In the figure, assume that the forward voltage drops of the PN di
In the figure, assume that the forward voltage drops of the PN diode D1 and Schottky diode D2 are 0.7 V and 0.3 V, respectively. If ON denotes conducting state of the diode and OFF denotes a non-conducting state of the diode, then in the circuit.
A. Both D<sub>1</sub> and D<sub>2</sub> are ON
B. D<sub>1</sub> is ON and D<sub>2</sub> is OFF
C. Both D<sub>1</sub> and D<sub>2</sub> are OFF
D. D<sub>1</sub> is OFF and D<sub>2</sub> is ON
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
A forward-biased Schottky diode behaves like a normal p-n diode which maintains a constant voltage when forward-biased.
The only difference is that the forward voltage drop for a Schottky-diode is lower than the normal diode.
Calculation:
Given:
\({\left( {{V_{{D_1}}}} \right)_{on}} = 0.7\;V,\;\;{\left( {{V_{{D_2}}}} \right)_{on}} = \;0.3\;V\)
Considering D1 → off and D2 → ON, we get the circuit as:
Applying KVL in the outer loop, we get:
10 = 1000 I + 20 I + 0.3
0.7 = 1020 I
\(\therefore I = \frac{{9.7}}{{1020}} = 9.5\;mA\)
Now, we calculate VD1 by applying KVL in loop 2, i.e.
\(10 - 9.5 - {V_{{D_1}}} = 0\)
\(\therefore {V_{{D_1}}} = 0.5\;V\)
Since \({V_{{D_1}\;}} < \left( {{V_{{D_1}}}} \right)\) ON. DiodeD1 will be in the OFF state and our assumption is correct.
