Example – 1:

Draw the transfer characteristic of the circuit shown in fig. 9.

Fig. 9

Solution:

When diode D₁ is off,  i₁ = 0, D₂ must be ON.

and        v_o = 10 - 5 x 0.25 = 7.5 V

v_p = v_o = 7.5 V

Therefore, D₁ is reverse biased only if v_i < 7.5 V

If D₂ is off and D₁ is ON, i₂ = 0

and        v_p = 10 ( 0.04 v_i - 0.1 ) + 2.5 = 0.4 v_i + 1.5

For D₂ to be reverse biased,

Between 7.5 V and 21.25 V both the diodes are ON.

Fig. 10

The transfer characteristic of the circuit is shown in fig. 10.

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