Question

In the circuit shown in Figure below, find the diode voltage $V_D$ and the supply voltage $V$ such that the current $I = 0.5$ mA. Assume the reverse-saturation current is $I_0 = 5 \times 10^{-12}$ A and the ideality factor ($\eta$) is 1. $R = 4.7 \text{ k}\Omega$ $V_s = 5V$ Hint: $I_D = I_0 \left[ \exp\left(\frac{V_D}{\eta V_T}\right) - 1 \right] \implies V_D \approx \eta V_T \ln\left(\frac{I_D}{I_0}\right)$

          In the circuit shown in Figure below, find the diode voltage $V_D$ and the supply voltage $V$ such that the current $I = 0.5$ mA. Assume the reverse-saturation current is $I_0 = 5 \times 10^{-12}$ A and the ideality factor ($\eta$) is 1.
$R = 4.7 \text{ k}\Omega$
$V_s = 5V$
Hint: $I_D = I_0 \left[ \exp\left(\frac{V_D}{\eta V_T}\right) - 1 \right] \implies V_D \approx \eta V_T \ln\left(\frac{I_D}{I_0}\right)$

In the circuit shown in Figure below, find the diode voltage VD and the supply voltage V such that the current I = 0.5 mA. Assume the reverse-saturation current is I0 = 5 × 10^-12 A and the ideality factor (η) is 1.
R = 4.7 kΩ
Vs = 5V
Hint: ID = I0 [ ((VD)/(η VT)) - 1 ] VD ≈η VT ln((ID)/(I0))

Added by Marc B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Narayan Hari

11/22/2021

Step 1

Given that I = 0.5 mA, Is = 5 x 10^-12 A, and n = 1, we can rearrange the equation to solve for Vo: Vo = n * Vt * ln(I/Is + 1) Show more…

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In the circuit shown in Figure below, find the diode voltage Vo and the supply voltage V such that the current I = 0.5 mA. Assume the reverse-saturation current is Io = 5 x 10^-12 A and the ideality factor (n) is 1. R = 4.7 kΩ Vs = 5 V Hint: I, = 1.1 Lexpl