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

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The discussion centers around identifying the type of op-amp used in a given circuit, with participants initially considering it a comparator before concluding it functions as a non-inverting amplifier with feedback. The circuit analysis involves calculating unknown resistors, with specific attention to the roles of resistors Rb and Rx, and the influence of a zener diode on the voltage levels. Participants utilize Kirchhoff's Voltage Law (KVL) to derive relationships between voltages and currents in the circuit, ultimately determining the values needed for the resistors to achieve the desired output. The conversation highlights the complexities of analyzing op-amp circuits, particularly regarding current flow and feedback mechanisms. Understanding the op-amp's properties is crucial for solving the circuit effectively.
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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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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.).
 
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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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.
 
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?
 
There are a few opamps with open collector outputs. This could be one of them.
 
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
 
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 ...
 
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. ;)
 
  • #36
jlvn911l.jpg
 
  • #37
Femme_physics said:
Oh yea, you'll get 2 flowers

Only 2? :redface:
 
  • #38
Femme_physics said:
I think I got it... full solution here:
It looks very well presented. Good work. https://www.physicsforums.com/images/icons/icon14.gif

Is the determination of Ra, Rb and Rx the full requirement for this exercise? Is this particular brand/number of OP-AMP one that you can tell us anything about, i.e., have you studied its data sheet? Otherwise, we are going to be left wondering what the exact purpose of it is.

P.S. To your classmates keeping watch on this thread...what will the teacher think when you all submit identical homework https://www.physicsforums.com/images/icons/icon5.gif
 
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  • #39
I still don't understand a few things. In the original drawing, we see a current that goes INTO the op-amp, but it is not an inverter, so what gives? This is still confusing to me.


It looks very well presented. Good work.
Thank you, but my classmate disagrees with me. He says that we don't know that "I" that goes down, which I marked as "Iamp", because it splits to the op-amp and to the ground. Do you see what I mean?

I didn't know what to tell him, maybe you can answer that?

Is the determination of Ra, Rb and Rx the full requirement for this exercise?

Yes.

Is this particular brand/number of OP-AMP one that you can tell us anything about, i.e., have you studied its data sheet?

I don't recall our teacher going over it. All I got in my notebook is inverter, follower, comparator, differential, and summing.

P.S. To your classmates keeping watch on this thread...what will the teacher think when you all submit identical homework

How do you know it's homework? :)

My classmates don't keep a watch of this thread...they're not even aware of the forum. Only my classmate who emailed me his solution that I posted here is aware of this forum (and that I'm using it) but he isn't registered here. Which, is a shame, because you guys are terrifically helpful, although I doubt my classmates would have the patience for the long, detailed, somewhat slow and thorough procedure that goes via solving exercises online.

Regardless, I doubt any of my classmates solved it, though they really should, but they have like 50+ exercises to hand out to the teacher so most if not all are going to neglect it. My teacher will probably excuse them if they haven't solved a few hard exercises in our textbook.

I like Serena said:
jlvn911l.jpg

LOL
 
  • #40
Now that you've worked out all the currents and voltages, do you have any clue what the circuit does? And what would you need to change to adapt it to a different output value?
 
  • #41
M Quack said:
Now that you've worked out all the currents and voltages, do you have any clue what the circuit does? And what would you need to change to adapt it to a different output value?

I actually didn't work out everything, as per the last post before that I made...still need to figure something out. I will answer this question as soon as I got it :)
 
  • #42
Femme_physics said:
I still don't understand a few things. In the original drawing, we see a current that goes INTO the op-amp, but it is not an inverter, so what gives?
How do you know it is not an inverter? Or how could you tell whether it is? It has either no inputs, or it has two, depending on how you want to view it. If you say it has two, then one input is non-inverting, the other is inverting.

In amplifiers, the term "inverting" has two meanings. One is that if the input is +x volts, the output will be -x volts. That behaviour isn't possible here, for the simple reason that the OP-AMP output cannot go negative because the OP-AMP isn't powered by a negative supply; it is powered by only a single positive supply, that's the 25v.

The other meaning of "inverting" is that if the input level should rise slightly it will cause the output to fall. So if the input rose, say, from 3.4 volts to 3.5, and the output was correspondingly noted to fall from 2.3 to 2.0 volts, we could say we have an inverting amplifer of gain x3.
Thank you, but my classmate disagrees with me. He says that we don't know that "I" that goes down, which I marked as "Iamp", because it splits to the op-amp and to the ground. Do you see what I mean?
No. But I conjecture that you are seeing the labelling arrows denoting Vout on the op-amp output and mistakenly concluding it represents a short circuit to ground https://www.physicsforums.com/images/icons/icon4.gif I ascribed to it a current of zero since we are told nothing about a load there. Consider it an open circuit; if it isn't, we would have been told.

What is the number written on the op-amp in your first schematic?
 
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  • #43
Why is Ra connected to the output of the op-amp, and not to the +25V supply? (obviously one would have to change the value of Ra to keep the current through the Zener at 2mA)
 
  • #44
No. But I conjecture that you are seeing the labelling arrows denoting Vout on the op-amp output and mistakenly concluding it represents a short circuit to ground I ascribed to it a current of zero since we are told nothing about a load there. Consider it an open circuit; if it isn't, we would have been told.

What is the number written on the op-amp in your first schematic?

You mean that Vout = 12V?

And well, yes, it could be a short circuit to the ground, it could be anything..-- we did make the assumption so figure it's illegal to touch it.
My classmate raised the point that we can't really tell so shouldn't assume (and also didn't like the fact the result is not a round number :P - but it's inconsequential right?), and thought it can't be solved that way. I thought it's better to make assumption if there appears to be no other way...but...well, that's why I ask you guys, just in case :)

How do you know it is not an inverter? Or how could you tell whether it is? It has either no inputs, or it has two, depending on how you want to view it. If you say it has two, then one input is non-inverting, the other is inverting.

In amplifiers, the term "inverting" has two meanings. One is that if the input is +x volts, the output will be -x volts. That behaviour isn't possible here, for the simple reason that the OP-AMP output cannot go negative because the OP-AMP isn't powered by a negative supply; it is powered by only a single positive supply, that's the 25v.

The other meaning of "inverting" is that if the input level should rise slightly it will cause the output to fall. So if the input rose, say, from 3.4

True, we can't really tell. Can we? It's rather confusing. BUT according to the fact the current goes into the op-amp, that's our key in telling it's an inverter, right?
 
  • #45
Femme_physics said:
You mean that Vout = 12V?

And well, yes, it could be a short circuit to the ground,
It definitely could not be a short circuit to ground; this would contradict the specification that Vout = 12V.
it could be anything..-- we did make the assumption so figure it's illegal to touch it.
It's obviously the circuit's output, but (in the absence of information to the contrary) just regard it as a node where we can measure a voltage.
BUT according to the fact the current goes into the op-amp, that's our key in telling it's an inverter, right?
Unless we can identify something to call an "input" we can't compare input vs. output to see whether there is a inversion or not! (Suppose you have an arrangement where the input current is negative so flows "out" of the input, then a non-inverting amplifier would have output current flowing "in" as here.)

The safest description is to just say the circuit uses an op-amp with negative feedback, and leave it at that. You can't lose marks for saying nothing wrong. :smile:
 
  • #46
NascentOxygen said:
What is the number written on the op-amp in your first schematic?
We needs to know!
 
  • #47
M Quack said:
Why is Ra connected to the output of the op-amp, and not to the +25V supply? (obviously one would have to change the value of Ra to keep the current through the Zener at 2mA)
It's not clear to whom you direct your question, M Quack. Are you offering to guide Femme_physics deeper into the workings of this intriguing circuit? Or are you the first of her classmates to discover you can tap into Femme_physics online homework solutions, and you're angling for bonus marks?? http://img593.imageshack.us/img593/2293/starwarssmiley010.gif
 
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  • #48
NascentOxygen said:
It's not clear to whom you direct your question, M Quack. Are you offering to guide Femme_physics deeper into the workings of this intriguing circuit? Or are you the first of her classmates to discover you can tap into Femme_physics online homework solutions, and you're angling for bonus marks?? http://img593.imageshack.us/img593/2293/starwarssmiley010.gif

I'm not on the same homework assignment :-) I am not an expert in electronics but I've been able to work out the resistances and currents, and I have a fair idea what the circuit is about. So for guiding deeper into the workings of the circuit, there a people better qualified.

But I have become curious (professional disease among physicists). I have a vague idea why one would set up Ra is show, but I am not sure. The way it is done now (ignoring the currents indicated), would there not be a second operating point at V+=V-=Vout=0V, not currents flowing except into the op-amp?
 
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  • #49
M Quack said:
But I have become curious (professional disease among physicists).
Ah, that old qurious disease. Is that what they call Q Fever? :smile:
would there not be a second operating point at V+=V-=Vout=0V, not currents flowing except into the op-amp?
You think that if the output of the op-amp were to be momentarily shorted to ground, it will stay there? Maybe you are right, or maybe not. How would you go about trying to show that 0v is a stable point, without actually constructing the circuit?
 
  • #50
Grrrr. now you've got me. I'll think about it. But now I have to help a friend move house.

My first idea is that the Zener will behave asymmetrically with respect to noise and might build up an input voltage that way. Btw, this was my point about feeding Ra from +25V. You have less control over the current through the Zener, though if there are variations on the +25V line.
 

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