*35. Determine v, for each network of Fig. 179 for the input shown. 3 V I k2 +8 V 2.2 kΩ + Si Si -8 V (a) (b) FIG. 179 Problem 35. 35. (a) Diode “on" for v, 2 4.7 V For v, > 4.7 V, V, = 4 V + 0.7 V = 4.7 V For v, < 4.7 V, diode “off" and v, = v; 4.7 V (b) Again, diode “on" for v; 2 3.7 V but v, now defined as the voltage across the diode For v, 2 3.7 V, v, = 0.7 V -8 V For v, < 3.7 V, diode “off", In = Ig = 0 mA and V22 kg = IR = (0 mA)R = 0 V Therefore, v, = v; – 3 V At v, = 0 V, v, =-3 V V; = -8 V, v, = -8 V – 3 V = -11 V 0.7 V -11 V

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Author:Robert L. Boylestad
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*35. Determine v, for each network of Fig. 179 for the input shown.
3 V
v; o
2.2 k2 +
+
1 k2
+8 V
Si
Si
-8 V
(a)
(b)
FIG. 179
Problem 35.
35.
(a) Diode “on" for v; 2 4.7 V
For v; > 4.7 V, V, = 4 V + 0.7 V = 4.7 V
For v; < 4.7 V, diode “off" and v, = v;
4.7 V
(b) Again, diode “on" for v; 2 3.7 V but v,
now defined as the voltage across the diode
For v; 2 3.7 V, v, = 0.7 V
-8 V
For v; < 3.7 V, diode “off", In = IR =0 mA and V221a = IR = (0 mA)R = 0 V
Therefore, v, = v; – 3 V
At v, = 0 V, v, = -3 V
V; = -8 V, v, = -8 V – 3 V = -11 v
0.7 V
-3
-11 V
Transcribed Image Text:*35. Determine v, for each network of Fig. 179 for the input shown. 3 V v; o 2.2 k2 + + 1 k2 +8 V Si Si -8 V (a) (b) FIG. 179 Problem 35. 35. (a) Diode “on" for v; 2 4.7 V For v; > 4.7 V, V, = 4 V + 0.7 V = 4.7 V For v; < 4.7 V, diode “off" and v, = v; 4.7 V (b) Again, diode “on" for v; 2 3.7 V but v, now defined as the voltage across the diode For v; 2 3.7 V, v, = 0.7 V -8 V For v; < 3.7 V, diode “off", In = IR =0 mA and V221a = IR = (0 mA)R = 0 V Therefore, v, = v; – 3 V At v, = 0 V, v, = -3 V V; = -8 V, v, = -8 V – 3 V = -11 v 0.7 V -3 -11 V
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