Analyzing internal circuit of an opamp

[nForce]

Hello,

I am trying to understand how does an internal simplified circuit of an op amp work.

This is only a picture of an simplified op amp. No text added, so can someone who has the knowledge describe where the current flows in various cases, Ui is the output and U means the voltage.

[Simon]

It's a very crude simplification. It looks like you have the two inputs into a differential amplifier circuit so that you get the difference of the two inputs that then run the output. It's very simplified and of no practical use.

[hamster_nz]

Have a look at the Evil  Scientist's XL741 - you dont just want one, you NEED one :-)

[c4757p]

It's a very crude simplification. It looks like you have the two inputs into a differential amplifier circuit so that you get the difference of the two inputs that then run the output. It's very simplified and of no practical use.

Flip T3 around, make it PNP, possibly add some compensation, and it will be of much greater practical use. It's not the simplification that's a problem, it's just not very well designed

[bson]

That's a voltage amplifier with limited drive.  It would need to be followed by a buffer, and in fact because it can't be used as a buffer it shouldn't really be thought of as an op amp.  It will be grossly non-linear without feedback.  To understand op amps you need to understand amplifiers in general, which is a big topic not easily explained here...  I'd recommend the first few chapters of Bob Cordell's "Designing Audio Amplifiers"; all the basic concepts of loop gain, gain bandwidth, bode plots, details like long-tailed pairs, current sources, biasing, transconductance stages, transimpedance stages, compensation, etc is all the same no matter what.

[Simon]

It might help if the OP can tell us the reson for the question.

[IanB]

Hello,

I am trying to understand how does an internal simplified circuit of an op amp work.

This is only a picture of an simplified op amp. No text added, so can someone who has the knowledge describe where the current flows in various cases, Ui is the output and U means the voltage.

To start getting an understanding of this kind of circuit you could begin with this video:

https://youtu.be/mejPNuPAHBY

[nForce]

Thanks IanB, for the video.

But I still don't understand the table, let's solve just for the first case:

If we apply positive voltage on the base of "T1" then the current will flow through "RC1", where positive voltage will develop on "RC1". On the "U-" where we apply 0V, the transistor will not conduct, and therefore 0V will be across RC2. 0V will be then on base of "T3", so "T3" will no conduct and output is 0V.

In the table Ui (which is the output) is a positive rise.

[Zero999]

Thanks IanB, for the video.

But I still don't understand the table, let's solve just for the first case:

If we apply positive voltage on the base of "T1" then the current will flow through "RC1", where positive voltage will develop on "RC1". On the "U-" where we apply 0V, the transistor will not conduct, and therefore 0V will be across RC2. 0V will be then on base of "T3", so "T3" will no conduct and output is 0V.

In the table Ui (which is the output) is a positive rise.

You've missed the point that all potential differences are relative to one another. If you connect both meter probes to the +V rail, then then it will read 0V, because both inputs are at the same voltage.

If there's 0V across RC2, that means both sides of RC2 have the same voltage on them and in this case one side is connected to +V, so the voltage on Tr3's base will be near +V.

[IanB]

If we apply positive voltage on the base of "T1" then the current will flow through "RC1", where positive voltage will develop on "RC1". On the "U-" where we apply 0V, the transistor will not conduct, and therefore 0V will be across RC2. 0V will be then on base of "T3", so "T3" will no conduct and output is 0V.

Remember that at the starting point both U+ and U- are at some positive voltage already. They don't start out at zero. This means that before any change in input T1 and T2 both have a current flowing through them.

[MatthewEveritt]

I think it's important to notice that the supply is bipolar, so if a point is at 0V (U- for instance) it will still be at a higher voltage that the bottom rail. As I think IanB was pointing out, if U- is held at 0V then there will still be current flowing through it, turning T2 on.

[nForce]

So in the second case, the current flows, like I marked with blue color. And now the output is connected with -V.

But in the last case, both transistors conduct equal amount of current, so all the current flows direct to the negative terminal of the battery. And the output is still conected to the -V, like for the second case.

[bson]

Thanks IanB, for the video.

But I still don't understand the table, let's solve just for the first case:

If we apply positive voltage on the base of "T1" then the current will flow through "RC1", where positive voltage will develop on "RC1". On the "U-" where we apply 0V, the transistor will not conduct, and therefore 0V will be across RC2. 0V will be then on base of "T3", so "T3" will no conduct and output is 0V.

The key isn't RC1,RC2 - those are just current limiters and in typical designs would be constant current sources.  Or Re1 can be a CCS (sink).  The key is RE1 which provides local feedback.  It controls the sum currents in T1,T2:

Ic1=hfe*(Vb1-Ve1-Vt)
Ic2=hfe*(Vb2-Ve1-Vt)
Ve1=(Ic1+Ic2)*Re1

Vt is the conductance threshold voltage for Vbe.
As you can see Ve1 controls the currents and the currents set Ve1.  Local feedback.  Increase one current and the other decreases.  Decrease one and the other increases because the feedback is a function of their sum.

[Simon]

I still don't understand why the OP wants to analyse this particular circuit. OPamps are usually treated as a block taking into account their characteristics. the equivalent circuit is not an exact representation of the chip. If your trying to build an amplifier coppying op amp functional diagrams won't guarantee success.

[Don Hills]

So in the second case, the current flows, like I marked with blue color. And now the output is connected with -V.

But in the last case, both transistors conduct equal amount of current, so all the current flows direct to the negative terminal of the battery. And the output is still conected to the -V, like for the second case.

You might find this interesting:

http://www.righto.com/2015/10/inside-ubiquitous-741-op-amp-circuits.html

[IanB]

I still don't understand why the OP wants to analyse this particular circuit. OPamps are usually treated as a block taking into account their characteristics. the equivalent circuit is not an exact representation of the chip. If your trying to build an amplifier coppying op amp functional diagrams won't guarantee success.

It is reasonable to suppose that it is for learning purposes. That kind of configuration is a common element in very many analog circuits. Therefore anyone curious should be supported in trying to understand the principle of operation of such fundamental building blocks.

[Richard Crowley]

Remember that most internal schematic diagrams for chips (both linear and digital) are approximate equivalents to show what the circuit is doing, and very roughly HOW it is doing it.
Remember also that we can create composite components and devices in monolithic silicon that have no simple direct equivalents in conventional schematic drawings with conventional components.
So you can't really take that diagram and implement it with conventional discreet components on a breadboard and expect it to perform the same.

[hamster_nz]

Remember that most internal schematic diagrams for chips (both linear and digital) are approximate equivalents to show what the circuit is doing, and very roughly HOW it is doing it.
Remember also that we can create composite components and devices in monolithic silicon that have no simple direct equivalents in conventional schematic drawings with conventional components.
So you can't really take that diagram and implement it with conventional discreet components on a breadboard and expect it to perform the same.

Which takes me back to my post... he just needs to get himself an XL741 (http://shop.evilmadscientist.com/productsmenu/762):

The XL741 circuit is a direct implementation of the "equivalent circuit" from the original Fairchild ?A741 datasheet, built up using discrete components like resistors and individual 2N3904 and 2N3906 transistors. It comes with terminal posts and solder points so that you can actually connect to it and build up classic and functional op-amp circuits. Using those terminal posts and solder points, you can hook up with with bare wires, lugs, alligator clips, and/or solder joints— however you see fit.
And unlike the chip version, you can even hook up probes to monitor what goes on inside the circuit.

The full schematic is in http://shop.emscdn.com/KitInstrux/741/741_datasheet_revB.pdf

And there us an 11 page document detailing the operation in http://shop.emscdn.com/KitInstrux/741/741_principles_RevA104.pdf



Edited by simon to correct faulty link

[Simon]

Yes, a real circuit that m imic an op amp will give you more realistic blocks to work with kand study.

[nForce]

I want to analyze this circuit because, it's a simplified version of an op amp. If I can't understand this circuit, I won't be able to understand xl741.

Why is in the last case, output zero?

[amyk]

It's not going to be 0, but ideally very close to. When both inputs change in the same direction, the circuit behaves like one transistor so the voltage across RE1 rises by that amount; RE1 should be a very high value, so small changes in the voltage across it produce an even smaller change in current. If it were a true constant-current sink, there would be no change in current. On the other hand, if you change only one input, the voltage across RE1 will also change, but since the other input hasn't changed, its transistor's Vbe changes and so does the current flowing through it.

[vodka]

I have got   resolved exams from Operational Amplifiers and  diffential pairs with bipolars and FET.  The only inconvenient is that it isn't at English.

[chris_leyson]

"Designing Analog Chips" by the late Hans Camenzind make a really good read and it's free to download http://www.designinganalogchips.com/_count/designinganalogchips.pdf. Got the hardback version on my bookshelf.

[Simon]

I want to analyze this circuit because, it's a simplified version of an op amp. If I can't understand this circuit, I won't be able to understand xl741.

Wrong, you just making your life more complicated. If you want to understand the internal workings then that is one matter. But if you want to use the device you don't need to understand the internal workings, you need to understand it as a block

[IanB]

Wrong, you just making your life more complicated. If you want to understand the internal workings then that is one matter. But if you want to use the device you don't need to understand the internal workings, you need to understand it as a block

No, this is not wrong at all. As an artisan or technician you may only need to understand the functional behavior as a block. But as a student or engineer it is useful to understand how things work. When you know how things work you can be much more sympathetic to how things can be used and what their limitations might be.