EE-203
Diode Circuits Analysis
Problem 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 2
Solution:
5 = 104 I D + VD | VD = 0 I D = 0.500mA | I D = 0 VD = 5V
Forward biased - VD = 0.5V ID =
4.5V
= 0.450 mA
104 Ω
1
EE-203
Diode Circuits Analysis
2 mA
iD
1 mA
Q-point
vD
1
2
3
4
5
Problem 2:
Find the Q-point for the circuit in Figure 3 using the ideal diode model and constant
voltage drop model with Von=0.6V.
Figure 3
Solution :
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)
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EE-203
Diode Circuits Analysis
Problem 3:
Find the Q-point for the diode in Figure 4 using (a) the ideal diode model and (b) the
constant voltage drop model with Von = 0.6 V. (c) Discuss the results. Which answer do
you feel is most correct?
Figure 4
Solution :
Using Thévenin equivalent circuits yields and then combining the sources
1.2 k Ω
1.6 V
I
-
V
+
I
1k Ω
-
+
+
-
-
2V
V
+
2.2 k Ω
+
-
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
= 0.182 mA . This current is greater than zero, which is consistent
yields I =
2.2k Ω
with the diode being "on". Thus the Q-pt is (0 V, +0.182 mA).
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EE-203
Diode Circuits Analysis
I
V
-
2.2 k Ω
+
+
-
Ideal Diode:
0.4 V
CVD:
V
on
- +
I
2.2 k Ω
0.6 V
+
-
0.4 V
(b) CVD model: The 0.4 V source appears to be forward biasing the diode so we will
assume it is "on". Substituting the CVD model with Von = 0.6 V yields
0.4V − 0.6V
= −90.9 µ A . This current is negative which is not consistent with
I=
2.2k Ω
the assumption that the diode is "on". Thus the diode must be off. The resulting Q-pt
is: (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 the
diode into significant conduction. For example, let us assume that IS = 10-15 A and
assume 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 ⎟⎠ ⎥⎦
⎣
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EE-203
Diode Circuits Analysis
Problem 4:
(a) Find I and V in the four circuits in Figure 5 using the ideal diode model. (b) Repeat
using the constant voltage drop model with Von = 0.7 V.
Figure 5
(a)
5 − ( −5)
= 0.500 mA
20k Ω
(b) Diode is reverse biased: I =0 | V=7 − 20k Ω ( I ) = 7 V | VD = −10 V
(a ) Diode is forward biased:V = − 5+0= − 5 V | I=
3 − ( −7)
= 0.500 mA
20k Ω
(d ) Diode is reverse biased: I =0 | V= − 5 + 20k Ω ( I ) = −5 V | VD = −10 V
(c) Diode is forward biased:V =3 − 0=3 V | I=
(b)
5 − ( −4.3)
= 0.465 mA
20k Ω
(b) Diode is reverse biased: I =0 | V=7 − 20k Ω ( I ) = 7 V | VD = −10 V
(a ) Diode is forward biased:V = − 5+0.7= − 4.3 V | I=
2.3 − ( −7)
= 0.465 mA
20k Ω
(d ) Diode is reverse biased: I =0 | V= − 5 + 20k Ω ( I ) = −5 V | VD = −10 V
(c) Diode is forward biased:V =3 − 0.7=2.3 V | I=
5
EE-203
Diode Circuits Analysis
Problem 5:
(a) Find I and V in the four circuits in Figure 5 using the ideal diode model if the resistor
values are changed to 100 kΩ. (b) Repeat using the constant voltage drop model with
Von = 0.6 V.
Solution :
(a)
5 − ( −5)
= 100 µ A
100k Ω
(b) Diode is reverse biased: I =0 A | V=7 − 100k Ω ( I ) = 7 V | VD = −10 V
(a ) Diode is forward biased:V = − 5+0= − 5 V | I=
3 − ( −7 )
= 100 µ A
100k Ω
(d ) Diode is reverse biased: I =0 A | V= − 5 + 100k Ω ( I ) = −5 V | VD = −10 V
(c) Diode is forward biased:V =3 − 0=3 V | I=
(b)
5 − ( −4.4)
= 94.0 µ A
100k Ω
(b) Diode is reverse biased: I =0 | V=7 − 100k Ω ( I ) = 7 V | VD = −10 V
(a ) Diode is forward biased:V = − 5+0.6= − 4.4 V | I=
2.4 − ( −7)
= 94.0 µ A
100k Ω
(d ) Diode is reverse biased: I =0 | V= − 5 + 20k Ω ( I ) = −5 V | VD = −10 V
(c) Diode is forward biased:V =3 − 0.6=2.4 V | I=
6
EE-203
Diode Circuits Analysis
Problem 6:
Find the Q-points for the diodes in the circuits in Figure 6 using the ideal diode model.
Figure 6
Solution :
Diodes are labeled from left to right
(a ) D1 on, D 2 off, D3 on: ID2 = 0 | ID1 =
10 − 0
= 1mA
3k Ω + 7 k Ω
0 − ( −5)
→ 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 )
I D3 + 1.00mA =
(b) D1 on, D 2 off, D3 off: I D2 = 0 | I D3 = 0
10 − ( −5)
= 0.500mA | VD2 = 5 − (10 − 8000 I D1 ) = −1.00V
8k Ω + 10k Ω + 12k Ω
VD3 = − ( −5 + 12000 I D1 ) = −1.00V
I D1 =
D1 : ( 0.500 mA, 0 V ) D2 : ( 0 A, − 1.00 V ) D3 : ( 0 A, − 1.00 V )
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EE-203
Diode Circuits Analysis
(c) D1 on, D 2 on, D3 on
0 − ( −10)
0 − ( 2)
= 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 )
I D1 =
(d ) D1 off , D2 off , D3 on: I D1 = 0, I D 2 = 0
12 − ( −5) V
= 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
I D3 =
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 drop
model with Von = 0.6 V.
Solution:
Diodes are labeled from left to right
(a ) D1 on, D 2 off, D3 on: ID2 = 0 | ID1 =
10 − 0.6 − ( −0.6)
= 1.00mA
3k Ω + 7 k Ω
−0.6 − ( −5)
→ 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 )
I D3 + 1.00mA =
(b) D1 on, D 2 off, D3 off: I D2 = 0 | I D3 = 0
10 − 0.6 − ( −5)
= 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
I D1 =
D1:( 0.480 mA, 0.600 V ) D 2 :( 0 A, − 0.560 V ) D3:( 0 A, − 0.760 V )
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EE-203
Diode Circuits Analysis
(c) D1 on, D 2 on, D3 on
−0.6 − ( −9.4) V
−0.6 − (1.4) V
= 1.10mA > 0 | I10 K =
= −0.200mA
8
kΩ
10
kΩ
1.4 − ( −5) V
I D 2 = I D1 + I10 K = 0.900mA > 0 | I12 K =
= 0.533mA | I D 3 = I12 K − I10 K = 0.733mA > 0
12
kΩ
D1:(1.10 mA, 0.600 V ) D 2 :( 0.900 mA, 0.600 V ) D3 :( 0.733 mA, 0.600 V)
I D1 =
(d ) D1 off , D2 off , D3 on: I D1 = 0, I D 2 = 0
11.4 − ( −5) V
= 547 µ A > 0 | VD1 = 0 − ( −5 + 10000 I D 3 ) = −0.467V < 0
30
kΩ
VD2 = 5 − (11.4 − 10000 I D 3 ) = −0.933V < 0
I D3 =
D1 : ( 0 A, − 0.467 V ) D2 : ( 0 A, − 0.933 V ) D3 : ( 547 µ A, 0 V )
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