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  1. Problems
  2. /A silicon diode follows the equation I = Is(e^(V/VT) - 1) with η = 1, Is = 1.0 pA and VT = 25.85 mV…
HardPhysicsHigh School (11–12)Electronics › Semiconductors and DiodesNumericalpythonJEE MainBoard ExamOlympiadAEEE (Amrita)AMUEEEAP EAPCET
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A silicon diode follows the equation I = Is(e^(V/VT) - 1) with η = 1, Is = 1.0 pA and VT = 25.85 mV…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A silicon diode follows the equation I=Is(eV/VT−1)I = I_s\left(e^{V/V_T} - 1\right)I=Is​(eV/VT​−1) with η=1\eta = 1η=1, Is=1.0I_s = 1.0Is​=1.0 pA and VT=25.85V_T = 25.85VT​=25.85 mV at 300 K. A designer wants to know how the diode behaves in a small circuit that drives the diode with 5.05.05.0 mA. Find the voltage that must be applied across the diode for a current of 5.05.05.0 mA. Find the dynamic resistance at this current. Find the current at 0.650.650.65 V. Last, find the extra voltage that raises the current by a factor of ten, for example from 5.05.05.0 mA to 505050 mA. This number is a standard rule: the forward voltage grows by a nearly fixed amount for each tenfold rise in current. The −1-1−1 in the equation is negligible in the forward direction.

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Need a reminder? The key idea

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. Is=1.0×10−12 AI_s = 1.0 \times 10^{-12}\ \text{A}Is​=1.0×10−12 A and VT=0.02585V_T = 0.02585VT​=0.02585 V.

Plan. Use the equation turned round for the voltage, the slope for the dynamic resistance, and the equation itself for the current.

Step 1. The voltage. V=VTln⁡ ⁣(IIs+1)=0.02585 V×ln⁡ ⁣(5.0×10−3 A1.0×10−12 A)=0.02585 V×22.33=0.577 VV = V_T \ln\!\left(\dfrac{I}{I_s} + 1\right) = 0.02585\ \text{V} \times \ln\!\left(\dfrac{5.0 \times 10^{-3}\ \text{A}}{1.0 \times 10^{-12}\ \text{A}}\right) = 0.02585\ \text{V} \times 22.33 = 0.577\ \text{V}V=VT​ln(Is​I​+1)=0.02585 V×ln(1.0×10−12 A5.0×10−3 A​)=0.02585 V×22.33=0.577 V.

Step 2. The dynamic resistance. rd=VTI=0.02585 V5.0×10−3 A=5.17 Ωr_d = \dfrac{V_T}{I} = \dfrac{0.02585\ \text{V}}{5.0 \times 10^{-3}\ \text{A}} = 5.17\ \Omegard​=IVT​​=5.0×10−3 A0.02585 V​=5.17 Ω.

Step 3. The current at 0.650.650.65 V. I=1.0×10−12 A×(e0.65/0.02585−1)=1.0×10−12 A×8.31×1010=0.0831 A=83.1I = 1.0 \times 10^{-12}\ \text{A} \times \left(e^{0.65/0.02585} - 1\right) = 1.0 \times 10^{-12}\ \text{A} \times 8.31 \times 10^{10} = 0.0831\ \text{A} = 83.1I=1.0×10−12 A×(e0.65/0.02585−1)=1.0×10−12 A×8.31×1010=0.0831 A=83.1 mA.

Step 4. A tenfold rise. If III becomes 10I10I10I, the voltage rises by VTln⁡10=0.02585 V×2.303=0.0595 V=59.5V_T \ln 10 = 0.02585\ \text{V} \times 2.303 = 0.0595\ \text{V} = 59.5VT​ln10=0.02585 V×2.303=0.0595 V=59.5 mV.

Answer: 0.5770.5770.577 V, 5.17 Ω5.17\ \Omega5.17 Ω, 83.183.183.1 mA and 59.559.559.5 mV.

Check. The diode drop of 0.580.580.58 V is close to the common value of 0.70.70.7 V. Because the voltage changes by only 606060 mV for a tenfold change in current, the drop stays between 0.60.60.6 V and 0.70.70.7 V over a wide range. That is why the constant drop model works.

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A p-n junction has a depletion layer of width WWW and a barrier V0V_0V0​. A forward bias VVV lowers the barrier to V0−VV_0 - VV0​−V, and a reverse bias VVV raises it to V0+VV_0 + VV0​+V. The diode equation is I=Is(eV/ηVT−1)I = I_s\left(e^{V/\eta V_T} - 1\right)I=Is​(eV/ηVT​−1), with VT=kTe=25.85 mVV_T = \dfrac{kT}{e} = 25.85\ \text{mV}VT​=ekT​=25.85 mV at 300 K. A conducting silicon diode drops about 0.70.70.7 V.

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