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.
  • #36
jlvn911l.jpg
 
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  • #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 [Broken]

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 [Broken]
 
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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 [Broken] 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 [Broken]
 
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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 [Broken]

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.
 
  • #51
M Quack said:
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.
A technique sometimes used in power supplies is to feed the zener from the regulated line so that the current into the zener is exceptionally stable. I don't know whether that thinking was behind the reason for doing it here.
 
  • #52
Femme_physics said:
You mean that Vout = 12V?

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?

The circuit is definitely a non-inverting amplifier, gain = +2.4V/V. If you were to change the input from +5V to say + 5.5V you would get 2.4*5.5 = +2.64V output.

If the output goes up when the input goes up, that's a non-inverting configuration. It has nothing to do with the direction of current at the input or output.
 
  • #53
Femme_physics, I notice in your calculation of V(-) you used an erroneous formula for the resistive divider involving Rb:

9N9AJ.jpg



So Rb and Rx will need to be recomputed. Circuit parameters do evaluate to nice round figures now. https://www.physicsforums.com/images/icons/icon6.gif [Broken]

http://img804.imageshack.us/img804/1388/writer.gif [Broken] "... so I think you should return the flowers..."
 
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  • #54
NascentOxygen said:
Femme_physics, I notice in your calculation of V(-) you used an erroneous formula for the resistive divider involving Rb:
Actually, your formula is perfectly correct! I was relying on memory and in my mind had Rb and R1 swapped around. :(
 
  • #55
rude man said:
The circuit is definitely a non-inverting amplifier, gain = +2.4V/V. If you were to change the input from +5V to say + 5.5V you would get 2.4*5.5 = +2.64V output.
True, but that's only half the story...

The circuit is also definitely an inverting amplifier, gain = -1.4 v/v. Using the lower end of R1 for input, if you change the input voltage from 0v (ground) to -0.5v, the output rises from +12v to +12.7v. Viewed this way, the zener functions to bias the op-amp so as to allow operation from a single polarity power supply.
 
  • #56
Q-Fever...OK, this is an attempt to see if the working point at V+=V-=Vout=0V is stable.

Note that my knowledge about electronics is very limited and that this may be a completely wrong way of looking at the problem.

Let's assume the following:

The op-amp as a huge open-loop gain G such that Vout=G(V+ - V-) + Vs/2. G will be something like 10^4 or 10^6. Vs/2 is the midpoint between the supply and ground.

Looking at the characterstic curve of the Zener diode: Around 0V, the Zener diode behaves like a resistor with a fairly large impedance Rz. I guess Rz>RI/4 (you will see later why). Let's say Rz=6.5 kΩ just to have a number.

Now assume V+=ε, a small voltage and V-=0 (it works the same way with V=-ε or V+=ε1 and V-=ε2).

You then get Vout = G ε + Vs/2, which leads to

V-=RI/(Rb+RI)*(Gε+Vs/2)=0.42 (Gε+Vs/2)
V+=Rz/(Ra+Rz)*(Gε+Vs/2)=0.65(Gε+Vs/2)

V+ - V- = (0.65-0.42)(Gε +Vs/2) which is larger than the starting point (which was ε), so the situation is unstable and Vout will drift up until the Zener no longer behaves as a
resistor. For a very small value of ε the constant offset Vs/2 on the output voltage dominates. It does not even matter if ε is positive or negative

For this to be the case RI cannot be too large, RI < Rb/Ra Rz.

For those who wonder if the assumption Vout=G(V+ - V-) + Vs/2 is compatible with the "normal" operation of the circuit:

Assume V+=5V and V-=5v-ε, Vout=12V and G=10000.
Then you get G(5V - (5V-ε))+Vs/2 = Gε + 12.5V = 12V, i.e. ε=-0.5V/G=-50μV which can be neglected.

At the operating point with Vz=5V, the Zener diode has a very steep characteristic curve, i.e. a very small impedance. If Vout changes, the change in V+ will be much smaller than that in V-, so that the negative feedback dominates and Vout is stabilized.

So, NascentOxygen, am I thinking along the right lines here? (I've just made it up, don't be too polite if it's all BS).
 
  • #57
M Quack said:
The op-amp as a huge open-loop gain G such that Vout=G(V+ - V-) + Vs/2.
I guess that's right. Op-amps usually are powered by double supplies, so your Vs/2 term is unusual, but I can't see it could be anything else.
looking at the characterstic curve of the Zener diode: Around 0V, the Zener diode behaves like a resistor with a fairly large impedance Rz. I guess Rz>RI/4 (you will see later why).
That's right.

situation is unstable and Vout will drift up until the Zener no longer behaves as a
resistor.
So there is no stable state with Vout=0. In practice, you usually won't get the op-amp delivering an output equal to either supply rail, in any case, so you wouldn't be able to obtain Vout=0.

You analysis is spot on. Below the zener breakdown, the amplifier has more positive feedback than negative. This could send Vout towards 0 but nothing happens that would keep it there, and any noise that sends Vout towards +Vs will succeed, owing to a nett positive feedback.

Well done!
 
  • #58
Thanks for checking.
 
<h2>1. What is an op-amp and how does it work?</h2><p>An op-amp, short for operational amplifier, is an electronic component that amplifies electrical signals. It consists of several transistors and other components that work together to take an input signal and produce an output signal that is a larger version of the input. Op-amps are commonly used in electronic circuits to perform tasks such as amplification, filtering, and signal conditioning.</p><h2>2. What are the different types of op-amps?</h2><p>There are several types of op-amps available, each with its own unique characteristics and applications. Some common types include the inverting op-amp, non-inverting op-amp, differential op-amp, and instrumentation op-amp. Each type has its own specific circuit configuration and is designed for different purposes.</p><h2>3. How can I determine which type of op-amp is being used in a circuit?</h2><p>The type of op-amp being used in a circuit can usually be determined by examining the circuit diagram or by looking at the op-amp's part number. Different types of op-amps have different symbols in circuit diagrams, and each type has a specific part number that can be found on the op-amp itself or in the circuit's documentation.</p><h2>4. What factors should be considered when choosing an op-amp for a circuit?</h2><p>When selecting an op-amp for a circuit, several factors should be considered, including the required gain, bandwidth, input and output voltage range, and power supply requirements. It is also essential to consider the op-amp's input and output impedance, as well as its noise and distortion characteristics, to ensure optimal circuit performance.</p><h2>5. Are there any common problems that can occur with op-amps in a circuit?</h2><p>Yes, there are several common problems that can occur with op-amps in a circuit, such as instability, oscillation, and saturation. These issues can be caused by incorrect circuit design, improper power supply, or component failure. It is crucial to carefully select and configure the op-amp in a circuit to avoid these problems and ensure proper functionality.</p>

1. What is an op-amp and how does it work?

An op-amp, short for operational amplifier, is an electronic component that amplifies electrical signals. It consists of several transistors and other components that work together to take an input signal and produce an output signal that is a larger version of the input. Op-amps are commonly used in electronic circuits to perform tasks such as amplification, filtering, and signal conditioning.

2. What are the different types of op-amps?

There are several types of op-amps available, each with its own unique characteristics and applications. Some common types include the inverting op-amp, non-inverting op-amp, differential op-amp, and instrumentation op-amp. Each type has its own specific circuit configuration and is designed for different purposes.

3. How can I determine which type of op-amp is being used in a circuit?

The type of op-amp being used in a circuit can usually be determined by examining the circuit diagram or by looking at the op-amp's part number. Different types of op-amps have different symbols in circuit diagrams, and each type has a specific part number that can be found on the op-amp itself or in the circuit's documentation.

4. What factors should be considered when choosing an op-amp for a circuit?

When selecting an op-amp for a circuit, several factors should be considered, including the required gain, bandwidth, input and output voltage range, and power supply requirements. It is also essential to consider the op-amp's input and output impedance, as well as its noise and distortion characteristics, to ensure optimal circuit performance.

5. Are there any common problems that can occur with op-amps in a circuit?

Yes, there are several common problems that can occur with op-amps in a circuit, such as instability, oscillation, and saturation. These issues can be caused by incorrect circuit design, improper power supply, or component failure. It is crucial to carefully select and configure the op-amp in a circuit to avoid these problems and ensure proper functionality.

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