(a) Calculate the value of $V_{0}$ and I if the Si diode and the Ge diode start conducting at $0.7 V$ and $0.3 V$ respectively, in the given circuit.
(b) If the Ge diode connection be reversed. What will be the new values of $V_{0}$ and I?

Question

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Answer 1

As in the given circuit,
Voltage, $V = 12$ V
Resistance, $R = 5$ k $Ω$
$∴$ , current, $I = \frac{1 2 - 0 . 3}{5 k Ω} = \frac{1 1 . 7 V}{5 \times 1 0 ^{3} Ω} = 2.34$ mA
Output voltage $V_{0} = R I = (5 \times 10^{3}) \times (2.34 \times 10^{- 3}) = 11.7$ V
When the connections of Ge diode are reversed, then current will bw through silicon.
In this case, current, $I = \frac{( 1 2 - 0 . 7 ) V}{5 \times 1 0 ^{- 3} Ω} = 2.26$ mA
$∴$ , Output voltage, $V_{0} = I R$
$= (2.26 \times 10^{- 3}) \times (5 \times 10^{3}) = 11.3$ V

Answer 2

As in the given circuit,

Voltage,

V = 12

V

Resistance,

R = 5

k

Ω

∴

, current,

I = \frac{1 2 - 0 . 3}{5 k Ω} = \frac{1 1 . 7 V}{5 \times 1 0 ^{3} Ω} = 2.34

mA

Output voltage

V_{0} = R I = (5 \times 10^{3}) \times (2.34 \times 10^{- 3}) = 11.7

V

When the connections of Ge diode are reversed, then current will bw through silicon.

In this case, current,

I = \frac{( 1 2 - 0 . 7 ) V}{5 \times 1 0 ^{- 3} Ω} = 2.26

mA

∴

, Output voltage,

V_{0} = I R

= (2.26 \times 10^{- 3}) \times (5 \times 10^{3}) = 11.3

V