Question

Consider the circuit shown in the figure below .

A diode $D_1$ is connected in parallel with a diode $D_2$ , with reverse saturation current equal to ${10}^{-12}A$ and ${10}^{-10}A$ respectively. The diodes are connected across a voltage source ($V_s$) in series with a resistance of $1k\Omega$. Then the value of voltage ' ($V_s$) ' is approximately equal to (Assuming $\eta =1and{V}_{Th}=26mV)$

• A. 2.004 V
• B. 5.241 V
• C. 2.436 V
• D. 4.444 V

Answer 1

The correct option is C 2.436 V

$V_S=V_x+V_{D1}(\because V_{D1}=V_{D2})$

and $I=I_1+I_2$

thus $2\times {10}^{-3}={10}^{-12}[e^{\frac{V_{D1}}{26\times {10}^{-3}}}-1]+{10}^{-10}[e^{\frac{V_{D1}}{26\times {10}^{-3}}}-1]$

$2\times {10}^{-3}={10}^{-10}\left(1.01\right).e^{\frac{V_{D1}}{26\times {10}^{-3}}}$

$\frac{V_{D1}}{26\times {10}^{-3}}=\ln (1.9801\times {10}^7)=16.801$

$V_{D1}=0.437V$

$V_x=2\times {10}^{-3}\times 1\times {10}^3=2V$

$V_S=V_x+V_{D1}=2+0.437$

$=2.437V$