Talk:Wilson current mirror

Starting the discussion
Created the article. Please check if it is ok. Thanks. Rohitbd 12:13, 22 March 2006 (UTC)

How do we reveal the secrets of Wilson current mirror?
''I dedicate all my insights about this legendary circuit that I have exposed in this discussion to my love who inspires me to continue revealing the secrets of electronic circuits. Circuit-fantasist (talk) 18:35, 6 September 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 (including the present Wikipedia page) 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 some 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 reinventing 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


BJT. 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!

MOSFET. 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 (see below).



How the Wilson current mirror equalizes the input and output currents
In order to justify the name current mirror, in this circuit the output current has to follow exactly the input current. So, regarding to the current magnitudes, a current mirror is actually a current follower. The Wilson current mirror meets closely this requirement. Circuit-fantasist (talk) 15:19, 21 September 2008 (UTC)

The current difference in the BJT simple current mirror
Let's first show, in the beginning of the discussion, how the Wilson current mirror eliminates 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 in this section. 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). You can see on the left picture (Fig. 3a) where currents flow and particularly how the transistors "suck" two base currents. The currents are represented by closed loops (every current finishes where it has started). The lines on the right picture (Fig. 3b) are actully sections of the closed current loops from Fig. 3a. 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 the 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. 7a)!

Wonderful, now it sinks the current IB from IOUT and injects the same current IB into IIN (Fig. 7b)! 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, there is something wrong in this connection because we can't change the input current in the left leg (by varying the resistor or the voltage). If we try to do that, the transistor T3 will resist to our intervention thus keeping up a constant current (see the next section). Let's continue thinking.





Well, the transistor T3 has to adjust its base current so that its collector current to remain unchanged as we want. 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. Instead, there is a negative feedback implemented by T3 and T1 that are connected between the collector and the emitter of T2 in the output part of the circuit (see again Fig. 7a)! What do we do then?

Eureka! We may swap the two circuit legs: the output part can serve as an input one and the input part - as an output one. Thus we obtain finally a true Wilson current mirror (Fig. 8a)!



How the Wilson current mirror keeps up a constant output current
We may consider the behavior of Wilson current mirror from two aspects. From one hand, when we vary the input quantities (the input voltage, the resistance R or actually, the input current), a current mirror behaves as a current follower. In this case we have considered above, it is important the output current to follow exactly the input current. From the other hand, if we vary the output quantities (the supply voltage, the load resistance or voltage), the current mirror behaves as a constant current source. In this case, it is important the circuit to keep up a steady current. Well, let's discuss the second aspect here. Circuit-fantasist (talk) 15:19, 21 September 2008 (UTC)

The second problem of the simple current mirror


The output part of the simple BJT current mirror exploits the basic property of the bipolar transistor to behave as a current-stable resistor if we keep up steady its base voltage or current. Actually, in combination with the power supply, the transistor constitutes a current source or sink. How does the transistor do this magic? If, for example, the load resistance RL varies, the transistor changes its present resistance RT between the collector and the emitter so that to keep up a constant total resistance Rtot = RL + RT = const (Fig. 9).

Only, due to Earley effect, the output part does not behave as a perfect current source; this is the second imperfection of the simple current mirror that we have to improve now.



Making the transistor keep a constant current by negative feedback:


We can make the output transistor keep up a constant current by applying various clever "tricks" but negative feedback is the more reliable of them. Then, how do we introduce such a "current-keeping" negative feedback in the simple transistor stage? Let's begin thinking...

We have a transistor that controls the current through the load by changing its present resistance between the collector and the emitter... What does it mean to introduce a "current-keeping" negative feedback? What does have to happen if the current tries to change because of RL or VCC variations? Obviously, if the load current tries to increase, the transistor has to close more (to increase its present collector-emitter resistance RT) so that to restore the previous magnitude of the current. And v.v., if the load current tries to decrease, the transistor has to open more (to decrease its present collector-emitter resistance) - Fig. 10. For simplicity, let's consider only the first case (the load current increases) from now on.

In order to make the transistor do this magic, we have to close the negative feedback loop from the output circuit where the load current flows to the transistor input. The base-emitter junction (gate-drain part) serves as a differential voltage input; but, regarding to the ground, the transistor has two single-ended voltage inputs - the emitter and the base. So, we might introduce two kinds of negative feedbacks by applying a voltage that is proportional to the load current to the emitter and to the base.



...driving the transistor from the emitter...




First, we may fix the base voltage and to drive the transistor from the emitter. For this purpose, we connect a reference voltage source VREF to the base and a current-to-voltage converter in the emitter producing a voltage proportional to the load current IL (Fig. 11).

A bare resistor RE can serve as a simple current-to-voltage converter (Fig. 12). Since the output quantity (the voltage drop VRe) is applied in series to the input quantity (the voltage VREF) this popular technique for keeping up a constant current is named series negative feedback or "emitter degeneration".

Only, the voltage drop VRe limits the maximum voltage drop across the load (the so called voltage appliance). What do we do then?



General idea
But don't you think that, with the same success, we may fix the emitter voltage and drive the transistor from the base? Let's try it! Maybe, it will lead us to the desired Wilson current mirror... In order to impelement this idea, we need again an element that produces a voltage proportional to the load current, i.e. current-to-voltage converter. But now the current flows in one place (the emitter) while the voltage has to be applied to other place (the base)! So, we need not a bare "resistance" I-to-V converter; we need a kind of "transresistance" I-to-V converter. How do we make it? How have Wilson solved this problem?





Eureka! We may "copy" the load current IL to the desired place where to pass it through a resistor, in order to create a voltage drop proportional to IL (Fig. 13). Then let's do it! For this purpose, we connect an I-to-V converter in the emitter that drives a reverse V-to-I converter. As you can see, the direct and the reversed converter constitute actually the well-known simple current mirror. It produces a "copy" of IL that flows through a conventional resistive I-to-V converter connected to the base (Fig. 14). In this way, by means of a current mirror and a current-to-voltage converter we apply again a current-keeping negative feedback.

The Wilson current mirror consists of a simple current mirror and a current-to-voltage converter connected in the feedback loop.



Implementation


As above, a bare resistor R can act as a simplest current-to-voltage converter (Fig. 15). Actually, it serves as the emitter "degeneration" resistor RE from Fig. 12. Only, here not the "original" load current IL flows through the resistor R but a "copy" IR = IL of this current. The voltage drop VR across the resistor R or, more strictly speaking, its supplemental to VCC voltage is the input voltage for the transistor T3.

Operation
Let's investigate how the circuit will react if we change the load resistance RL. In the beginning, suppose equal currents IL = IR flow through the two circuit legs.

If we increase RL, the load current IL tries to decrease. This current is the input quantity of the simple current mirror T1, T2; so, its output quantity IR decreases also. As a result, the voltage drop VR decreases and its supplement VCE2 increases. The input voltage VBE3 of the transistor T3 increases; it begins opening more until the load current restores its previous magnitude.

What does the transistopr T3 actually keep? It keeps up a constant base-emitter voltage VBE. If we look at this circuit as a negative feedback stabilizer, VBE is its input reference quantity and the transistor T3 keeps constant this quantity. Doing that, it keeps actually a constant voltage drop VR across the steady resistor R; so, the current IR and IL are constant too.



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...."

Date of invention
Is there any indication about when the circuit was invented exactly? The book by Sedra & Smith does not say anything about that.

ICE77 (talk) 17:05, 13 June 2011 (UTC)