# Circuit Idea/How the Wilson current mirror keeps the current

Circuit idea: Realizing a current-keeping feedback from the emitter to the base by "copying" the load current and converting it to voltage.

## The two aspects of the Wilson current mirror

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 in the previous story, 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.

## 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 R$_L$ varies, the transistor changes its present resistance R$_T$ between the collector and the emitter so that to keep up a constant total resistance R$_{tot}$ = R$_L$ + R$_T$ = const (Fig. 1).
Only, due to Early 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 R$_L$ or V$_{CC}$ variations? Obviously, if the load current tries to increase, the transistor has to close more (to increase its present collector-emitter resistance R$_T$) 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. What is 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 V$_{REF}$ to the base and a current-to-voltage converter in the emitter producing a voltage proportional to the load current I$_L$ (Fig. 3).
A bare resistor R$_E$ can serve as a simple current-to-voltage converter (Fig. 4). Since the output quantity (the voltage drop V$_{Re}$) is applied in series to the input quantity (the voltage V$_{REF}$) this popular technique for keeping up a constant current is named series negative feedback or "emitter degeneration".
Only, the voltage drop V$_{Re}$ limits the maximum voltage drop across the load (the so called compliance voltage). What do we do then?

### ...driving the transistor from the base
#### 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 implement 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 I$_L$ to the desired place where to pass it through a resistor, in order to create a voltage drop proportional to I$_L$ (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 I$_L$ 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 R$_E$ from Fig. 4. Only, here not the "original" load current I$_L$ flows through the resistor R but a "copy" I$_R$ = I$_L$ of this current. The voltage drop V$_R$ across the resistor R or, more strictly speaking, its supplemental to V$_{CC}$ voltage is the input voltage for the transistor T$_3$.

#### Operation

In order to ascertain if this odd circuit behaves as a constant current source, we have somehow to "provoke" it and see its reaction to our "intervention". Well, how may we "provoke" it? Let's for instance, investigate how the circuit will react if we change the load resistance R$_L$. In the beginning, suppose equal currents I$_L$ = I$_R$ flow through the two circuit legs what is normal circuit condition.
If we increase R$_L$, the load current I$_L$ tries to decrease. This current is the input quantity of the simple current mirror T$_1$, T$_2$; so, its output quantity I$_R$ decreases also. As a result, the voltage drop V$_R$ decreases and its supplement V$_{CE2}$ increases. The input voltage V$_{BE3}$ of the transistor T$_3$ increases; it begins opening more until the load current restores its previous magnitude.
What does the transistor T$_3$ actually keep? It keeps up a constant base-emitter voltage V$_{BE}$. If we look at this circuit as a negative feedback stabilizer, V$_{BE}$ is its input reference quantity and the transistor T$_3$ keeps constant this quantity. Doing that, it keeps actually a constant voltage drop V$_R$ across the steady resistor R; so, the current I$_R$ and I$_L$ are constant too.

#### How many feedbacks are there?

The most resources about Wilson current mirror do not notice any feedback in the circuit. Some of them have noticed two positive feedbacks, other - two negative ones. But actually there are as many as three negative feedbacks in this sophisticated circuit! Let's try to see them.
1. A local feedback - the wire between the T$_1$'s collector and its base. It reverses T$_1$ making it act as a logarithmic current-to-voltage converter instead as an antilogarithmic voltage-to-current converter what is inherent for BJT. This is a parallel constant-voltage keeping negative feedback.
2. A more global feedback - the whole active diode (T$_1$ + the wire between its collector and the base) that is connected in the T$_3$'s emitter. This is a series constant-current keeping feedback (an emitter degeneration). Only, it is negligible for keeping the constant current because the active diode behaves as a voltage-stable element (significant current variations cause insignificant voltage changes).
3. A global negative feedback - it is constituted by the simple current mirror (T$_1$ and T$_2$) and the resistor R acting as a simple current-to-voltage converter. It is the important negative feedback that keeps the current constant; we have already discussed it above.

## References

* ↑ Reinventing constant current source reveals the philosophy behind constant current sources.
* ↑ Current mirrors shows how Wilson current mirror keeps up an almost constant output current. It is interesting the author prefers to say "positive" instead "negative" feedback.
* ↑ Understanding Wilson current mirror via the negative feedback approach shows the presence of two negative feedbacks.
* ↑ How to Reverse Current Direction is our story about the simple BJT current mirror.

## Resources
### Trying to explain the circuit by intuition

Paul Horowitz, Winfield Hill. The art of electronics, second edition, p 89. ISBN 0521370957.
Designing analog chips
Mechkov C., Eliminating the Early effect in Wilson current mirror, Fourth international conference - Computer science'2008, 18–19 September 2008, Kavala, Greece. This is the same story as the presented here but it is written a bit more formal.

### "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...."