Skip to content
Additional Exercises · 14.10

Q.In a p-n junction diode, the current II can be expressed as
[!FORMULA] I=I0[exp⁡(eVkBT)−1]I = I_0 \left[\exp\left(\dfrac{eV}{k_B T}\right) - 1\right]
where I0I_0 is called the reverse saturation current, VV is the voltage across the diode and is positive for forward bias and negative for reverse bias, and II is the current through the diode, kBk_B is the Boltzmann constant (8.6×10−5 eV/K8.6 \times 10^{-5}\ \text{eV/K}) and TT is the absolute temperature. If for a given diode I0=5×10−12 AI_0 = 5 \times 10^{-12}\ \text{A} and T=300 KT = 300\ \text{K}, then

(a) What will be the forward current at a forward voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Yanam CbseNCERTSubjective· 3mImportance★★★★★
38% · 14/37 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Plug V=0.6VV=0.6\text{V} and V=0.7VV=0.7\text{V} into I=I0[exp⁡(eV/kBT)−1]I=I_0[\exp(eV/k_BT)-1] to get the two forward currents, take their difference for part (b), the ratio ΔV/ΔI\Delta V/\Delta I for the dynamic resistance in part (c), and note that in reverse bias the current saturates at −I0-I_0 regardless of voltage for part (d).

Setup: I0=5×10−12 AI_0 = 5\times10^{-12}\ \text{A}, T=300 KT = 300\ \text{K}, kB=8.6×10−5 eV/Kk_B = 8.6\times10^{-5}\ \text{eV/K}, so kBT/e=8.6×10−5×300=0.0258 Vk_BT/e = 8.6\times10^{-5}\times300 = 0.0258\ \text{V}. Since VV is in volts and kBT/ek_BT/e works out in volts here, the exponent is simply eV/(kBT)=V/0.0258eV/(k_BT) = V/0.0258.

(a) Forward current at V=0.6 VV = 0.6\ \text{V}

VkBT/e=0.60.0258≈23.26\frac{V}{k_BT/e} = \frac{0.6}{0.0258} \approx 23.26

I=I0[e23.26−1]≈5×10−12×1.26×1010≈0.063 AI = I_0\left[e^{23.26} - 1\right] \approx 5\times10^{-12} \times 1.26\times10^{10} \approx 0.063\ \text{A}

(The −1-1 is utterly negligible next to e23.26e^{23.26}.)

(b) Increase in current at V=0.7 VV = 0.7\ \text{V}

0.70.0258≈27.13,I′=I0 e27.13≈5×10−12×6.07×1011≈3.03 A\frac{0.7}{0.0258} \approx 27.13, \qquad I' = I_0\, e^{27.13} \approx 5\times10^{-12} \times 6.07\times10^{11} \approx 3.03\ \text{A}

ΔI=I′−I≈3.03−0.063≈2.97 A\Delta I = I' - I \approx 3.03 - 0.063 \approx 2.97\ \text{A}

(c) Dynamic resistance

rd=ΔVΔI=0.7−0.62.97=0.12.97≈0.0337 Ω≈0.034 Ωr_d = \frac{\Delta V}{\Delta I} = \frac{0.7 - 0.6}{2.97} = \frac{0.1}{2.97} \approx 0.0337\ \Omega \approx 0.034\ \Omega

This is very small — a forward-biased diode presents almost no opposition to further current increase, since the current grows exponentially with voltage.

(d) Reverse bias from 1 V to 2 V …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.