Question 4.20: If two similar Germanium diodes are connected back to back a...

The arrangement is shown in the Fig. 4.20.

As D_{1} is reverse biased, the total current flowing in the circuit is I_{0} = 1  \mu A. The diode D_{2} is forward biased and its forward current is equal to the reverse current I_{0} = 1  \mu A, which can flow as D_{1} is reverse biased.

For diode D_{2},  I =  I_{0} = 1  \mu A and voltage across D_{2}  is  V_{D2}.

Therefore,                          I = I_{0} [ e^{ V/ \eta V_{T}} – 1]

or                                         I_{0} = I_{0}  [ e ^{V_{D_{2}} / \eta V_{T}} – 1]

Hence,                       e^{V_{D_{2}}/ \eta V_{T}}= 1 + 1 = 2

\frac{V_{D2} }{\eta V_{T} }= 1n  2

V_{D2} = \eta V_{T} × ln  2 = 1 × 26 × 10^{ – 3} × 0.6931 = 0.01802  V

Therefore,                       V_{D1} = V – V_{D2} =  V – 0.01802  V

The current through the diodes D_{1}  and  D_{2}  is  I = I_{0} = 1  \mu A