Ge and Si diodes conduct at \(0.3 \,V\) and \(0.7\, V\) respectively. In the following figure if Ge diode connection are reversed, the valve of \(V0\) changes by.....\(V\)

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 \(i = \frac{{12 - 0.3}}{{5\,k\Omega }} = 2.34\,mA\)
 Out put voltage \(V_0 = Ri = 5 k \Omega ×2.34 \,mA = 11.7 \,V\)
Now consider the case when diode connection are reversed. In this case voltage drop across the diode's combination \(= 0.7 \,V\)
==> Current \(i = \frac{{12 - 0.7}}{{5\,k\Omega }} = 2.26\,mA\)
 \({V_0} = iR = 2.26\,mA \times 5\,k\Omega = 11.3\,V\)
Hence charge in the value of \(V_0 = 11.7 -11.3 = 0.4 V\)