Problem 4.5

The 10kW resistor is used to decrease current level (Widlar current source). With this configuration, we get a lower V_BE value for Q₂ than for Q₁.

To investigate b effects, we assume b effect of Q₂ is small; of course, we need to validate this assumption after solving this problem, that is that the current in Q₂ is much smaller than for Q₁.

We also know that neglecting the base current for Q₁ results in a percent error of 1/(1 + 1 / b ) which is roughly 1% error. Therefore, we will ignore b effects throughout the remainder of this problem.

Therefore, I_in = I_s e^{^{V / U_T}} = (15V - V ) / 20kW, where V is the middle node of the current source. Note, we have the same problem as in Prob 4.1; therefore we get V.

Then writing a KVL loop including V_BE of Q₁ and Q₂, and the 10kW resistor, we get

I_out = (U_T / 10kW) ln( I_in / I_out ) = 2.5mA ln( I_in / I_out ).

Knowing I_in = , we get I_out by iteration as .