Talk:Circuit Idea/How the Wilson current mirror equalizes the currents

I have copied the discussion below from the related Wikipedia page, in order to start the discussion about the legendary Wilson current mirror. Circuit-fantasist (talk) 16:29, 2 August 2008 (UTC)

How do we reveal the secrets of Wilson current mirror?
''I dedicate all my insights about this legendary circuit that I will expose in this discussion to my love who inspires me to continue revealing the secrets of electronic circuits. Circuit-fantasist (talk) 14:18, 19 July 2008 (UTC)''



The great challenge
It was a challenge to reveal the idea behind the popular BJT current mirror   ; but it is a great challenge to disclosure the mystery of the legendary Wilson current mirror (Fig. 1)! Maybe, there is no so simple (containing only three transistors) and, at the same time, so incomprehensible and misunderstood circuit as Wilson current mirror. There are many resources that have tried to explain this sophisticated, ingenious and elegant circuit solution by using formal methods. But they do not give us what we need, first and foremost, as human beings - the basic idea(s) behind this odd, strange and exotic circuit. It is a great paradox to calculate circuit without knowing the basic idea behind it!?! So, before showing in detail how to calculate the electronic circuit in this Wikipedia page we have first to show what the very basic idea is behind the circuit.

The purposes of this discussion
I start this discussion with a few purposes. First of all, I would like to provoke creative Wikipedians:) who are bold enough not only to convey blindly "cut & paste" knowledge from "reputable sources" but to process it first and even to add their own viewpoint (as far as NOR allows such a liberty:) Frankly, judging by my previous attempts, I do not cherish great hopes to make Wikipedians join this discussion; but yet I try it again with the hope that someone will use this "wisdom" to improve the main article. I would rather believe that some reputable author (e.g., Tony Kuphaldt) would join this discussion. Second, I will use this discussion as a base of a few Circuit idea stories dedicated to the legendary circuit. Finally, I will use these materials to compose two papers for Computer science 2008 conference that I will join in September (I do not cherish hope that I will impress my computer-oriented colleagues by these low-level circuit insights but I will try:) Circuit-fantasist (talk) 16:07, 28 July 2008 (UTC)

Questions to be answered
Looking at the circuit diagram (Fig. 1), we need to answer dozens of questions that are never answered.

''What does the transistor Q3 do in this circuit? What is its function there? Why this current mirror contains another simpler current mirror Q1 and Q2 (why an additional simple current mirror is nested in the main current mirror)?!?! What is its function? But why this simple current mirror is reversed (why the transistors Q1 and Q2 are swapped)? Why the Q1's collector current serves as an input quantity and the Q2's collector current as an output one (we thought the Q2's collector current was the input quantity and the Q1's collector current was the output quantity)? Are there negative feedbacks in Wilson current mirror? If there are, what are they? How many negative feedbacks there are? What are their functions (why they are included)? What and how do they control - voltage or current? What are the advantages of Wilson current mirror versus current mirror with emitter degeneration? How does Wilson current mirror keep up an almost constant current (why it has an almost infinite output resistance)? Why the input and output currents are almost equal (what is the trick)? Has a MOSFET Wilson current mirror any advantage versus the simple MOSFET current mirror?

Heuristic approach
Since there are no satisfactory answers to the questions above, let's try to disclose the mystery of the famous circuit by ourselves; let's answer these questions relying mainly on our human intuition, imagination and common sense. Please, just forget all kinds "cut-and-dried" citations and begin thinking by yourself to make an exciting discussion here! The best way of understanding and presenting electronic circuits is by reinventing them, by showing the circuit evolution. So, let's imagine how Wilson has invented his current mirror by "inventing" and building it, in order to grasp the basic ideas behind the circuit and then to present them in an attractive manner to readers. Of course, it would be wonderful if the very Wilson, if he is still alive, would expose how he has invented the famous circuit. Only, it is a well-known truth that, as a rule, due to variety of reasons, inventors do not show willigness for disclosing the process of invention.

The imperfections of the simple current mirror


The simple BJT current mirror (Fig. 2) has two main imperfections: first, the output current differs from the input one because of the two base currents that the transistors Q1 and Q2 "suck" from the input current; second, the output current varies when the output (load) voltage changes because of the Early effect. These are completely different problems; there is no any connection between them. Wilson was a lucky man; in the early 60's, he "killed two birds with one stone" - adding only one transistor to the humble current mirror, he managed to remove the both imperfections of the simple BJT current mirror!

The simple MOSFET current mirror has only the second imperfection because there are no gate currents. And even in this case the Wilson idea is beneficial; it enables the MOSFET Wilson current source to keep up a constant output current.



The current difference in the BJT simple current mirror
Let's first show, in the beginning of the discussion, how the Wilson current mirror eleiminates the difference between the input and output currents. I intend to create a relevant Circuit idea story about this clever trick and will copy this discussion there.





Current map. In order to grasp the ideas behind circuits, we may visualize electrical quantities voltage and current by voltage bars and current loops. In the case of current mirrors, it is extremely interesting to show where currents flow and to visualize their magnitudes. So, I suggest to draw two kinds of figures. The first, placed on the left, will represent the respective circuit diagram with superimposed voltage and current "maps". The second, placed on the right, will visualize the magnitudes of the currents by fat lines whose thickness is proportional to the magnitude of the corresponding current.

Let's begin with presenting the simple BJT current mirror in such an attractive way (Fig. 3). The lines on the right figure are actully sections of closed current loops. The transistors are shown with extremely low ß (on this figure ß = 4), in order to present the base currents by thick enough lines.



Equalizing the currents by "pushing" 2IB to IIN...
How do we solve the poblem of the two IB in the simple BJT current mirror? Let's begin thinking relying on our human common sense... We know from daily routine the great idea of compensation: if there are some losses of something, we might compensate them by adding the same quantity of this "thing". So, let's put in practice this powerful idea!





Since the transistors "suck" two IB, the first idea that might dawn on us is, of course, to add the same two IB. Then IOUT = IIN - 2IB + 2IB = IIN. For this purpose, we have to connect an injecting current source 2IB to the T1's collector (more precisely speaking, this is rather a current-stable resistor than a source). Only, this has to be not an ordinary constant current source but a "following" current source that copies the current 2IB. What an idiocy! It turns out that we need another current mirror?!?



...by "sucking" 2IB from IOUT...




Above, we have added the two compensating base currents by injecting them into the input current (the "original" quantity). But with the same success we might add them to the output current (the "copy") by "sucking" 2IB from it. Now IIN = IOUT - 2IB + 2IB = IOUT. In this case, we have to connect a sinking current source 2IB to the T2's collector. As before, this has to be not an ordinary constant current source but a "following" current source that copies the current 2IB.





...by "sucking" IB from IOUT and "pushing" IB to IIN...
We have almost reached the great Wilson's current equalizing idea... Well, let's continue thinking. It is inconvenient to create 2IB; it is easier to produce (source or sink) only IB. What do we do then?





Eureka! We might connect a current source between the two collectors that sinks a current IB from IOUT and injects the same current IB into IIN (Fig. 6a)! In this way, we add one base current to the input current and another base current to the output current. As a result, the two currents become equal:

IIN + IB = IOUT + IB, IIN = IOUT

Here is the great Wilson's idea! In order to equalize the two currents, he has "moved" one base current from the one to the other leg!

Don't you think this connection resembles a bridge circuit (the current source serves as a "bridge" between the two circuit legs)?



Realizing the Wilson's current equalizing idea
Once we revealed the brilliant Wilson's idea we have only to implement it. Let's continue thinking...





What is this mysterious element that can consume IB from one part of the circuit and can add it to other part? Of course, there is nothing more natural for a bipolar transistor to do that "donkey work"! It "sucks" IB from the point where its base is connected and adds it to the emitter current. Then let's connect a transistor T3 in the left leg of our circuit (Fig. 6a)!

Wonderful, now it sinks the current IB from IOUT and injects the same current IB into IIN! So, we have managed to reveal the role of the mysterious transistor T3!

In the circuit of Wilson current mirror, the transistor T3 "moves" one base current from the right to the left leg.



Reversing the circuit to obtain the true Wilson current mirror
Only, this connection mustn't change the current in the left leg. For this purpose, the transistor T3 has to adjust its base current so that its collector current to remain unchanged. The only "thing" that can do this magic is the ubiquitous negative feedback that keeps up an almost constant voltage (this should be the same kind of negative feedback as this applied to the transistor T1). But there isn't a feedback in this input part of the circuit; there is no connection between the T3's collector and base. What do we do then?

(to be continued) Circuit-fantasist (talk) 15:53, 2 August 2008 (UTC)







Trying to explain the circuit by intuition
Paul Horowitz, Winfield Hill. The art of electronics, second edition, p 89. ISBN-10: 0521370957. Current mirrors shows how Wilson current mirror keeps up an almost constant output current (the author prefers to say "positive" instead "negative" feedback) Designing analog chips

"Explaining" the circuit by formal methods
This Wikipedia article is a typical example of formal approach in circuitry Analogue Electronics from Wikibooks uses the same approach

Only showing the circuit solution
Analogue and mixed signal integrated circuit design gives interesting facts about the George Wilson invention

Are Op Amps Really Linear? in this article, Barrie Gilbert admits: "...the first reported monolithic JFET op amp was designed by my good friend George Wilson that threw in a new type of BJT current mirror, now widely known as the Wilson mirror...."

current mirror question
hi,

i have no idea on how and where to ask a question and/or join a "discussion" in wikibooks, if this posting is not appropriate, please delete, thanks!

great philosophy and books/articles, very much needed and appreciated!

question on the current mirror:

Q3's base takes one Ib from one leg of the current mirror circuit with its base current Ib, and adds one Ib to the other leg through its Ic. but doesn't Q3 have a current gain? would the current added to the other leg be (beta)x(Ib) instead of just Ib and therefore make the current flowing though the two legs unequal? where did i go wrong in reasoning this way?

i realize that, in nature, this question may be the same as my question on emitter follower. since Ic=Ie almost, Ib should have the same influence on both Ic and Ie, to the same degree almost, why then there is no amplification if we take the output from the emitter instead of the collector?

thank you very much for helping!

Modernjackass (discuss • contribs) 14:30, 22 July 2014 (UTC)