Analyzing internal circuit of an opamp - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: January 17, 2016, 09:18:11 pm
content: 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.

author: Simon
author_rank: Unknown
timestamp: January 17, 2016, 09:20:33 pm
content: 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.

author: hamster_nz
author_rank: Super Contributor
timestamp: January 17, 2016, 09:27:05 pm
content: Have a look at the Evil Scientist's XL741 - you dont just want one, you NEED one :-)

author: c4757p
author_rank: Super Contributor
timestamp: January 17, 2016, 09:39:31 pm
content: Quote from: Simon on January 17, 2016, 09:20:33 pmFlip T3 around, make it PNP, possibly add some compensation, and it will be ofmuchgreater practical use. It's not the simplification that's a problem, it's just not very well designed

author: bson
author_rank: Unknown
timestamp: January 17, 2016, 09:57:46 pm
content: 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.

author: Simon
author_rank: Unknown
timestamp: January 17, 2016, 10:01:06 pm
content: It might help if the OP can tell us the reson for the question.

author: IanB
author_rank: Super Contributor
timestamp: January 17, 2016, 10:20:59 pm
content: Quote from: nForce on January 17, 2016, 09:18:11 pmTo start getting an understanding of this kind of circuit you could begin with this video:https://youtu.be/mejPNuPAHBY

author: nForce
author_rank: Frequent Contributor
timestamp: January 18, 2016, 03:08:53 pm
content: 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.

author: Zero999
author_rank: Super Contributor
timestamp: January 18, 2016, 03:24:30 pm
content: Quote from: nForce on January 18, 2016, 03:08:53 pmYou'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.

author: IanB
author_rank: Super Contributor
timestamp: January 18, 2016, 03:36:51 pm
content: Quote from: nForce on January 18, 2016, 03:08:53 pmRemember 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.

author: MatthewEveritt
author_rank: Unknown
timestamp: January 18, 2016, 03:40:45 pm
content: 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.

author: nForce
author_rank: Frequent Contributor
timestamp: January 18, 2016, 08:12:09 pm
content: 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.

author: bson
author_rank: Unknown
timestamp: January 18, 2016, 09:45:47 pm
content: Quote from: nForce on January 18, 2016, 03:08:53 pmThe 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)*Re1Vt 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.

author: Simon
author_rank: Unknown
timestamp: January 18, 2016, 09:52:59 pm
content: 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.

author: Don Hills
author_rank: Regular Contributor
timestamp: January 18, 2016, 10:43:50 pm
content: Quote from: nForce on January 18, 2016, 08:12:09 pmYou might find this interesting:http://www.righto.com/2015/10/inside-ubiquitous-741-op-amp-circuits.html

author: IanB
author_rank: Super Contributor
timestamp: January 18, 2016, 10:46:31 pm
content: Quote from: Simon on January 18, 2016, 09:52:59 pmIt 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.

author: Richard Crowley
author_rank: Super Contributor
timestamp: January 18, 2016, 10:50:11 pm
content: 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.

author: hamster_nz
author_rank: Super Contributor
timestamp: January 19, 2016, 12:26:10 am
content: Quote from: Richard Crowley on January 18, 2016, 10:50:11 pmWhich takes me back to my post... he just needs to get himself an XL741 (http://shop.evilmadscientist.com/productsmenu/762):QuoteThe full schematic is inhttp://shop.emscdn.com/KitInstrux/741/741_datasheet_revB.pdfAnd there us an 11 page document detailing the operation inhttp://shop.emscdn.com/KitInstrux/741/741_principles_RevA104.pdfEdited by simon to correct faulty link

author: Simon
author_rank: Unknown
timestamp: January 19, 2016, 06:52:23 am
content: Yes, a real circuit that m imic an op amp will give you more realistic blocks to work with kand study.

author: nForce
author_rank: Frequent Contributor
timestamp: January 19, 2016, 09:59:31 am
content: 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?

author: amyk
author_rank: Super Contributor
timestamp: January 19, 2016, 11:33:20 am
content: 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.

author: vodka
author_rank: Frequent Contributor
timestamp: January 19, 2016, 03:15:55 pm
content: 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.

author: chris_leyson
author_rank: Super Contributor
timestamp: January 19, 2016, 03:43:17 pm
content: "Designing Analog Chips" by the late Hans Camenzind make a really good read and it's free to downloadhttp://www.designinganalogchips.com/_count/designinganalogchips.pdf. Got the hardback version on my bookshelf.

author: Simon
author_rank: Unknown
timestamp: January 19, 2016, 05:42:06 pm
content: Quote from: nForce on January 19, 2016, 09:59:31 amWrong, 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

author: IanB
author_rank: Super Contributor
timestamp: January 19, 2016, 08:37:44 pm
content: Quote from: Simon on January 19, 2016, 05:42:06 pmNo, 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.

author: Richard Crowley
author_rank: Super Contributor
timestamp: January 19, 2016, 08:48:43 pm
content: But those "equivalent circuits" are not practical. They are only a general guideline to what the REAL circuit is in silicon.Actually understanding how the actual internal circuit works would require the actual schematic diagram (not published) PLUS understanding of how monolithic circuits are designed. None of that is necessary to use all the features of the IC product. The fundamental knowledge here is understanding how differential amplifiers work, and then some additional knowledge of how monolithic power amplifiers are implemented.

author: Simon
author_rank: Unknown
timestamp: January 19, 2016, 08:55:17 pm
content: Quote from: IanB on January 19, 2016, 08:37:44 pmIf I am not mistaken the OP think he needs to understand the inner workings of the IC just to make a circuit with it. This defeats the whole point of why an opamp exists. The equivalent circuit he is working to is so dumbed down that a decent descrete design would teach more and yield better results. It's like needing to know about how a uC is made up at the logic gate level or even being over concerned about the internals of a specific flip flop are made.If the Op wnts to understand how opamps work internally then that is a whole other subject and not one that can be covered by looking at a mere representation that a manufacturer conceeds in a datasheet. The OP seems to think that he can make an opamp following the internal diagram in the datasheet.

author: Zero999
author_rank: Super Contributor
timestamp: January 19, 2016, 09:01:38 pm
content: It's a very good idea to study the basic differential amplifier building block, before attempting to understand the full op-amp circuit.It's actually quite important to understand how op-amps work because it makes them easier to use and avoids typical beginner traps, such as not taking bias currents into account. In fact go far as to say, it's this sort of knowledge and understanding which sets apart the formally educated from the self-taught.The next step is to add a constant current sink

author: c4757p
author_rank: Super Contributor
timestamp: January 19, 2016, 09:15:16 pm
content: Quote from: Hero999 on January 19, 2016, 09:01:38 pmYou and I have clearly had different formal educations

author: Simon
author_rank: Unknown
timestamp: January 20, 2016, 07:09:35 am
content: It's not a case of formal education versus self taught. The OP seems to think he needs to understand what every last transistor on an inacurate diagram is doing in order to just use the chip. There is enough information around to use an opamp without knowing about the internals. If he then wants to understand the internals then there is plenty of information about that like the linked videos where he can "self teach" himself about how the functional blocks work and gain a better understanding.

author: Zero999
author_rank: Super Contributor
timestamp: January 20, 2016, 09:24:17 am
content: He bumped into these kinds of problems in another thread.https://www.eevblog.com/forum/beginners/circuit-with-op-amp-lm339/I suppose what I meant being the difference between formally educated and self-taught is, this is the kind of thing taught in college but isn't the type of thing most hobbyists will study of their own volition.I don't know why some people seem to be discouraging learning. Fair enough, it's not necessary to know what every transistor inside an op-amp does, just to use one, but understanding how they work can help a lot.Anyway, I've always found the differential pair easier to understand with a constant current sink, in place of the emitter resistor because it starts to behave more like a proper op-amp.Watching videos is good but how about building some of these circuits? I have a full discrete op-amp in a draw somewhere.

author: nForce
author_rank: Frequent Contributor
timestamp: January 20, 2016, 12:07:19 pm
content: Quote from: amyk on January 19, 2016, 11:33:20 amIf I understand it correctly, all the current will flow, straight to ground, and because of that there will be no voltage on the base of T3?

author: IanB
author_rank: Super Contributor
timestamp: January 20, 2016, 04:31:40 pm
content: Quote from: Simon on January 20, 2016, 07:09:35 amWhy do you think that? I have read back this thread from the beginning and all I see is an inquiry about the behaviour of a simplified differential pair circuit. That looks to me like a purely educational endeavour. I really don't get all the negativity in this thread.

author: IanB
author_rank: Super Contributor
timestamp: January 20, 2016, 04:34:37 pm
content: Quote from: Richard Crowley on January 19, 2016, 08:48:43 pmWho said that a simplified example circuit has to be practical? The circuit from the first post doesn't even look like it's from a data sheet, it looks more like it is from a textbook.As a student you begin at the beginning and start with simplified circuits. It would be madness to try and understand the full details of actual silicon when you are still learning basic concepts. You would get hopelessly lost.

author: Simon
author_rank: Unknown
timestamp: January 20, 2016, 05:37:47 pm
content: It was the mention that the OP wanted to understand the circuit in order to use a particular op amp part number. basic op amp circuits can be built by understanding the op amp as a black box with it's regular rules.Obviously if one wants to understand the internals in the form of descrete circuits that is fine but they are descrete circuit blocks that exist outside of opamps as well and can be studied as such.

author: MatthewEveritt
author_rank: Unknown
timestamp: January 20, 2016, 06:01:06 pm
content: The XL741 he mentions is the palm sized discrete model of an LM741 hamster_nz suggested the OP look at.I have to agree actually, I think the use of resistors here is confusing. I'd suggest to the OP that it might help to learn about current mirrors, then look at a model op-amp using them (such as the XL741).

author: nForce
author_rank: Frequent Contributor
timestamp: January 20, 2016, 06:24:22 pm
content: I am going to analyze more circuits, because the theory is important before practice.By the way, no one has answered me, if I understand the circuit correctly.Quote from: nForce on January 20, 2016, 12:07:19 pmThanks EEVblog, for helping me.

author: MatthewEveritt
author_rank: Unknown
timestamp: January 20, 2016, 10:18:06 pm
content: On the subject of the third case:Quote from: nForce on January 20, 2016, 12:07:19 pmNo, the current has to go through RE1, so there will be some voltage at the base of T3. Quite a lot, if RE1 is the highest value resistor.At the risk of muddying the waters, here's my (likely inaccurate) take on things.Assume that the two inputs are both at 0V relative to ground, and conducting roughly in the middle of their linear region. Also assume that the transistors never saturate.Perhaps the biggest assumption is statement that the current through RE1 is constant, whatever the rest of the circuit is doing. As others have pointed out that isn't really the case, but it's pretty close if RE1 is large compared to all the other resistors. This means that a lot of voltage is dropped over RE1, meaning that the base of T3 can never get close to the negative rail (just like a lot of real op-amps)This also means that the voltage at the emitters of transistors T1 and T2 is constant.Case the first : The voltage applied to the base of T1 increases.The current through T1 increases, making the T1 path lower impedance than the path through T2. As we have a constant current through RE1 the reduced impedance of T1 means less current flows through RC2. This reduces the voltage dropped over it, bringing the base of T3 higher, which in turn brings the output higher.Case the Second : The voltage applied to the base of T2 increases.This case is the conjugate of case one. The T2 path impedence decreases, increasing the current through RC2, decreasing the voltage at the base of T3 and thus the output.Case the third : Both inputs are increased.The impedance of both paths decreases equally, but as RE1 is limiting the total current and the ratio of T1/T2 path impedances doesn't change the current through RC2 doesn't change, keeping the voltage at the base of T3 constant.

author: nForce
author_rank: Frequent Contributor
timestamp: January 21, 2016, 07:38:06 pm
content: Quote from: MatthewEveritt on January 20, 2016, 10:18:06 pmSorry, I can't grasp this. So if it's a constant voltage (above 0V), let's say 5 V or 3V or 1V, there is a positive voltage, so T3 would be conducting.

author: IanB
author_rank: Super Contributor
timestamp: January 21, 2016, 09:15:27 pm
content: Quote from: nForce on January 20, 2016, 12:07:19 pmQuote from: nForce on January 21, 2016, 07:38:06 pmI think you have some fundamental misconceptions in your understanding of electricity.Firstly, there is no such thing as "a voltage" at a point in a circuit. Voltage in a circuit exists as potential difference between two points. For example, there may be a voltage difference between the base and the emitter of a transistor. This voltage difference affects what the transistor is doing.Secondly, there is no such thing as "ground" and current does not flow to it. Current flows in closed loops, driven by voltage differences between points in the loop. A given circuit may have one or more loops, and the currents sum to zero at each node where the loops intersect.Once you grasp these concepts, you can begin to understand better the workings of the simple differential pair circuit.

author: Zero999
author_rank: Super Contributor
timestamp: January 21, 2016, 10:02:34 pm
content: The voltage at Tr3's base does change because the current through the emitter resistor is not constant. As I said before, adding a constant current sink makes it easier to understand. I admit I didn't understand this circuit until I analysed its behaviour with a constant current sink. A constant current has a near infinite impedance.The LED drops 1.8V, VBEis 0.6V so the voltage across the 1k resistor is around 1.2V, making the current equal to 1.2mA, irrespective of the collector voltage.If decent/modern transistors are used (Hfe >200) the tiny current used to bias the bases can be ignored for the purposes of this exercise.When the voltage at both inputs is equal, the current through both of the transistors and their 10k collector resistors will be equal. Half of the current will flow through one collector resistor and the other half through the other. In this case, the constant current sink takes 1.2mA so the current though the 10k collector resistors will be 0.6mA and the voltage across them 6V. Relative to the 0V point of the power supply, the output voltage will be equal to 9 - 6 = 3V.As long as the voltage at both inputs is below 3V + VBE= 3.6V and VBEabove the minimum voltage required for the current sink to regulate properly (the voltage across the 1k resistor, plus a little bit): -9 + 1.2V + 0.6V = -7.2V, then the output voltage will remain at 3V. This is known as the common mode range.

author: MatthewEveritt
author_rank: Unknown
timestamp: January 22, 2016, 08:50:49 pm
content: Quote from: nForce on January 21, 2016, 07:38:06 pmIt will be conducting, it's always conducting to some extent.I get the feeling that you're thinking of a transistor as a switch, which isn't the case here. The current through T3 (collector to emitter) will be directly proportional to the current into the base.See the attached image showing the effect of base current on a transistor (it could be any of the three). Resistor RE2 will be chosen so that the current into the base is always less than 6 microamps in this example, averaging about 3.

author: dom0
author_rank: Super Contributor
timestamp: January 23, 2016, 12:41:14 am
content: I didn't read the entire thread, but if this hasn't been said before: You'll have averyhard time to understand many if not most linear circuits using bipolar transistors when you think of them as a current controlled (Ic = hFE * Ib) device*. The voltage model (Ie ~ eUbe) is needed to understand them. It also models the physical reality of BJTs much better ; they are charge-controlled devices, hence terminal voltages directly relate to charges, and to produced function.So for example analyzing a long-tailed pair, what this thread seems to be about, is very hard with the simple current-control view, on the other hand, it's quite straightforward from a voltage-control view.

author: nForce
author_rank: Frequent Contributor
timestamp: January 23, 2016, 12:26:07 pm
content: Ok, thanks.I have one question regarding op amps:http://www.calvin.edu/~svleest/circuitExamples/OpAmps/soln6.htmLook at "KCL at Vd". Why is in the equation here "(Vd - Vout)/R6" and not "(Vout - Vd)/R6"? Because on the output it will be bigger voltage because of the gain. So the current will flow back.

author: PChi
author_rank: Frequent Contributor
timestamp: January 23, 2016, 12:39:38 pm
content: Operational Amplifiers Design and Applications by Tobey, Graeme and Huelsman is worth reading both on the design of and use of Op Amps. It's available for next to nothing because it is rather old.

author: Zero999
author_rank: Super Contributor
timestamp: January 23, 2016, 02:52:12 pm
content: Quote from: nForce on January 23, 2016, 12:26:07 pmWhere's the schematic?

author: nForce
author_rank: Frequent Contributor
timestamp: January 23, 2016, 02:58:12 pm
content: Sorry Hero999.It's here, the last scheme on the page:http://www.calvin.edu/~svleest/circuitExamples/OpAmps/

author: amyk
author_rank: Super Contributor
timestamp: January 23, 2016, 03:34:37 pm
content: Quote from: dom0 on January 23, 2016, 12:41:14 amI think both are necessary to see how the feedback works, as well as viewing the constant-current characteristic as more of a currentlimiter--- increasing the Vbe wouldallowmore current to flow through the transistor, but due to the resistor/current sink at the emitter, this raises the voltage at the emitter, which decreases Vbe and thus the emitter current. The opposite effect happens when decreasing Vbe, such that the emitter voltage is always kept one diode-drop below the base.When both inputs change in the same direction, the common emitter voltage changes accordingly. Since this voltage is applied across the emitter resistor, if it is high in value, then the overall change in current through the circuit is small, and the higher the value, the smaller the change. A constant-current sink always passes the same current regardless of voltage, so there is effectively no change. However, when only one input changes, or alternatively if they change in opposite directions, the change in emitter voltage affects the Vbe of the other transistor, causing the current through it to also change.

author: IanB
author_rank: Super Contributor
timestamp: January 23, 2016, 05:29:16 pm
content: Quote from: nForce on January 23, 2016, 12:26:07 pmThe node equation given in the solution is this:Vd / R5 + (Vd - Vout) / R6 = 0This says that all the currents at node Vd sum to zero.In this particular case you could also write the node equation like this:Vd / R5 = (Vout - Vd) / R6Notice that both equations are equivalent.However, in the general case with more than two branches you don't know which branches might have currents flowing in or out, so it is better to use the first form of the equation where all the currents sum to zero.

author: nForce
author_rank: Frequent Contributor
timestamp: January 23, 2016, 06:17:17 pm
content: Ok, thanks.I have another question, which isn't really on topic, but it's a beginners question:Which symbols is that? And how do connect the leads in lt spice? One lead is an arrow which points in the resistor.

author: Richard Crowley
author_rank: Super Contributor
timestamp: January 23, 2016, 06:57:41 pm
content: Quote from: nForce on January 23, 2016, 06:17:17 pmThat is a potentiometer.QuoteThe arrow designates the "wiper" or the moving contact.QuoteIt is just two resistors. One from the bottom to the wiper.And the other resistor between the wiper and the top.

author: Zero999
author_rank: Super Contributor
timestamp: January 23, 2016, 07:00:44 pm
content: It's a potentiometer. There is no model for it in the default LTSpice configuration. It can be modelled using two resistors, forming a potential divider.

author: nForce
author_rank: Frequent Contributor
timestamp: January 23, 2016, 08:05:17 pm
content: Thank you.I am analyzing different op amp configuration, and I found this video, where I have a question.How do we know when, to use superposition theorem when analyzing op amp circuits?

author: Zero999
author_rank: Super Contributor
timestamp: January 24, 2016, 11:14:22 pm
content: I just use whatever method I find to be the quickest and easiest at solving the problem.Forgive me, it's been years since I did this at college. I'm spoilt. I use SPICE a lot of the time now.