Wilson Current Source

Another current source transistor configuration that provides a very large parallel resistance is the Wilson current source which uses three transistors and provides this capability an the output is almost independent of the internal transistor characteristics. The Wilson current source as shown in fig. 3, uses the negative feedback provided by Q₃ to raise the output impedance

Fig. 3

The difference between the reference current and I_C1 is the base current of Q₂.

I_E2 =(β + 1) I_B2 = I_C3      (E-9)

Since the base of Q₁ is connected to the base of Q₃, the currents in Q₁ are approximately independent of the voltage of the collector of Q₂. As such, the collector current of Q₂ remains almost constant providing high output impedance.

Let us now see that I_C2 is approximately equal to I_REF. Applying Kirchhoff's current law at the emitter of Q₂ yields

I_E2 = I_C3 + I_B3 + I_B1      (E-10)

Using the relationship between collector and base currents

        (E-11)

Since all three transistors are matched, V_BE1= V_BE2 = V_BE3 and β₁ = β₂ = β₃

With identical transistors, current in the feedback path splits equally between the bases of Q₁ and Q₃ leading so that I_B1 = I_B3 and therefore I_C1 = I_C3. Thus, the emitter current of Q₂ becomes

     (E-12)

The collector current of Q₂ is

      (E-13)

Solving for I_C3 yields

      (E-14)

Summing currents at the base of Q₂,

      (E-15)

    (E-16)

Since I_C1 = I_C3, we substitute I_C3 to obtain

    (E-17)

and solving for I_C2

    (E-18)

Equation (E-10) shows that β has little effect upon I_C2 since, for reasonable values of β.

   (E-19)

Therefore, I_C2 = I_REF

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