If two similar Germanium diodes are connected back to back and the voltage V is impressed upon, calculate the voltage across each diode and current through each diode. Assume similar value of Io = 1 μA for both the diodes and η = 1.

Question

If two similar Germanium diodes are connected back to back and the voltage V is impressed upon, calculate the voltage across each diode and current through each diode. Assume similar value of [latex]I_{o}[/latex] = 1 μA for both the diodes and η = 1.

Accepted Answer

The arrangement is shown in the Fig. 4.20.

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

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

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

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

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

[latex]\frac{V_{D2} }{\eta V_{T} }= 1n 2 [/latex]

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

Therefore, [latex]V_{D1} = V – V_{D2} = V – 0.01802 V[/latex]

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

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

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