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- (a) I (^) D = I (^) R = 30 V^ 0.7 V 1.5 k
E V D
R
= 19.53 mA VD = 0.7 V , VR = E VD = 30 V 0.7 V = 29.3 V
(b) I (^) D = 30 V^ 0 V 1.5 k
E V D
R
(^) = 20 mA
VD = 0 V , VR = 30 V
Yes, since E VT the levels of I (^) D and VR are quite close.
- (a) I = 0 mA ; diode reverse-biased.
(b) V 20 = 20 V 0.7 V = 19.3 V (Kirchhoff’s voltage law) I (20 Ω) = 19.3 V 20
= 0.965 A
V (10 Ω) = 20 V 0.7 V = 19.3 V
I (10 Ω) = 19.3 V
= 1.93 A
I = I (10 Ω) + I (20 Ω)
= 2.895 A
(c) I = 10 V 10
= 1 A ; center branch open
- (a) Diode forward-biased, Kirchhoff’s voltage law (CW): 5 V + 0.7 V Vo = 0 Vo = 4.3 V I (^) R = I (^) D = 4.3 V 2.2 k
V o R
= 1.955 mA
(b) Diode forward-biased, I (^) D = 8 V + 6 V^ 0.7 V 1.2 k 4.7 k
= 2.25 mA Vo = 8 V (2.25 mA)(1.2 kΩ) = 5.3 V
- (a) Vo = 10 k (12 V 0.7 V 0.3 V) 2 k 10 k
= 9.17 V
(b) Vo = 10 V
- (a) Determine the Thevenin equivalent circuit for the 10 mA source and 2.2 k resistor.
ETh = IR = (10 mA)(2.2 k) = 22 V RTh = 2. 2k
Diode forward-biased I (^) D = 22 V^ 0.7 V 2.2 k 2.2 k
= 4.84 mA Vo = I (^) D (1.2 k) = (4.84 mA)(1.2 k) = 5.81 V (b) Diode forward-biased I (^) D = 20 V + 20 V^ 0.7 V 6.8 k
(^) = 5.78 mA
Kirchhoff’s voltage law (CW):
- Vo 0.7 V + 20 V = 0 Vo = 19.3 V
- (a) Vo (^) 1 = 12 V – 0.7 V = 11.3 V
Vo (^) 2 = 1.2 V (b) Vo (^) 1 = 0 V Vo (^) 2 = 0 V
- (a) Both diodes forward-biased Si diode turns on first and locks in 0.7 V drop. 12 V 0.7 V R 4.7 k
I
= 2.4 mA I (^) D = I (^) R = 2.4 mA Vo = 12 V 0.7 V = 11.3 V
(b) Right diode forward-biased: I (^) D = 20 V + 4 V^ 0.7 V 2.2 k
(^) = 10.59 mA
Vo = 20 V 0.7 V = 19.3 V
- (a) Si diode “on” preventing GaAs diode from turning “on”:
I = 1 V^ 0.7 V^ 0.3 V 1 k 1 k
(^) = 0.3 mA
Vo = 1 V 0.7 V = 0.3 V
(b) I = 16 V^ 0.7 V^ 0.7 V + 4 V^ 18.6 V 4.7 k 4.7 k
(^) = 3.96 mA
Vo = 16 V 0.7 V 0.7 V = 14.6 V
- The Si diode requires more terminal voltage than the Ge diode to turn “on”. Therefore, with 5 V at both input terminals, assume Si diode “off” and Ge diode “on”. The result: Vo = 5 V 0.3 V = 4.7 V The result supports the above assumptions.
- V dc = 0.318 Vm Vm = dc^ 2 V 0.318 0.
V = 6.28 V
I (^) m = 6.28 V 2 k
V m R
= 3.14 mA
- Using V dc 0.318( Vm VT ) 2 V = 0.318( Vm 0.7 V) Solving: Vm = 6.98 V 10:1 for Vm : VT
- Vm = dc^ 2 V 0.318 0.
V = 6.28 V
I (^) L max =
6.28 V
10 k = 0.628 mA
