Question

. please help me from a to e

Recall the Offset Nulling Circuit that you used to zero out the offset of the LM741 OpAmp in L2002B, shown in Figure 1(a). An external 10 k2 potentiometer was applied across the emitter terminals of the current mirror formed by Q5 and Q6 in Figure 1(b), which can be thought of as making an adjustment to collector currents Ics and Ics such that Ics-Ics exactly A simplified version of that current mirror circuit is shown in Figure 1(c), where Q7 and R3 are replaced with a simple diode-connection on Q5, and the effect of the 10 kΩ potentiometer is shown as making an adjustment to the values of R1 and R2 in opposite directions. (a) For the current mirror shown in Figure 1(c), find an expression for the difference in base-emitter voltages AVBEVBES VBE in terms of lout, Iref, Vr and the ratio of emitter areas N- AE6l AEs. Ignore the effect of base currents (b) Now assume that there is a size mismatch in this current mirror such that N-1.1 (i.e AE6 was made 10% bigger than AEs). If RI and R2 are adjusted to balance low,-hf exactly, then use your answer from part (a) to find the difference AR - R2 - Rl given Vr 25mV and Iref 0.5mA. Again, ignore the effect of base currents. (c) If a 10 kΩ external potentiometer as shown in Figure 1 (a) is used to balance lout-Inf as found in part (b), then the average value of R1 and R2 can be found as <R> - (RI+R2)/2 -5 k2 // 1 k2 833.32. Using your answer for AR in part (b), solve for the values of R1 and R2. (d) Draw the small signal equivalent circuit for calculating Rout as shown in Figure 1(c) (e) Given B- 100, VA-50V, Vcc 10V and using the same DC bias condition in part (b), calculate the value of output resistance Rout for the current mirror shown in Figure 1(c) R. Q7 Offset Nulling Circuit 06 LM741 Q5 area-AEs area 10 k0 OFFSET NULL OFFSET NULL R2 R1 R3 50K1 K RI re

Answer 1

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