A junction diode is connected to a 10 V source and 103 Ω resistor as shown in figure. The slope of load line on the characteristic curve of diode will be

Question

A junction diode is connected to a 10 V source and 103 Ω resistor as shown in figure. The slope of load line on the characteristic curve of diode will be (A) 10-2 A V-1 (B) 10-3 A V-1 (C) 10-4 A V-1 (D) 10-5 A V-1

Tardigrade Admin · Accepted Answer

If V is the voltage across the junction and I is the circuit current, then V + IR = E or I = (E/R) -(V/R) = -(V/R) +(E/R) Slope of load line = â (1/R) = (1/1000) = 10-3 A V-1

Q. A junction diode is connected to a 10 V source and $1 0^{3} Ω$ resistor as shown in figure. The slope of load line on the characteristic curve of diode will be

$1 0^{- 2} A V^{- 1}$

$1 0^{- 3} A V^{- 1}$

$1 0^{- 4} A V^{- 1}$

$1 0^{- 5} A V^{- 1}$

If V is the voltage across the junction and $I$ is the circuit current, then
$V + I R = E$ or $I = \frac{E}{R} - \frac{V}{R} = - \frac{V}{R} + \frac{E}{R}$
Slope of load line $= - \frac{1}{R} = \frac{1}{1000} = 1 0^{- 3} A V^{- 1}$

Q. A junction diode is connected to a 10 V source and 103Ω resistor as shown in figure. The slope of load line on the characteristic curve of diode will be

10−2AV−1

10−3AV−1

10−4AV−1

10−5AV−1