DESIGN OF ELECTRONIC
SYSTEMS
Course Code : 11-EC201
DEPARTMENT OF
ELECTRONICS & COMPUTER ENGINEERING
The two modes of operation of ideal diodes and the use of an
external circuit to limit the forward current (a) and the reverse
voltage (b).
A Simple Application: The Rectifier
A fundamental application of the diode, one that makes use of its
severely nonlinear i-v curve, is the rectifier circuit. The circuit
consists of the series connection of a diode D and a resistor R .
Figure 3 (a) Rectifier circuit. (b) Input waveform.
Another Application: Diode Logic Gates
Diodes together with resistors can be used to implement
digital logic functions.
Diode logic gates: (a) OR gate; (b) AND gate (in a positive-logic system).
A silicon diode said to be a 1-mA device displays a forward voltage
of 0.7 V at a current of 1 mA. Evaluate the junction scaling constant
7; in the event that n is either 1 or 2. What scaling constants would
apply for a 1-A diode of the same manufacture that conducts 1 A at
0.7 V? example
Solution
v / nV
 v / nV
T
T
Since i  I s e
then I s  ie
For the 1-mA diode:
If n = 1: Is = 10 -3 e-700/25 = 6.9 x 10-16 A, or about 10 -15 A
If n = 2: Is = 10 -3 e-700/50 = 8.3 x 10 -10 A, or about 10 -9 A
The diode conducting 1 A at 0.7 V corresponds to one-thousand 1-mA diodes in
parallel with a total junction area 1000 times greater.
Thus IS is also 1000 times greater, being 1pA and 1µA, respectively
for n=1 and n=2.
Example-1
For the circuits shown in Fig. below using ideal diodes, find the
values of the voltages and currents indicated
Solution
a) In this case the diode is conducting, such that the
voltage at the top end of the diode equals that at the
bottom end, so V = −3V. Then, I = 6V/10 kΩ = 0.6mA.
b) The diode is not conducting, so I = 0, and V = 3V.
c) The diode is conducting so I = 0.6mA as before, but now
V = 3V.
d) The diode is not conducting, so I = 0 and V = −3V.
Ex:2 :- For the circuits shown in Fig. below using ideal diodes,
find the values of the voltages and currents indicated
Solution
(a) For D1: The diode is conducting so VD1 = +1V;
For D2: The diode is conducting so VD2 = +3V;
I = (3-(-3))/2kΩ = 3mA;
Two voltages (1V, 3V) try to appear at the point V, but at a
node no more than one voltage can appear.
Hence, V = 3V and I = 3mA.
(b) For D1: VD1 = +1V;
For D2: VD2 = +3V;
I = (3-(1))/2kΩ = 1mA;
Hence V = 1V; I = 1mA.
Ex:3 :- In each of the ideal-diode circuits shown in Fig.
P3.4, v1 is a 1 kHz, 10V peak sine wave. Sketch the
waveform resulting at v0. What are its positive and
negative peak values?
Cont…
Solution
(a) Piecewise expression for V0 is:

0
V 
0 V
1
for V  0
1
for V  0
1
Minimum value: 0V
Maximum value: 10V
(b) Piecewise expression for V0 is:

V1
V 
0 0

for V  0
1
for V  0
1
Minimum value: -10V
Maximum value: 0V
(c) Current never flows, so V0 = 0 always.
(e) Current always flows so V0 = V1
Minimum values: −10V.
Maximum value: 10V.
Cont…
(g) Piecewise expression for V0 is:

V1
V 
0 0

for V  0
1
for V  0
1
Minimum value: -10V
Maximum value: 0V
(h) The output is always connected to ground, V0 = 0 always.
(i) When V1 < 0 this acts as a voltage divider. Otherwise they
two are equal.
V1
Piecewise expression is: V   2
0
V
 1
for V  0
1
for V  0
1
Minimum value: −5V.
Maximum value: 10V.
Cont…
(k) The current source causes 1V voltage drop across the
resistor, such that V0 is always one volt higher than the
voltage at the bottom of the resistor. When V1 > 0 the voltage
at the bottom of the resistor is ground. When V1< 0 the
voltage at the bottom of the resistor is V1 .
The piecewise expression for V0 is:

V1  1V for V1  0
V 
0 V
for V  0
 1
1
Ex:4 :- The circuits shown in Fig. below can function as logic gates
for input voltages are either high or low. Using “1" to denote the
high value and “0” to denote the low value, prepare a table with
four columns including all possible input combinations and the
resulting values of X and Y. What logic function is X of A and S?
What logic function is Y of A and B? For what values of A and 5 do
X and F have the same value? For what values of A and B do X and
Y have opposite values?
Sol:
X=A.B
Sol:
Y=A+B
Ex:5(a) :- Assuming that the diodes in the circuits of Fig. below
are ideal, find the values of the labeled voltages and currents.
Solution
(a) Assume that both the diodes are conducting.
ID2= (10-0)/10 = 1mA;
At node B:
I+1 = (0-(-10))/5 = 2mA;
Results in I = 1mA. Thus D1 is conducting as originally assumed,
and the final result is I = 1mA and V = 0V.
(b) For the circuit in Fig.(b). If we assume that both diodes are
conducting, then VB=0 and V=0. The current in D2 is obtained from
ID2= (10-0)/5 = 2mA;
At node B:
I+2 = (0-(-10))/10 = 1mA;
Which yields I = -1mA. Since this is not possible, our original
assumption is not correct. We start again, assuming that D1 is off
and D2 is on. The current ID2 is given by
ID2= (10-(-10))/15 = 1.33mA;
Ex:5(b) :- Assuming that the diodes in the circuits of Fig. below
are ideal, find the values of the labeled voltages and currents.
I1
I1
2
2
I3
D1& D2 Conducting
I1=1mA; I3=0.5 mA; I2=0.5 mA
V= 0 V
I3
D1=off, D2=On
I1= I3=0.66 mA
V = -1.7 V
Ex:6 :- Assuming that the diodes in the circuits of Fig. below
are ideal, utilize Thevenin's theorem to simplify the circuits
and thus find the values of the labeled currents and
voltages.
Solution
(a) D conducting
I=0.225 mA
V=4.5V
(b) D is not conducting
I=0A
V=-2V
Ex:7 :- The circuit of Fig. below can be used in a signaling
system using one wire plus a common ground return. At any
moment, the input has one of three values: +3 V, 0 V, - 3 V.
What is the status of the lamps for each input value? (Note
that the lamps can be located apart from each other and that
there may be several of each type of connection, all on one
wire!).
Solution:
V
3V
Red Green
On Off
0 V Off
-3 V Off
Off
On
D1 conducts
D2 conducts
Ex:8 :- Listed below are the results of measurements taken
on several different junction diodes. For each diode, the data
provided are the diode current /, the corresponding diode
voltage V, and the diode voltage at a current 1/10. In each
case, estimate Is, n, and the diode voltage at 10/.
(a) 10.0 mA, 700 mV, 600 mV
(b) 1.0 mA, 700 mV, 600 mV
(c) 10 A, 800mV,700mV
(d) 1 mA, 700 mV, 580 mV
(e) 10 µA, 700 mV, 640 mV
Ex:9 :- The circuit in Fig. below utilizes three identical diodes
having n = 1 and Is = 10-14 A. Find the value of the current I
required to obtain an output voltage V0 = 2 V Assume n=1. If
a current of 1 mA is drawn away from the output terminal by
a load, what is the change in output voltage? Assume n=1
Solution
The circuit shown utilizes three identical diodes having n=1
and Is= 10 -14 A.
To find the value of the current I required to obtain an output
voltage Vo=2V : we know that n  1, I S  10 14 A,V0  2V
The voltage across each diode is
 I DX  I S e
vDX
ηVT
v0
2
 vDX 
3
3
2
 10 14 e
3
0 .025
 3.81mA
Solution
If a current of 1mA is drawn away from the output terminal
by a load, what if the change in the output voltage.
Load current = 1mA, therefore IDγ = 2.81mA
I DY
e
I DX
( vDY v DX )
0 .025
e
( vDY  2 / 3 )
0 .025
Δv 0Y  v02  v01  22.8mV
Ex:10 :- For the circuit shown in Fig. below, both diodes are
identical, conducting 10 mA at 0.7 V and 100 mA at 0.8 V.
Find the value of R for which V = 80 mV.
Solution
Both the diodes are identical, therefore Is, η, VT are same
For diode 1 : VD1 = 0.7V; ID1 = 10 mA.
For diode 1 : VD2 = 0.8V; ID2 = 100 mA.
To find η:V D 2  V D1
I D2
 VT ln
I D1
100
0.8  0.7    0.025  ln
10
  1.739
To find R if V=80mV:V  V D 2  V D1
80
R 
 57.1
1.4
I D2
 VT ln
I D1
0.01  I D1
0.08  1.737  0.025  ln
I D1
I D1  1.4mA
Ex: 11 :- A designer of an instrument that must operate over
a wide supply-voltage range, noting that a diode's junction
voltage drop is relatively independent of junction current,
considers the use of a large diode to establish a small
relatively constant voltage. A power diode, for which the
nominal current at 0.8 V is 10 A, is available. Furthermore,
the designer has reason to believe that n = 2. For the
available current source, which varies from 0.5 mA to 1.5
mA, what junction voltage might be expected? What
additional voltage change might be expected for a
temperature variation of +25°C?
As an alternative to the idea suggested in Problem (Ex:10),
the designer considers a second approach to producing a
relatively constant small voltage from a variable current
supply: It relies on the ability to make quite accurate copies of
any small current that is available (using a process called
current mirroring). The designer proposes to use this idea to
supply two diodes of different junction areas with the same
current and to measure their junction-voltage difference. Two
types of diodes are available; for a forward voltage of 700 mV,
one conducts 0.1 mA while the other conducts 1 A. Now, for
identical currents in the range of 0.5 mA to 1.5 mA supplied
to each, what range of difference voltages result? What is the
effect of a temperature change of ―25°C on this
arrangement? Assume n=1.
Ex: 12 :- A “1−mA” diode (i.e. one that has vD = 0.7V at iD =
1mA is connected in series with a 200 resistor to a 1.0 − V
supply.
1. Provide a rough estimate of the diode current you would
expect.
Solution:
1. For the rough estimate we assume the voltage drop across
the diode is vD = 0.7V.
In that case the voltage drop across the resistor is
vR = 1.0 − 0.7 = 0.3V.
The curent is then iD = VR /R = 0.3 200 = 1.5mA.