Question

In the diode circuit shown below, the value of voltage for each diode



$V_{A}is(Assume,V_{\gamma }=0.7V)$

• A. -2.15 V
• B. 0 V
• C. 1.75 V
• D. 2 V

Answer 1

The correct option is A -2.15 V

Initially assume each diode is in its conducting state

$V_B=-0.7V$

and $V_A=-0.7+0.7=0V$

Applying KCL at node A,

$\frac{5-V_A}{5k}=I_{D_2}+\frac{(V_A-0.7)-(-10)}{5k}$

Since, $V_A=0$

$I_{D2}=-0.86mA$

Which is in consistant with the assumption that all diodes are ON.
Now assume that $D_{1}and{D}_{3}areONand{D}_2$ is OFF

Applying KVL in the loop,

$-5+(I_{D1}\times 5k)+0.7+(I_{D1}\times 5k)-10=0$

$I_{D1}=\frac{15-0.7}{10k}=1.43mA$

$=0.7+(1.43\times 5)-10$

$V_A=-2.15V$