Talk:Circuit Idea/How the Wilson current mirror keeps the current

I have copied the discussion below from the related Wikipedia page, in order to start the discussion about the second main property of the legendary Wilson current mirror. Circuit-fantasist (talk) 13:14, 22 September 2008 (UTC)

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. 1).

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. 2. 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. 3).

A bare resistor RE can serve as a simple current-to-voltage converter (Fig. 4). 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. 5). 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. 6). 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. 7). Actually, it serves as the emitter "degeneration" resistor RE from Fig. 4. 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...."