Diode Solutions
Tutorial 2 – Sept. 20, 21, and 25, 2006 The following pertain to the 8 questions on diodes. 1.
4 equations, 4 unknowns (I, V1, V2, m)
V1 + V2 = 1.657V VOUT − I (1.67 k ) = 5 − (1.67 k ) I = 1.657V
I = ISe
V1 nVT
V2
I = I S e nVT m Eqn. 2 yields I = 2mA directly multiply
m=
x
I2 : = I S2 e m
I2 1.657 2 1.25( 25 m )
V1 +V2 nVT
, where V1+ V2 is known from
…= 37.56 (pick a real integer : 37 or 38)
ISe
now solve for V1, V2 with eqns 3 and 4 V1 = 0.885V, V2 = 0.772V
2. we must make assumptions and test them is D3 off? Try it! get
8 − Vout (Vout − 0.7) + 8 = 8k 1k Vout = −5.6V , so node B = -6.3V
D2 OFF would be obviously inconsistent D2 ON puts node A at ground – not good for D3 OFF! Therefore, D3 is ON node A is at Vout Try D2 OFF:
(Vout − 0.7) + 8 8 − Vout 8 − Vout 8 − Vout = + = 3× 1k 4k 8k 8k is D2 OFF? YES – solution is consistent
3. a)
Vout = −3.127V
b) By symmetry, ID1,2,3,4 = Iref / 2 and rd = (2nVT)/ Iref c)
2rd||2rd = rd vOUT/vS = RL/( RS + rd + RL )
4. a) D2, D3 OFF by inspection D1 and D4 cannot both be ON D5 is most likely ON
VO = 1.3V
is D1 ON?
Yes: I300 = (4 – (1.3-0.7))/300 = 11.3mA
not good, D4 is off by
assumption! No: D4 must be ON Check D1: 1.3V b) I = 1.3V/400
= 3.25mA
I300 = (4 – 2)/(300 + 700) = 2mA 3.4V = OFF as assumed 4V - 300Ohm * 2ma
c) I300 = 2mA
5. a) D1, D5 OFF by inspection ID2 = I1 , ID3 = I2 , ID4 = I1 + I2
these 3 diodes are ON
b)
rd2 = nVT/ I1 = 100 ,
rd4 = nVT/(I1 + I2)= 25
c) vout/vin = (R1||rd4)/ (R1||rd4 + rd2) = 0.196V/V
6. a) ID3 = I If D7 is ON then D8 is off (has 0.2V reverse bias) D8 ON is impossible… turns off D7 D7 ON requires D4, D5 and D6 ON too D1 and D2 are OFF b)
c) vout/vin = 0.294V/V by analysis
7. slope is ½ by voltage divider
no current flow in D8! ID4,5,6,7 = I/2
limited to -3V when D2 is ON, to 3V when D1 is ON, and gain of ½ when both are OFF
8. DC analysis: IR1 = 5V/R1 IR2 = 4.3V/R2 rd = nVTR2/4.3 Rin = R1|| R2(1+ nVT/4.3)
4 equations, 4 unknowns (I, V1, V2, m)
V1 + V2 = 1.657V VOUT − I (1.67 k ) = 5 − (1.67 k ) I = 1.657V
I = ISe
V1 nVT
V2
I = I S e nVT m Eqn. 2 yields I = 2mA directly multiply
m=
x
I2 : = I S2 e m
I2 1.657 2 1.25( 25 m )
V1 +V2 nVT
, where V1+ V2 is known from
…= 37.56 (pick a real integer : 37 or 38)
ISe
now solve for V1, V2 with eqns 3 and 4 V1 = 0.885V, V2 = 0.772V
2. we must make assumptions and test them is D3 off? Try it! get
8 − Vout (Vout − 0.7) + 8 = 8k 1k Vout = −5.6V , so node B = -6.3V
D2 OFF would be obviously inconsistent D2 ON puts node A at ground – not good for D3 OFF! Therefore, D3 is ON node A is at Vout Try D2 OFF:
(Vout − 0.7) + 8 8 − Vout 8 − Vout 8 − Vout = + = 3× 1k 4k 8k 8k is D2 OFF? YES – solution is consistent
3. a)
Vout = −3.127V
b) By symmetry, ID1,2,3,4 = Iref / 2 and rd = (2nVT)/ Iref c)
2rd||2rd = rd vOUT/vS = RL/( RS + rd + RL )
4. a) D2, D3 OFF by inspection D1 and D4 cannot both be ON D5 is most likely ON
VO = 1.3V
is D1 ON?
Yes: I300 = (4 – (1.3-0.7))/300 = 11.3mA
not good, D4 is off by
assumption! No: D4 must be ON Check D1: 1.3V b) I = 1.3V/400
= 3.25mA
I300 = (4 – 2)/(300 + 700) = 2mA 3.4V = OFF as assumed 4V - 300Ohm * 2ma
c) I300 = 2mA
5. a) D1, D5 OFF by inspection ID2 = I1 , ID3 = I2 , ID4 = I1 + I2
these 3 diodes are ON
b)
rd2 = nVT/ I1 = 100 ,
rd4 = nVT/(I1 + I2)= 25
c) vout/vin = (R1||rd4)/ (R1||rd4 + rd2) = 0.196V/V
6. a) ID3 = I If D7 is ON then D8 is off (has 0.2V reverse bias) D8 ON is impossible… turns off D7 D7 ON requires D4, D5 and D6 ON too D1 and D2 are OFF b)
c) vout/vin = 0.294V/V by analysis
7. slope is ½ by voltage divider
no current flow in D8! ID4,5,6,7 = I/2
limited to -3V when D2 is ON, to 3V when D1 is ON, and gain of ½ when both are OFF
8. DC analysis: IR1 = 5V/R1 IR2 = 4.3V/R2 rd = nVTR2/4.3 Rin = R1|| R2(1+ nVT/4.3)
