Question

Consider the two diode circuit as shown in the figure below:

$\text{The diode}{D}_1\text{and}{D}_2\text{are identical except for that the area of diode}{D}_1\text{is}{A}_1\text{and area of diode}{D}_2\text{is}{A}_2.\text{If the input current supplied by the voltage source}{V}_{in}\text{is equal to}{I}_{in},\text{then the value of current}{I}_{D_2}\text{passing through the diode}{D}_2\left(I_{D_2}\right)\text{is equal to}$

• A. $I_{D_2}=\frac{I_{\text{in}}}{1+\frac{A_1}{A_2}}$
• B. $I_{D_2}=\frac{I_{\text{in}}}{1-\frac{A_2}{A_1}}$
• C. $I_{D_2}=\frac{I_{\text{in}}}{\frac{A_2}{A_1}-1}$
• D. $I_{D_2}=\frac{I_{\text{in}}}{1+\frac{A_2}{A_1}}$

Answer 1

The correct option is A $I_{D_2}=\frac{I_{\text{in}}}{1+\frac{A_1}{A_2}}$

$\text{Now,}{I}_{in}=I_{D_1}+I_{D_2}$

$\text{and}{V}_{D_1}=V_{D_2}(\because \text{diodes are in parallel})$

$\text{thus,}{V}_{T}ln\left(\frac{I_{D_1}}{I_{s_1}}\right)=V_{T}ln\left(\frac{I_{D_2}}{I_{s_2}}\right)$

$\Rightarrow \frac{I_{D_1}}{I_{s_1}}=\frac{I_{D_2}}{I_{s_2}}$

$I_{D_1}=\left(\frac{I_{D_2}}{I_{s_2}}\right).I_{s_1}$

$\text{Now,}{I}_{\text{in}}=I_{D_2}+I_{D_2}\left(\frac{I_{s_1}}{I_{s_2}}\right)$

$I_{\text{in}}=(1+\frac{I_{s_1}}{I_{s_2}})I_{D_2}$


$\Rightarrow I_{D_2}=\frac{I_{\text{in}}}{1+\frac{I{s}_1}{I_{s_2}}}$

$\text{Now,}{I}_{s_2}\text{and}{I}_{s_2}\text{and reverse saturation currents of the diodes and}{I}_s\infty A$

Thus the above equation can be written as,

$I_{D_2}=\frac{I_{in}}{1+\frac{I_{s_1}}{I_{s_2}}}=\left(\frac{I_{in}}{1+\frac{A_1}{A_2}}\right)$