Diode circuits-analysis

EE-203 Diode Circuits Analysis

1

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:

4D D D D

D D 4

5 10 | V 0 I 0.500 | I 0 V 54.5Forward biased - V 0.5 I 0.450 10

D DI V mA VVV mA

= + = = = =

= = =Ω


2

1 2 3 4

1 mA

2 mA

Q-point

i D

vD

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)


3

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

0.4 V

+-

+-

VI

2.2 k Ω

(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

yields 0.4 0.182 2.2

VI mAk

= =Ω

. This current is greater than zero, which is consistent

with the diode being "on". Thus the Q-pt is (0 V, +0.182 mA).

+-

+-+-

VI 1 k Ω1.2 k Ω

2 V1.6 V


4

Ideal Diode:

0.4 V

+-

+-

V

I

2.2 k Ω

CVD:

0.4 V+-

0.6 VI

2.2 k Ω

+-on

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

0.4 0.6 90.9 2.2V VI A

kµ−

= = −Ω

. This current is negative which is not consistent with

the assumption that the diode is "on". Thus the diode must be off. The resulting Q-pt is: (0.4 V, 0 mA).

0.4 V+-

I=0 2.2 k Ω- +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

15 0.410 exp 1 8.890.025D

Vi A nAV

− ⎡ ⎤⎛ ⎞= − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦, a very small current.


5

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)

( )

( )( )

D

5 5( ) Diode is forward biased: = 5+0= 5 | I= 0.500

20( ) Diode is reverse biased: =0 | V=7 20 7 | V 10

3 7( ) Diode is forward biased: =3 0=3 | I= 0.500

20( ) Diode is reverse

a V V mAk

b I k I V V

c V V mAk

d

− −− − =

Ω− Ω = = −

− −− =

Ω( ) Dbiased: =0 | V= 5 20 5 | V 10 I k I V V− + Ω = − = −

(b) ( )

( )( )

D

5 4.3( ) Diode is forward biased: = 5+0.7= 4.3 | I= 0.465


2.3 7( ) Diode is forward biased: =3 0.7=2.3 | I= 0.465

20( ) Diode

a V V mAk

b I k I V V

c V V mAk

d

− −− − =

Ω− Ω = = −

− −− =

Ω( ) D is reverse biased: =0 | V= 5 20 5 | V 10 I k I V V− + Ω = − = −


6

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)

( )

( )( )

D

5 5( ) Diode is forward biased: = 5+0= 5 | I= 100

100( ) Diode is reverse biased: =0 A | V=7 100 7 | V 10

3 7( ) Diode is forward biased: =3 0=3 | I= 100

100( ) Diode is reverse

a V V Ak

b I k I V V

c V V Ak

d

µ

µ

− −− − =

Ω− Ω = = −

− −− =

Ω( ) D biased: =0 A | V= 5 100 5 | V 10 I k I V V− + Ω = − = −

(b) ( )

( )( )

D

5 4.4( ) Diode is forward biased: = 5+0.6= 4.4 | I= 94.0


2.4 7( ) Diode is forward biased: =3 0.6=2.4 | I= 94.0

100( ) Diod

a V V Ak

b I k I V V

c V V Ak

d

µ

µ

− −− − =

Ω− Ω = = −

− −− =

Ω( ) De is reverse biased: =0 | V= 5 20 5 | V 10 I k I V V− + Ω = − = −


7

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

( ) ( )( ) ( ) ( )

1 2 3 D2 D1

D3 D3 D2 1

1 2 3

10 0( ) on, D off, D on: I 0 | I 13 7

0 5I 1.00 I 1.00 | V 5 10 3000 2

2.5D : 1.00 mA, 0 V D : 0 mA, 2 V D : 1.00 mA, 0 V

D

a D mAk k

mA mA I Vk

−= = =

Ω + Ω− −

+ = → = = − − = −Ω

−

( ) ( )

( )( ) ( ) ( )

1 2 3 D2 D3

D1 D2 1

D3 1

1 2 3

( ) on, D off, D off: I 0 | I 010 5

I 0.500 | V 5 10 8000 1.008 10 12

V 5 12000 1.00

: 0.500 , 0 : 0 , 1.00 : 0 , 1.00

D

D

b D

mA I Vk k k

I V

D mA V D A V D A V

= =

− −= = = − − = −

Ω + Ω + Ω= − − + = −

− −


8

( ) ( )

( )

( ) ( ) ( )

1 2 3

D1 10K 2 1 10

12K 3 12 10

1 2 3

( ) on, D on, D on0 10 0 2

I 1.25 0 | I 0.200 | 1.05 08 10

2 5I 0.583 | 0.783 0

12: 1.25 , 0 : 1.05 m , 0 : 0.783 m , 0

D D K

D K K

c D

mA mA I I I mAk k

mA I I I mAk

D mA V D A V D A V

− − −= = > = = − = + = >

Ω Ω− −

= = = − = >Ω

( ) ( )( )

( ) ( ) ( )

1 2 3 1 2

D3 D1 3

D2 3

1 2 3

( ) , , : 0, 012 5

I 567 0 | V 0 5 10000 0.667 030

V 5 12 10000 1.33 0

: 0 , 0.667 : 0 , 1.33 : 567 , 0

D D

D

D

d D off D off D on I I

V A I Vk

I V

D A V D A V D A V

µ

µ

= =

− −= = > = − − + = − <

Ω= − − = − <

− −

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 ( )

( ) ( )( ) ( ) ( )

1 2 3 D2 D1

D3 D3 D2 1

1 2 3

10 0.6 0.6( ) on, D off, D on: I 0 | I 1.00

3 70.6 5

I 1.00 I 0.760 | V 5 10 0.6 3000 1.402.5

D : 1.00 mA, 0.600 V D : 0 mA, 1.40 V D : 0.760 mA, 0.600V

D

a D mAk k

mA mA I Vk

− − −= = =

Ω + Ω− − −

+ = → = = − − − = −Ω

−

( ) ( )

( )( ) ( ) ( )

1 2 3 D2 D3

D1 D2 1

D3 1

1 2 3

( ) on, D off, D off: I 0 | I 010 0.6 5

I 0.480 | V 5 10 0.6 8000 0.5608 10 12

V 5 12000 0.760

D : 0.480 mA, 0.600 V D : 0 A, 0.560 V D : 0 A, 0.760 V

D

D

b D

mA I Vk k k

I V

= =

− − −= = = − − − = −

Ω + Ω + Ω= − − + = −

− −


9

( ) ( )

( )

( )

1 2 3

1 10

2 1 10 12 3 12 10

1 2

( ) on, D on, D on0.6 9.4 0.6 1.4

1.10 0 | 0.2008 10

1.4 50.900 0 | 0.533 | 0.733 0

12D : 1.10 mA, 0.600 V D : 0.900 mA, 0.60

D K

D D K K D K K

c D

V VI mA I mAk k

VI I I mA I mA I I I mAk

− − − − −= = > = = −

Ω Ω− −

= + = > = = = − = >Ω

( ) ( )30 V D : 0.733 mA, 0.600 V

( ) ( )( )

( ) ( ) ( )

1 2 3 1 2

D3 D1 3

D2 3

1 2 3

( ) , , : 0, 011.4 5

I 547 0 | V 0 5 10000 0.467 030

V 5 11.4 10000 0.933 0

: 0 , 0.467 : 0 , 0.933 : 547 , 0

D D

D

D

d D off D off D on I I

V A I Vk

I V

D A V D A V D A V

µ

µ

= =

− −= = > = − − + = − <

Ω= − − = − <

− −

Diode circuits-analysis

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