to my advantage, and I thought I got Rx, but when I tried calculating it turns out I got the same equation despite the fact I picked two different loops...my equations cancel out Rx therefor!

http://img29.imageshack.us/img29/396/17609068.jpg
Correct for V+. But you are forgetting the important thing about op-amps (on which I expounded at length in your other thread) that V_ = V+ when operating as an amplifier. So V_ is not 7v.

Oh right, as long as they're not used as comparators! I remember that explanation!

So V- = 5 V! Great. I can redo the calculation setting 5 instead of 7, but am still stuck with Rx...
 
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  • #28
There are a number of theories accounting for the inclusion of Rx in this circuit. Fortunately, we don't need to understand any of them in order to determine what value the designer must have chosen for Rx. :smile: We can come back and consider the finer points of his design philosophy later.

You have noted there is a potential difference across Rx of 13v. Good.

The current that flows from the 25v supply through Rx must get to ground somehow, otherwise no current would flow. 3mA of it goes into the output of the op-amp. I can see that you account for about another 2mA thorough the zener, to keep the zener's voltage firmly at 5v. Any other routes for Rx's current to get to ground that you can see?

Once you have accounted for all the current through Rx you can use Ohm's Law to determine what value Rx must be.

Good luck! :wink:
 
  • #30
Femme_physics said:
What do you think?

Flowers...
Or...
More flowers...
:!)
 
  • #31
Yea?!? I got it?! YESS! YSES! WEEPIE! I love my life. Really was a great walkthrough. I really appreciate your help, this feels orgasmic to finally get it done! Now, there's that other annoying exercise to conquer and hopefully this is the last op-amp that would give me troubles!

THANK YOU! Nascent, you're great. ILS, thanks a bunch as usual!
 
  • #32
Now can I get a drawing?
Preferably with some flowers? And Or?
 
  • #33
girl%2Bw%2Bflower%2BLR.jpg
 
  • #34
Oh yea, you'll get 2 flowers, and 1 Or! Tomorrow after a good sleep. It's 1 A.M. and Or doesn't look so great after a marathon of HW. :)

EDIT: OOOOOOOOOH! Thanks :) I'll do one myself too.
 
  • #35
Promises, promises. ;)
 

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