TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success)
CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES

EQUIVALENT CIRCUIT OF PN-JUNCTION DIODE

Ideal diode : An ideal diode is a two terminal device which has the
following properties :

PERSONAL REMARK :
Refer figure 1 for Q. 1 and
Q. 2. Assume D1 and D2 to
be ideal diodes.
(i) conducts with zero resistance when forward-biased, and
(ii) offers an infinite resistance when reverse-biased.
–2V
The ideal diode acts like an automatic switch. The switch is closed when
the diode is forward bias and is open when it is reverse biased as shown
in figure 1.
(Anode)
+
(Cathode)
–
D1
V0
2V
D2
Closed

I
Forward bias
(Anode)
–
(Cathode)
+
Open

–9V
Reverse bias
Figure 1
Figure 1
Which one of the following
statements is true ?
(a) Both D1 and D2 are ON
(b) Both D1 and D2 are OFF
(c) D1 is ON and D2 is OFF
(d) D2 is ON and D1 is OFF
Sol.(d)Here the diode D 2 is ON
while D1 is OFF. Since first off
all D2 turn ON which makes
V0 = 2 V(because diodes are
ideal) Now, the anode of the
diode D1 is connected to – 2 V
while cathode is connected to
+ 2 V which makes the diode
D1 OFF.
1.

Practical Diode : Practical diode is a two terminal device which has the
following properties:
(i) conducts with low resistance (of the order of few ) when forwardbiased, and
(ii) offers an very high resistance (of the order of M) when reversebiased.
The Equivalent Circuit of Practical Diode Shown in Figure 2.
(Anode)
+
(Cathode)
–

Rf
Forward bias
(Anode)
–
(Cathode)
+
Reverse bias

VB
Rr
Values of V 0 and I,
respectively, are
(a) 2 V and 1.1 mA
(b) 0 V and 0 mA
(c)  2 V and 0.7 mA
(d) 4 V and 1.3 mA
Sol.(a)The given circuit can be
redrawn according to
the question
2.
Figure 2
 Table given below illustrates the comparision between Ge and Si
diode :
Parameter
Ge - diode
Si - diode
Material used
Germanium
Silion
Cut-in-voltage (Vg)
0.2 V
0.6 V
D1
O.C.
– 2V
Knee voltage
0.3 V
0.7 V
Reverse saturation
current
Breakdown voltage
in µA
in nA
Lower
Applications
Low voltage low
2V
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+
–
V0
I
– 9V
+
– 9V
Higher
Rectifiers, clippers,
temperature applications
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D2
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clampers etc.
or I = 1.1 mA and
V0 = 2V
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CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES
PROBLEMS BASED ON GATE/IES/PSUs
PERSONAL REMARK :
1. Determine the current I for each of the configurations of figure 1 using the
approximate equivalent model for the diode.
30 V
100 
Si
–
I
+
Si
–
10 
I
12 V
+
20 

Ex. The values of I & V for the
circuit shown is (Assume
forward drop across each
diode is 0.7 V)
(a) 6 mA, 2.3 V
(b) 0.3 mA, 1.3 V
(c) 1.3 mA, 2.3 V
(d) 2.3 mA, 2.3 V
D3
+3V
(a)
D2
(b)
I
D1
Si
+
10 V
I
1 k
+1V
10 
–
V
+2V
Si
(c)
Figure 1
Sol. (a) I = 0 mA , since diode is reverse-biased
Ans.
(b) V20  = 30 V – 0.7 V = 29.3 V (Kirchhoff's Voltage Law)
29.3 V
= 1.465 A
20
Ans.
10 V
= 1 A , since center branch is open
10 
Ans.
I=
(c) I =
Determine Vo and ID for the networks of figure 2
2.
ID
–5V
+ 12 V
V0
V0
1.2 k
Si
4.7 k
2.2 
ID
Sol.(d) From the given circuit
Three voltage (2.3 V, 1.3 V and
0.3 V) try to appear at the pant
V, but at a node more than one
voltage can appear. Hence one
voltage will appear forcing the
other two not to appear through
diodes D1 and D2.
Hence, V = 2.3 Volts and
I
2.3V
 2.3 mA
1 k
Ex. Obtain the values of V and I
for the circuit shown in figure
(Assume forward drop across
each diode is 0.7 V)
(a) 1.3 mA (b)  1.3 m A
(c) 6.3 mA (d) None of these
+5V
Si
I
(a)
(b)
Figure 2
Sol. (a) Diode forward-biased,
Sol.(c)V =  1.3 V and
0  V0
0  (4.3 V)

 1.95 mA
R
2.2 k
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V
+1V
V0 = – 4.3 V
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2V
Kirchhoff's Voltage Law : – 5 V + 0.7 V – V0 = 0
IR = ID =
1K
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I
5  1.3
 6.3 mA
1
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CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES
(b) Diode forward-biased,
PERSONAL REMARK :
Ex. The value of V & I for the
circuit shown assuming the drop
across the diode of 0.7 V
+5 V
12 V – 0.7 V
= 1.915 mA
.2 k  4.7 k
ID =
V 0 = V4.7k  + VD = (1.915 mA) (4.7 k ) + 0.7 V = 9.7 V
3.
Determine the level of Vo for each network figure 3
+ 20 V
Si
Ge
2 k
+ 10 V
2.5 K
Si
2 k
V0
D1
4.7 k
(a)
(b)
Figure 3
4 k (20 V – 0.7 V – 0.3 V)
2
=
(20 V – 1 V)
4 k  2 k
3
I=
5  0.7 5  0.7  0.7

2.5K
5K
 1.72  1  0.72 mA
Ans.
(10 V  4 V – 0.7 V)
13.3 V

= 1.985 mA
2 k  4.7 k
6.7 k
Applying KVL,
0.7  0.7  V = 0 .Gives,V = 0 V
Hence alternative (b) is the
correct choice.
V' = IR = (1.985 mA) (4.7 k ) = 9.33 V
V0 = V' – 4 V = 9.33 V – 4 V = 5.33 V
4.
5V
I
2
or V0 =
(19 V) = 12.67 V
3
(b)
V
D2
(a) 0 V,  0.72 mA
(b) 0 V, 0.72 mA
(c) 1.4 V, 0.72 mA
(d)  1.4 V, 0.72 mA
Sol. Assume that both the diode are
'ON' .From the given circuit
–4 V
V0 =
I
V0
4 k
Sol. (a)
Ans.
Ex. The values of I and V assuming
drop across the diode of 0.7 V
D1
Determine V0 and ID for the networks of figure 4.
ID
V0
1V
Si
I
10 mA
2.2 k
2.2 k

ID
V0
+ 20 V
D2
–4V
6.8 k
Si
(a)
(a)
(b)
(c)
(d)
(b)
Figure 4
Sol.(a) Applying source transformation for the 10 mA source and 2.2 k
resistor as shown in figure 5.
V
I
3V
2.7 mA, 2.3 V
2.3 m A, 2.3 V
7.3 mA, 2.3 V
7.3 mA, 1.7 V
1K
5V
Sol.(c) V = 2.3 V and
V = IR = (10 mA) (2.2 k  ) = 22 V
I
R = 2.2 k 
2.3  ( 5)
 7.3  7.3 mA
1K
Diode forward-biased
ID =
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22 V – 0.7 V
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2.2 k  2.2 k
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CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES
V0 = ID (2.2 k  )
PERSONAL REMARK :
= (4.841 mA) (2.2 k  ) = 10.65 V
Ans.
(20 V – 0.7 V + 4 V)
= 3.43 mA
6.8 kΩ
(b) Diode forward-biased, ID =
Ans.
Ex. The states of the diodes D1 and
D2 for the figure shown under
extremely high negative
Vs >> V is
D1
+ V0 – 0.7 V + 4 V = 0
or
V0
V0 = – 3.3 V
Ans.
R1
5. Determine V01 and V02 for the network of figure 6.
Si
+ 12 V
4.7 k
– 10 V
Ge
Vs
Si
V02
V01
V02
V01
3.3 k
(a)
(b)
Figure 6
(b)
V01 = 12 V – 0.7 V = 11.3 V
Ans.
V02 = 0.3 V
Ans.
Ex. For the circuit shown below,
using ideal diode, the values of
voltage and current are
[IES-EC-2011]
+ 3V
10 k
10 V – 0.7 V – 0.3 V
9V

= 2 mA
1.2 k  3.3 k
4.5 k
V=?
V02 = – (2 mA) (3.3 k ) = – 6.6 V
Ans.
6. Determine V0 and ID for the network of figure 7
ID
ID
Si
Si
V0
Si
Si
I=?
– 3V
(a)
(b)
(c)
(d)
15 V
+ 15 V
R3
(a) Both D1 and D2 OFF
(b) D1 OFF and D2 ON
(c) Both D1 and D2 ON
(d) D1 ON and D2 ON
Sol.(d) Since Vs is extremely high
negative therefore diode D1 will
'ON' and D2 will 'OFF'.
V01 = – 10 V + 0.3 V + 0.7 V = – 9 V
I=
R2
D2
V
1.2 k
Ge
Sol. (a)

– 3 V and 0.6 mA
3 V and 0.0 mA
3 V and 0.6 mA
–3 V and 0.0 mA
61(a) Let the diode is ON
+ 3V
Si
V0
4.7 k
10k
2.2 k
V
–5V
(a)
I
(b)
– 3V
Figure 7
Sol.(a) Both diodes forward-biased , IR =
15 V – 1.4 V
= 2.894 mA
4.7 k
I
2.894 mA
Assuming identical diodes , ID = R 
= 1.45 mA
2

V0 = 15 V
– 0.7 V – 0.7 V = 13.6 V
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I=
3 – (–3)
= 0.6 mA
10 K
Direction of current is from p to n-
Ans.
side, So the diode is ON.
Clearly V = –3 V
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CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES
(b) Right diode forward-biased : ID =
PERSONAL REMARK :
15 V  5 V – 0.7 V
= 8.77 mA
2.2 k
V0 = 15 V – 0.7 V = 14.3 V
Ans.
Ex. Two Ge diodes are connected
in series opposition across a
5 V battery as shown in figure.
Determine Vo and I for the networks shown in figure 8.
7.
+ 10 V
V1
+ 16 V
I
Ge
Si
+
D2
I
+
Si
Si
V2
+
D1
I
Si

5V
V0
V0
1 k
4.7 k
+8V
(a)
(b)
Figure 8
Sol. (a) Ge diode "on" preventing Si diode from turning "on":
10 V – 0.3 V
9.7 V

= 9.7 mA
1 k
1 k
Ans.
V0 = I × 1 k  = 9.7 mA × 1 k  = 9.7 V
Ans.
I=
(b) I =
16 V – 0.7 V – 0.7 V – 8 V
6.6 V

= 1.404 mA
4.7 k
4.7 k
V0 = 8 V + (1.404 mA) (4.7 k ) = 14.6 V
Ans.
Ans.
Determine V01, V02, and I for the network of figure 9.
Both diodes forward-biased : Vg (si) = 0.7 V and Vg (Ge) = 0.3 V
8.
Sol.
 V – 0.7 V
I2 k =

 k
I (Si diode) = I2 k
V01 = 0.7 V
V02 = 0.3 V
(a) Find voltage across each diode
assuming breakdown voltage of
diodes is greater than 5 V.
(b) Find current if Vz = 4.9 V and
I0 = 5 A
Sol. Since diodes are in series, same
current I flows in D1 and D2 .
From the given circuit we
observe that diode D 2 in
forward direction while diode
D 1 in reverese direction
,therefore current in the circuit
is limited by diode D1.
Since D2 is forward biased, V2
will be very small and therefore,
V1 (i.e. 5  V2) will be much
larger that VT (i.e. 0.026 V).
So ,the current will be Consider
.7 V – 0.3 V
diode D2 : I  I 0 (e V/VT  1)
Since, I = I0 & V = V2
= 9.65 mA , I0.47 k =
= 0.851 mA

. k
– I0.47 k = 9.65 mA – 0.851 mA = 8.799 mA Ans.


Ans.
Ans.
We get , I0  I0 (eV/VT  1)
or
V2  VT In2  0.026  0.693  0.018 V
V1  5  V2  5  0.018  4.892 V
9. Determine V0 and ID for the network of figure 10.
Vo1
(b)
2 k
0.47 k
Vo2
ID
I
Vo
+10 V
20 V
Si
Si
Ge
2 k
–
Figure 9
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2 k
Si
+
e V /VT  1  1
If V z is 4.9 V, then D 1 will
breakdwon.So,V1 = 4.9 V.Therefore ,
V2 = 5  V1 = 5  4.9 = 0.1 V
Putting , I0 = 5 A, V2 = 0.1 V
We get, I  5(e 0.1/ 0.026  1)  229 A
2 k
Figure 10
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CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES
Sol. For the parallel Si – 2 k branches a Thevenin equivalent will result (for
"on" diodes) in a single series branch or 0.7 V and 1 k resistor as
shown in figure 11.
2 k (10 V – 0.7 V)
2

V0 =
(9.3 V) = 6.2 V
 k  2 k
3
I2 k =
6.2 V
= 3.1 mA
2 k
ID =
PERSONAL REMARK :

V0
10V
0.7V 1 k 
2K
Ans.
Eth , R th
I 2k
3.1 mA

= 1.55 mA Ans.
2
2
Figure 11
10. Determine V0 for the network of figure 12.
5V
Si
V0
0V
Si
2.2
Figure 12
5V
Sol. The Si diode with –5 V at the cathode is "on" while the other is "off".
The result is , V0 = 0 – 0.7 V = – 0.7 V
11. Determine V0 for the network of figure 13.
–5V
Si
0V
Si
V0
Figure 13
Sol. 0 V at one terminal is "more positive" than – 5 V at the other input
terminal. Therefore assume lower diode "OFF" and upper diode
"ON".The result : V0 = – 5 V + 0. 7 V = – 4.3 V
12. Determine the level of V0 for the gate of figure 14.
10 V
Si
10 V
V0
Si
Figure 14
Sol. Since all the system terminals are at 10 V the required difference of 0.7
V across either diode cannot be established. Therefore, both diodes are
"OFF" and V0 = + 10 V as established by 10 V supply connected to
1 k resistor..
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13. Assuming that the diodes are ideal in figure, the current in diode D1 is
(GATE-EE-2004)
1 k
1 k
PERSONAL REMARK :

(a) 8 mA
D1
(b) 5 mA
(c) 0 mA
D2
I
5V
8V
(d) –3 m A
Sol.(c) Let Both the diodes are ‘ON’ and replaced by short circuit
1 k
VA 1 k 
I
5V
8V
From the diagram VA = 0 V and Apply nodal at VA
5 – VA
V 8
I A
or 5 – VA = I + VA + 8 or 5 = I + 8 or I = –3 mA
1
1
The direction of current flow is against the diode convention. The diode
is reverse bised  I = 0 mA.
14. Assume that D1 and D2 in figure are ideal diodes. The value of current
I is
(GATE-EE-2005)
(a) 0 mA
D1
2k 
(b) 0.5 mA
1mA
(DC)
(c) 1 mA
I
2k 
(d) 2 mA
D2
Sol.(a) The diode D1 is forward biased.
The diode D2 is reverse biased as its convention is opposite to the direction
of current.
D2 is reverse biased and is replaced by open-circuit.
Therefore, I = 0 mA
15. What are the states of the three ideal diodes of the circuit shown in
figure ?
(GATE-EE-2006)
1
1
+
10V
D2
D1
1
5A
D3
(a) D1 ON, D2 OFF, D3 OFF
(b) D1 OFF, D2 ON, D3 OFF
(c) D1 ON, D2 OFF, D3 ON
(d) D1 OFF, D2 ON, D3 ON
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Sol.(a) Let all the diodes be reverse biased.
PERSONAL REMARK :

1
Aply KCL at node V1
1 V
1
V1 – 10 V1 – V2

0
1
1
or 2V1 – V2 = 10 ...(1)
Apply KCL at node V2
V1 – V2
+ 5 = V2
1
V2
D2
1
10V
5A
or V1 – 2V2 = –5
... (2)
20
25
V or V1 =
V
3
3
Clearly, D1 is forward biased and D3 is reverse biased
Solve (1) and (2) , V2 =
16. Assuming that the diodes in the given circuit are ideal, the voltage V0 is
D1
(GATE-EE-2010)
10k D 2
(a) 4 V
10k
(b) 5 V
V0
10V
15V
(c) 7.5 V
10k 
(d) 12.12 V
Sol.(b) Diode D1 is always ‘ON’
and D2 is ‘OFF’
V0 =
10k
V0
10V
10k 
  10
10  R 2
=
=5V
20
R1  R 2
+6V
10k
The voltage at V1 and V2 of the arrangement
D2
shown in the figure will be respectively (IES-EE-1998)
+6V
( Given the cut-in voltage of diode is 0.7 V )
D1
(a) 6 V and 5.4 V (b) 5.4 V and 6 V
10k
(c) 3 V and 5.4 V (d) 6 V and 3 V
+3V
Sol.(a)Clearly D1 is forward biased and D2 is reverse biased.
The n-side of D2 is connected to more positive terminal ,V1 = 6 V
Taking cut-in voltage of diode 0.6 V we get ,V2 = 6  0.6 = 5.4 V
17.
V1
V2
19. For the circuit shown in figure the diode D is ideal. The power dissipated by the 300  resistor is
[GATE-IN- 2004]
100, 0.5W
(a) 0.25 W
200
(b) 0.50 W
0.5W
(c) 0.75 W
+
(d) 1.00 W
60 cos314t –
Sol.(c) From the circuit, I rms 
or I rms 
Im
(Since the given circuit represents HWR)
2
Vm
60
1


2(100  200  300) 2  600 20
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Now, the power dissipated by the 300  resistor is
PERSONAL REMARK :

2
2
P300   I rms R 300 
300
1 
 0.75 W
    300 or P300  
 20 
400
20. A voltage 1000 sin t Volts is applied across YZ. Assuming ideal diodes,
the voltage measured across WX in Volts is [GATE-EC-EE- 2013]
1 k
(a) sin t
(b)
(c)
(d)
W
( sin t + | sin t |)/ 2
( sin t  | sin t |)/2
0 for all t
Y X
Z
1 k
Sol.(d) The given circuit
1 k
W
Z
Y X
1 k
 During positive half cycle, the equivalent
circuit is shown
1 k
Y
X
W
Z
+
From the equivalent circuit ,VWZ = 0
 During negative half cycle , the equivalent
circuit is shown
1 k
W Y X
Z
+
22. Which of the following is NOT
associated with a p-n junction?
(GATE-EC-2008)
From the equivalent circuit ,VWZ = 0
21. In a forward biased pn junction diode the sequence of events that best
describes the mechanism of current flow is
[GATE-EC-2013]
(a) injection and subsequenct diffusion and recombination of minority
carriers
(b) injection and subsequenct diffusion and generation of minority carriers
(c) extraction, and subsequent diffusion and generation of minority carriers
(d) extraction and subsequent drift and recombination of minority carriers
LUCKNOW GORAKHPUR
9919526958
0522-6563566
LUCKNOW
ALLAHABAD
AGRA
9919751941
9451056682
Sol.(a)
In a forward biased pn junction
diode the sequence of events
that best describes the
mechanism of current flow is injection and subsequenct
diffusion and recombination of
minority carriers
PATNA
9919751941
(a) Junction Capacitance
(b) Charge Storage Capacitance
(c) Depletion Capacitance
(d) Channel Length Modulation
Sol.(d)Channel length modulation
takes place in MOSFET
NOIDA
SUMMER CRASH COURSE ONLINE TEST SERIES
9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES
66