Chap.4-handout

Chap.4-handout - 9/28/2010 4 Read sections 4.1 thru 4.5...

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9/28/2010 1 Diodes 1 Read sections 4.1 thru 4.5 section 4.6 skipped section 4.7 skipped Chapter Expectation: know Diode I-V Characteristics : Forward, Reverse, and reverse Breakdown know various Diode Models and apply them to circuit analysis Know to analyze and design simple diode circuits: rectifier, regulator, filtered rectifier, . . 1 4 Various Semiconductor Products 2
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9/28/2010 2 Diode I-V Characteristics 3 Diode I-V Summary 4
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9/28/2010 3 1. Ideal Diodes Circuit Symbol: forward reverse ( short circuit) I-V Characteristics: open short V < 0 V > 0 I-V : Model: Application Example: 5 6 Application: (1) rectifier using ideal diode For V D > 0 (forward) input waveform output waveform For V D < 0 (reverse)
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9/28/2010 4 Ex.3.3) V P =10V R=1 k W Peak diode current 7 = 3.2V 3.2V 10 V Application: (1) rectifier - Ex.3.3 DC component of v O q =0 q = p q =2 p 2. Diode I-V Characteristics (terminal char.) -- REAL DIODE ) 1 ( / T nV v s e I i i n =ideality factor=between 1 and 2 V T =thermal voltage=kT/q I S =saturation current parameter Forward and Reverse, but not Breakdown 8
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9/28/2010 5 2. Diode I-V Characteristics (terminal char.) -- REAL DIODE Forward Expand x-axis i 2.1 Forward Bias Region i ≈ 0 for v < 0.5V 0.5 V = Cut-in voltage Diode is fully on at v = 0.7 V = Turn-On voltage Cut-in voltage Turn-On voltage 9 2. Diode I-V Characteristics (terminal char.) -- REAL DIODE Expand y-axis i 2.2 Reverse-Bias Region : v < 0V I = I s [exp(v/nV T ) – 1] ≈ - I s I S = Reverse Saturation Current ~ typically 10 pA/cm 2 Real device: reverse current >> I S I S 10
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9/28/2010 6 2.1 Forward Bias Region ) 1 ( / T nV v s e I i Where, V T = kT/q =25 mV at T=300K, k=Boltzmann const.=1.38x10 -23 joules/K, and n = ideality factor = between 1 and 2 1= most ideal, IC device; 2= least ideal, many traps, discrete device Ex) For nearly all forward bias regions, i >> I s i = I S exp(v/nV T ) [ i vs. v ] v = nV T ln(i/I s ) = 2.3 nV T log(i/I s ) [ v vs. i ] Ex) Q: you measured the diode voltages at two different current values how are they related? Ans: v 2 -v 1 = 2.3 nV T log(i 2 /I s ) - 2.3 nV T log(i 1 /I s ) = 2.3 nV T log(i 2 /i 1 ) Ex.3.6) For a silicon diode with n=1.5, find the change in voltage for a current change from 0.1 mA to 10 mA Answer) i 2 /i 1 = 10 mA/0.1 mA = 100 v 2 -v 1 = 2.3 n V T log(100) = 2.3*1.5*25 mV*2 = 172.5mV v 2 -v 1 = 2.3 nV T log(i 2 /i 1 ) 11 Modeling the Forward Characteristics 12
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9/28/2010 7 3. Forward Characteristics – Diode Models Piecewise Linear Model Const. Voltage Drop Model Small-signal model
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This note was uploaded on 10/31/2010 for the course EE 423 taught by Professor Mitin during the Spring '10 term at SUNY Buffalo.

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Chap.4-handout - 9/28/2010 4 Read sections 4.1 thru 4.5...

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