Question

When an external voltage is applied across a semiconductor diode such that the p-side is connected to the positive terminal of the battery and the n-side to the negative terminal, it is said to be forward biased. The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. When an external voltage is applied across the diode such that the n-side is positive and the p-side is negative, it is said to be reverse biased. The applied voltage mostly drops across the depletion region.

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

1. 0.2 V
2. 0.4 V
3. 0.6 V
4. 0.8 V

II. The values of V_o and I_d for the network are:

1. 13 V, 2.32 mA
2. 11.7 V, 2.08 mA
3. 11.3 V, 2.01 mA
4. 11 V, 1.96 mA

III. The circuit shown below contains two ideal diodes, each with a forward resistance of 50 Ω. If the battery voltage is 6 V, the current through the 100 Ω resistance (in amperes) is:

1. 0.036
2. 0.020
3. 0.030
4. 0.027

IV. In the figure, potential difference between A and B is:

1. Zero
2. 5 V
3. 10 V
4. 15 V

Answer 1

I. 0.4 V

Explanation:

The voltage drop across the diode will change from 0.3 to 0.7 V.

The value of V_o changes by 0.4 V.

II. 11 V, 1.96 mA

Explanation:

V_o = E − V_si − V_Ge

= 12 − 0.7 − 0.3

= 11 V

I_d = `V_0/R`

= `11/(5.6 xx 10^3)`

= 1.96 mA

III. 0.020

Explanation:

I = `6/(50 + 150 + 100)`

= `6/300 A`

= 0.02 A

IV. 10 V

Explanation:

Here, the diode is forward biased, so we replace it with a connecting wire.

After replacement, the two 10 kΩ resistors between A and B become parallel.

`"R"_"parallel" = (10 k xx 10 k)/(10 k + 10 k)`

= `(100 k^2)/(20 k)`

= 5 kΩ

Now the circuit reduces to a 10 kΩ resistor on the top branch and an equivalent 5 kΩ resistor between A and B, forming a voltage divider from the 30 V supply.

`V_A - V_B = "R"_"parallel"/("R"_"top" + "R"_"parallel") xx 30`

= `5/(10 + 5) xx 30`

= `5/15 xx 30`

= 5 × 2

= 10 V