Q3 VB Find Figure 3.18 (a) Circuit symbol of the pr junction diode; (b) voltage -current character- istic of a silicon R₁ 33052 diode with I, = 10-12 A and n = 1. For negative values of up, the diode current is too small to plot precisely on the graph. 3R₂ 27052 a) iD for VB = 2 V b) is for Up NB = 6 V is and no it -2.0 + diode is reversed. (a) ip=-1₁ VD J -1.0 Reverse-biased region ip (mA) 20 15 10 5 Given the diode has the characteristic given in the vi plot 12 Is = 10-1¹²A Use load-line technique VB = 6V and the I Forward-biased region A 1.0 Turn-on voltage 2.0 UD (V)

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Q3
VB
R₁
33052
3R₂
27052
Figure 3.18
(a) Circuit symbol
of the pr junction
diode; (b) voltage
-current character-
istic of a silicon
diode with I, =
10-12 A and n = 1.
For negative values
of up, the diode
current is too small
to plot precisely on
the graph.
b) is for
LD
Find a) is for VB = 2 V
+
UD
-2.0
c) is and VD if
diode is reversed.
XVD
VB = 6 V
(a)
iD
ip=-1₁
-1.0
Reverse-biased region
} Use load-line technique
VB = 6V and the
ip (mA)
20
15
10
Given the diode
has the characteristic
given in the vi plot.
Is = 10-1²2 A
5
Forward-biased
region
1.0
Turn-on voltage
2.0
UD (V)
Transcribed Image Text:Q3 VB R₁ 33052 3R₂ 27052 Figure 3.18 (a) Circuit symbol of the pr junction diode; (b) voltage -current character- istic of a silicon diode with I, = 10-12 A and n = 1. For negative values of up, the diode current is too small to plot precisely on the graph. b) is for LD Find a) is for VB = 2 V + UD -2.0 c) is and VD if diode is reversed. XVD VB = 6 V (a) iD ip=-1₁ -1.0 Reverse-biased region } Use load-line technique VB = 6V and the ip (mA) 20 15 10 Given the diode has the characteristic given in the vi plot. Is = 10-1²2 A 5 Forward-biased region 1.0 Turn-on voltage 2.0 UD (V)
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