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- 1. EE-203 Diode Circuits AnalysisProblem 1:Plot the load line and find the Q-point for the diode circuit in Figure 1 if V = 5 V and R =10 kΩ. Use the i-v characteristic in Figure 2. Figure 1 Figure 2Solution:5 = 104 I D + VD | VD = 0 I D = 0.500mA | I D = 0 VD = 5V 4.5VForward biased - VD = 0.5V ID = = 0.450 mA 104 Ω 1
- 2. EE-203 Diode Circuits Analysis iD 2 mA 1 mA Q-point vD 1 2 3 4 5Problem 2:Find the Q-point for the circuit in Figure 3 using the ideal diode model and constantvoltage drop model with Von=0.6V. Figure 3Solution : Ideal diode model: ID = 1V/10kΩ = 100 µA; (100 µA, 0 V) Constant voltage drop model: ID = (1-0.6)V/10kΩ = 40.0 µA; (40.0 µA, 0.6 V) 2
- 3. EE-203 Diode Circuits AnalysisProblem 3:Find the Q-point for the diode in Figure 4 using (a) the ideal diode model and (b) theconstant voltage drop model with Von = 0.6 V. (c) Discuss the results. Which answer doyou feel is most correct? Figure 4Solution :Using Thévenin equivalent circuits yields and then combining the sources I I 1.2 k Ω 1k Ω 2.2 k Ω - V + - V + + + + 1.6 V 2V 0.4 V - - - (a) Ideal diode model: The 0.4 V source appears to be forward biasing the diode so we will assume it is "on". Substituting the ideal diode model for the forward region 0.4V yields I = = 0.182 mA . This current is greater than zero, which is consistent 2.2k Ω with the diode being "on". Thus the Q-pt is (0 V, +0.182 mA). 3
- 4. EE-203 Diode Circuits Analysis I - V + 2.2 k Ω + 0.4 V - Ideal Diode: CVD: V on - + I 0.6 V 2.2 k Ω + 0.4 V -(b) CVD model: The 0.4 V source appears to be forward biasing the diode so we willassume it is "on". Substituting the CVD model with Von = 0.6 V yields 0.4V − 0.6V I= = −90.9 µ A . This current is negative which is not consistent with 2.2k Ωthe assumption that the diode is "on". Thus the diode must be off. The resulting Q-ptis: (0.4 V, 0 mA). - V + I=0 2.2 k Ω + 0.4 V -(c) The second estimate is more realistic. 0.4 V is not sufficient to forward bias thediode into significant conduction. For example, let us assume that IS = 10-15 A andassume that the full 0.4 V appears across the diode. Then ⎡ ⎛ 0.4V ⎞ ⎤ iD = 10−15 A ⎢exp ⎜ − 1 = 8.89 nA , a very small current. ⎣ ⎝ 0.025V ⎟ ⎥ ⎠ ⎦ 4
- 5. EE-203 Diode Circuits AnalysisProblem 4:(a) Find I and V in the four circuits in Figure 5 using the ideal diode model. (b) Repeatusing the constant voltage drop model with Von = 0.7 V. Figure 5 (a) 5 − ( −5) (a ) Diode is forward biased:V = − 5+0= − 5 V | I= = 0.500 mA 20k Ω (b) Diode is reverse biased: I =0 | V=7 − 20k Ω ( I ) = 7 V | VD = −10 V 3 − ( −7) (c) Diode is forward biased:V =3 − 0=3 V | I= = 0.500 mA 20k Ω (d ) Diode is reverse biased: I =0 | V= − 5 + 20k Ω ( I ) = −5 V | VD = −10 V (b) 5 − ( −4.3) (a ) Diode is forward biased:V = − 5+0.7= − 4.3 V | I= = 0.465 mA 20k Ω (b) Diode is reverse biased: I =0 | V=7 − 20k Ω ( I ) = 7 V | VD = −10 V 2.3 − ( −7) (c) Diode is forward biased:V =3 − 0.7=2.3 V | I= = 0.465 mA 20k Ω (d ) Diode is reverse biased: I =0 | V= − 5 + 20k Ω ( I ) = −5 V | VD = −10 V 5
- 6. EE-203 Diode Circuits AnalysisProblem 5:(a) Find I and V in the four circuits in Figure 5 using the ideal diode model if the resistorvalues are changed to 100 kΩ. (b) Repeat using the constant voltage drop model withVon = 0.6 V.Solution :(a) 5 − ( −5) (a ) Diode is forward biased:V = − 5+0= − 5 V | I= = 100 µ A 100k Ω (b) Diode is reverse biased: I =0 A | V=7 − 100k Ω ( I ) = 7 V | VD = −10 V 3 − ( −7 ) (c) Diode is forward biased:V =3 − 0=3 V | I= = 100 µ A 100k Ω (d ) Diode is reverse biased: I =0 A | V= − 5 + 100k Ω ( I ) = −5 V | VD = −10 V (b) 5 − ( −4.4) (a ) Diode is forward biased:V = − 5+0.6= − 4.4 V | I= = 94.0 µ A 100k Ω (b) Diode is reverse biased: I =0 | V=7 − 100k Ω ( I ) = 7 V | VD = −10 V 2.4 − ( −7) (c) Diode is forward biased:V =3 − 0.6=2.4 V | I= = 94.0 µ A 100k Ω (d ) Diode is reverse biased: I =0 | V= − 5 + 20k Ω ( I ) = −5 V | VD = −10 V 6
- 7. EE-203 Diode Circuits AnalysisProblem 6:Find the Q-points for the diodes in the circuits in Figure 6 using the ideal diode model. Figure 6Solution : Diodes are labeled from left to right 10 − 0 (a ) D1 on, D 2 off, D3 on: ID2 = 0 | ID1 = = 1mA 3k Ω + 7 k Ω 0 − ( −5) I D3 + 1.00mA = → I D3 = 1.00mA | VD2 = 5 − (10 − 3000 I D1 ) = −2V 2.5k Ω D1:(1.00 mA, 0 V ) D 2 :( 0 mA, − 2 V ) D3:(1.00 mA, 0 V ) (b) D1 on, D 2 off, D3 off: I D2 = 0 | I D3 = 0 10 − ( −5) I D1 = = 0.500mA | VD2 = 5 − (10 − 8000 I D1 ) = −1.00V 8k Ω + 10k Ω + 12k Ω VD3 = − ( −5 + 12000 I D1 ) = −1.00V D1 : ( 0.500 mA, 0 V ) D2 : ( 0 A, − 1.00 V ) D3 : ( 0 A, − 1.00 V ) 7
- 8. EE-203 Diode Circuits Analysis (c) D1 on, D 2 on, D3 on 0 − ( −10) 0 − ( 2) I D1 = = 1.25mA > 0 | I10K = = −0.200mA | I D 2 = I D1 + I10 K = 1.05mA > 0 8k Ω 10k Ω 2 − ( −5) I12K = = 0.583mA | I D 3 = I12 K − I10 K = 0.783mA > 0 12k Ω D1 : (1.25 mA, 0 V ) D2 : (1.05 mA, 0 V ) D3 : ( 0.783 mA, 0 V ) (d ) D1 off , D2 off , D3 on: I D1 = 0, I D 2 = 0 12 − ( −5) V I D3 = = 567µ A > 0 | VD1 = 0 − ( −5 + 10000 I D 3 ) = −0.667V < 0 30 kΩ VD2 = 5 − (12 − 10000 I D 3 ) = −1.33V < 0 D1 : ( 0 A, − 0.667 V ) D2 : ( 0 A, − 1.33 V ) D3 : ( 567 µ A, 0 V )Problem 7:Find the Q-points for the diodes in the circuits in Figure 6 using the constant voltage dropmodel with Von = 0.6 V.Solution: Diodes are labeled from left to right 10 − 0.6 − ( −0.6) (a ) D1 on, D 2 off, D3 on: ID2 = 0 | ID1 = = 1.00mA 3k Ω + 7 k Ω −0.6 − ( −5) I D3 + 1.00mA = → I D3 = 0.760mA | VD2 = 5 − (10 − 0.6 − 3000 I D1 ) = −1.40V 2.5k Ω D1:(1.00 mA, 0.600 V ) D 2 :( 0 mA, − 1.40 V ) D3 :( 0.760 mA, 0.600V ) (b) D1 on, D 2 off, D3 off: I D2 = 0 | I D3 = 0 10 − 0.6 − ( −5) I D1 = = 0.480mA | VD2 = 5 − (10 − 0.6 − 8000 I D1 ) = −0.560V 8k Ω + 10k Ω + 12k Ω VD3 = − ( −5 + 12000 I D1 ) = −0.760V D1:( 0.480 mA, 0.600 V ) D 2 :( 0 A, − 0.560 V ) D3:( 0 A, − 0.760 V ) 8
- 9. EE-203 Diode Circuits Analysis(c) D1 on, D 2 on, D3 on −0.6 − ( −9.4) V −0.6 − (1.4) VI D1 = = 1.10mA > 0 | I10 K = = −0.200mA 8 kΩ 10 kΩ 1.4 − ( −5) VI D 2 = I D1 + I10 K = 0.900mA > 0 | I12 K = = 0.533mA | I D 3 = I12 K − I10 K = 0.733mA > 0 12 kΩD1:(1.10 mA, 0.600 V ) D 2 :( 0.900 mA, 0.600 V ) D3 :( 0.733 mA, 0.600 V)(d ) D1 off , D2 off , D3 on: I D1 = 0, I D 2 = 0 11.4 − ( −5) VI D3 = = 547 µ A > 0 | VD1 = 0 − ( −5 + 10000 I D 3 ) = −0.467V < 0 30 kΩVD2 = 5 − (11.4 − 10000 I D 3 ) = −0.933V < 0D1 : ( 0 A, − 0.467 V ) D2 : ( 0 A, − 0.933 V ) D3 : ( 547 µ A, 0 V ) 9

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