Consider the case when Ge and Si diodes are connected as show in the given figure.
Equivalent voltage drop across the combination Ge and Si diode = 0.3 V
⇒ Current i = 12−0.35 kΩ = 2.34 mA
∴ Out put voltage V0 = Ri = 5 kΩ×2.34 mA = 11.7 V
Now consider the case when diode connection are reversed. In this case voltage drop across the diode's combination = 0.7 V
⇒ current i = 12−0.75 kΩ = 2.26 mA
∴V0 = iR = 2.26 mA×5 kΩ = 11.3 V
Hence charge in the value of v0 = 11.7 - 11.3 = 0.4 V