Question

P9.67, Sketch the transfer characteristic (v, versus if in) for the circuit shown in Figure P9.67 carefully labeling the breakpoint and slopes Allow vin to range from -5 V to +5 V and assume that the diodes are ideal. p37, Find the values of / and V for the circuits of igure P9.37 assuming that the diodes are deal. 515 1.2 kf Probles 3 kf voD 2 ks? 5 V Figure P9.67 DV Sre for 2 E ing ent the lle (c) Figure P9.37 D W 9+10

Answer 1

While the diode is reverse biased, let the voltage at node 1 be V₁

The equivalent circuit is given by

1.2 k? 3 k 2 kf SV HH

Since the segment of the circuit with the 5 V source is detached from the output due to the ideal diode being reverse biased, v_in = v₀

Applying KCL at node 1,

5-V1 — И и 2к З

10 2V13V{

5V110V

V15V

When v_in > 2 V the ideal diode becomes forward biased. The equivalent circuit is given by:

1.2 k? 3 kn SV 2 kfn H

Using the superposition theorem to find the voltage due to v_in at node 1, with the diode forward biased, we short the 5 V circuit (below figure).

1.2 k? 3 kn 2 kfn

The equivalent resistance of 2k || 3k = 2kx3k/(2k+3k) = 6k²/5k = 1.2k

From voltage division principle, the voltage at node 1 = v_in x (1.2k/(1.2k + 1.2k)) = v_in/2

Therefore, the relation between v₀ and v_in are:

Vin2 V Vin/2 ,vin> 2 V Vin

The range of v_in is : -5 V < v_in < 5 V

Following is the sketch of v₀ vs v_in

_{st C un n <---- (A) A}