Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of $$V_0$$ changes by: (assume that the Ge diode has large breakdown voltage)

Initial State:

Both diodes are forward-biased. The Ge diode turns on first at $$0.3\text{ V}$$ and clamps the parallel voltage drop, keeping the Si diode off.

$$V_{0, \text{initial}} = V_{\text{in}} - V_{\gamma, \text{Ge}} = 12 - 0.3 = 11.7\text{ V}$$

Reversed State:

The Ge diode connection is reversed, making it reverse-biased (open circuit). The Si diode is forward-biased and conducts.

$$V_{0, \text{reversed}} = V_{\text{in}} - V_{\gamma, \text{Si}} = 12 - 0.7 = 11.3\text{ V}$$

$$\Delta V_0 = V_{0, \text{initial}} - V_{0, \text{reversed}} = 11.7 - 11.3 = 0.4\text{ V}$$