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The Wilson Current Mirror

How the Wilson Current Mirror Works

The Wilson current mirror.

At first glance, the current mirror circuit shown to the right would seem to be backwards. In fact, however, it works quite well, although in a round-about fashion. A small amount of the current flowing through R provides base current for Q3, enabling Q3 to conduct current. The bulk of this current, as with any transistor, flows from emitter to collector. Of course, IC3 = IO.

At the same time, IE3 becomes the reference current for the current mirror formed by Q1 and Q2. As we have already seen when studying the basic current mirror, IC1 = IC2, and IE3 = IC2 + 2IB.

This arrangement provides a form of negative feedback loop. If for any reason IO begins to increase, IC2 must likewise increase, and IC1 will mirror that increase. This will increase the voltage drop across R, which will in turn reduce the base voltage on Q3. That will reduce IO again. In reality, IO cannot change that way, and the circuit does indeed operate as a current mirror.



Transistor Voltages and Currents

Before we can calculate the currents flowing through different parts of this circuit, we must know the voltages. Fortunately, that's very easy in this case. The emitters of Q1 and Q2 are grounded, and their bases are connected to each other as well as to the collector of Q2 and the emitter of Q3. Therefore this common point is at a fixed voltage of VBE. In addition, the base of Q3, which is connected to the collector of Q1, must be VBE above its emitter, which puts its voltage at 2VBE. The collector of Q3, which provides IO, can float to any voltage between +V and [VBE + VCE(SAT)].

Now that we know all of the fixed circuit voltages, we can determine the basic circuit currents within the mirror circuit:

IREF  =  +V – 2VBE

R
 
IE3  =  IC2  +  2IB2
 
IE3  =  β + 2  IC2

β
 
IB3  =  IE3  

β + 1
 
IB3  =  β + 2  IC2

β(β + 1)
 
IC2 = IC1  =  IREF  –  IB3
 
IREF  =  IC1 + IB3
 
IREF  =  IC2 + IB3
 
IO = IC3  =  β  IE3

β + 1
 
IO  =  β + 2  IC2

β + 1
 

From the above, it looks as if everything can be expressed in terms of IC2, which is central to the mirror circuit. We can use this fact to derive the relationship between IO and IREF entirely based on β, assuming all transistors have the same parameters. The final expression becomes

IO  =  β + 2


IREF β + 2 + 2/β
 
IO  =  β² + 2β


IREF β² + 2β + 2

That slight difference of is generally negligible with modern high-gain transistors. This expression shows that by splitting the base current between the two sides, the Wilson current mirror comes very close to a complete current match between IO and IREF. If the transistors have a β of 100, the error is less than 0.02%.



Output Resistance

The Wilson current mirror.

The use of two transistors in the output side of the Wilson current mirror (reproduced to the left for easy reference) affects the output resistance of the circuit, much as an emitter resistor does. The advantage of using Q2 in place of a resistor is that the voltage across Q2 will only be VBE regardless of the magnitude of IO, where an emitter resistor will generally drop a higher voltage in accordance with Ohm's Law.

Of course, the fact that Q2 is connected as a diode limits its effective resistance in this application. Nevertheless, it does help to significantly increase the effective output resistance of Q3. For the Wilson current mirror,

ROUT ≈ βrO/2

If the transistors have a β of 100 and a VA of 50 V, and we set IO to 1 mA, ROUT of a single transistor by itself would be 50,000Ω. Including Q2 in the emitter lead of Q3 we raise that to 2.5M. This represents a very large improvement in the behavior of this current mirror, over the basic mirror circuit.



Drawbacks

Adding collector voltage adjustment to the Wilson mirror. The modified Wilson current mirror.

The main drawback of using the Wilson current mirror has to do with the required circuit voltages. The minimum value of input voltage +V has been increased from VBE to 2VBE. Also, the minimum collector voltage for Q3 has increased from VCE(SAT) to VBE + VCE(SAT). This might preclude the use of the Wilson circuit in very low voltage applications, and always reduces the amount of voltage swing available to the output circuit.

The fact that VCE1 is twice the value of VCE2 creates a mismatch between the two transistors, due to the Early voltages of these devices. The transistors themselves are matched, so they have the same value of rO (= VA/IC). However, Q1 will lose twice as much current to this internal resistance as Q2. This causes the mismatch.

A practical way to remove the mismatch is to add a diode-connected transistor in the collector lead of Q1, as shown to the left. This reduces the collector voltage of Q1 to the same VBE as the collector of Q2, thus completely balancing Q1 and Q2 to eliminate this mismatch. We don't even change the base current mismatch of the overall circuit, because Q3's base current is returned through the emitter (IE = IC + IB).

If we redraw the schematic diagram slightly, we get the circuit shown to the right. It is electrically identical; simply more compactly drawn. This circuit is known as the modified Wilson current mirror. We also note that this circuit is very like the cascode current mirror, with the Q1-Q2 pair reversed as to which transistor is used as the reference.


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