Question

In the diode circuit shown below, the value of voltage $V_A$ will be (Assume $V_r=0.7V$ for each diode)

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

Answer 1

The correct option is D - 2.15 V

Initially assume each diode is in its conducting state
$V_B=-0.7V$
$V_A=0.7-0.7=0V$


Applying KCL at node - A

$\frac{5-V_A}{5k}$ = $I_{D2}+\frac{(V_A-0.7)-(-10)}{5k}$
Since , $V_A$ = 0

$I_{D2}=\frac{5-9.3}{5k}=-0.86mA$

As it was assumed initially all diodes were ON
Now we know,
Diodes $D_1$ and $D_3$ are ON and diode $D_2$ is OFF.
Apply KVL in the loop ,
$-5+(I_{D1}\times 5k)+0.7+(I_{D1}\times 5k)-10=0$

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

$V_A=0.7+(I_{D1}\times 5K)-10$
= 0.7 + (1.43 $\times 5$ ) - 10
= - 2.15 V