A diode has reverse saturation current Is 10 18 Aand nonideality factor 1.05. If diode has current of 70A, then diode voltage is(A) 0.63 V (B) 0.87 V(C) 0.54 V (D) 0.93 V21. An ideal pn junction diode is operating in theforward bais region. The change in diode voltage, thatwill cause a factor of 9 increase in current, is(A) 83 mV (B) 59 mV(C) 43 mV (D) 31 mV
The Correct Answer and Explanation is:
Question 1:
Given:
- Reverse saturation current, IS=10−18 AI_S = 10^{-18} \, \text{A}
- Non-ideality factor, n=1.05n = 1.05
- Current, I=70 μA=70×10−6 AI = 70 \, \mu\text{A} = 70 \times 10^{-6} \, \text{A}
- Temperature, T=300 KT = 300 \, \text{K} (assumed standard)
- Thermal voltage, VT=kTq≈25.85 mVV_T = \frac{kT}{q} \approx 25.85 \, \text{mV}
Using the diode equation: I=IS(eVnVT−1)I = I_S \left( e^{\frac{V}{nV_T}} – 1 \right)
Since I≫ISI \gg I_S, we can approximate: I≈ISeVnVTI \approx I_S e^{\frac{V}{nV_T}}
Solving for VV: V=nVTln(IIS)V = nV_T \ln\left( \frac{I}{I_S} \right)
Substitute values: V=1.05×25.85×ln(70×10−610−18)=27.14×ln(7×1013)V = 1.05 \times 25.85 \times \ln\left( \frac{70 \times 10^{-6}}{10^{-18}} \right) = 27.14 \times \ln(7 \times 10^{13})
Break it down: ln(7×1013)=ln(7)+ln(1013)≈1.9459+13ln(10)=1.9459+13×2.3026=31.8847\ln(7 \times 10^{13}) = \ln(7) + \ln(10^{13}) \approx 1.9459 + 13 \ln(10) = 1.9459 + 13 \times 2.3026 = 31.8847 V≈27.14×31.8847≈865 mV=0.87 V⇒Answer: (B)V \approx 27.14 \times 31.8847 \approx 865 \, \text{mV} = \boxed{0.87 \, \text{V}} \Rightarrow \text{Answer: (B)}
Question 2:
Given: Ideal diode. The current increases by a factor of 9.
We want to find the voltage change ΔV\Delta V.
From the ideal diode equation (n = 1), ignoring ISI_S again: I2I1=eΔVVT⇒ΔV=VTln(I2I1)\frac{I_2}{I_1} = e^{\frac{\Delta V}{V_T}} \Rightarrow \Delta V = V_T \ln\left( \frac{I_2}{I_1} \right)
Given: I2I1=9,VT=25.85 mV⇒ΔV=25.85×ln(9)≈25.85×2.1972≈56.8 mV\frac{I_2}{I_1} = 9, \quad V_T = 25.85 \, \text{mV} \Rightarrow \Delta V = 25.85 \times \ln(9) \approx 25.85 \times 2.1972 \approx 56.8 \, \text{mV}
Closest option: (B) 59 mV
✅ Final Answers:
- (B) 0.87 V
- (B) 59 mV
Explanation (Summary):
Diodes follow an exponential I-V relationship given by the Shockley diode equation. For forward-biased conditions, if the current is much larger than the reverse saturation current, we can ignore the “–1” term for simplicity. This leads to a simple logarithmic relationship between current and voltage. In the first question, we used this to find the voltage corresponding to a given forward current, factoring in non-ideality. In the second, the ideal case simplifies analysis further, and we used the thermal voltage to calculate the voltage increment that leads to a 9x current increase. The logarithmic and exponential nature of diode characteristics is the key to solving such problems accurately.
