Chapter 3
Diodes, Home Work Solutions
3.1
Problem 3.11
For the rectifier circuit of Figure (3.1) let the input sine wave have 120-V rms value and
assume the diode to be ideal. Select a suitable value for R so that the peak diode current
does not exceed 0.1 A. What is the greatest reverse voltage that will appear across the diode.
vD
iD
D
R
vI
Figure 3.1:
Solution
The peak diode current iD−peak is given by:
iD−peak =
vI−peak
R
vo
2
CHAPTER 3. DIODES, HOME WORK SOLUTIONS
To keep the current below 0.1 A we then have:
vI−peak
≤ 0.1
R
vI−peak
R ≥
0.1 √
vI−rms 2
R ≥
0.1
√
120 2
≥
0.1
≥ 1.7 kΩ
The largest reverse voltage vr−max is the peak input voltage:
vr−max = vI−peak
√
= 120 2
= 169.7 V
3.2. PROBLEM 3.25
3.2
3
Problem 3.25
In the circuit shown in Figure (3.2) both diodes have n = 2 but D1 has ten times the junction
area of D2 . What value of V results? To obtain a value of V of 50 mV, what current I2 is
needed?
Solution
Diode D1 carries a current iD1 of 1 mA then diode D2 will carry a current iD2 of 9 mA. Let
VD1 and VD2 be the voltage across D1 and D2 respectively, we then have for iD1 and iD2 :
iD1 =
=
=
iD2 =
1 × 10−3 =
9 × 10−3 =
Is1 eVD1 /nVT
Is1 eVD1 /2×0.025
Is1 eVD1 /0.05
Is2 eVD2 /0.05
Is1 eVD1 /0.05
Is2 eVD2 /0.05
Noting that Is1 = 10 Is2 and using the above equations
we get for VD1 and VD2 :
I1 = 10 mA
D2
D1
1 × 10−3
10 × Is2
9 × 10−3
= 0.05 ln
Is2
V
VD1 = 0.05 ln
VD2
I2 = 1 mA
The voltage V is given by:
V
= VD2 − VD1
9 × 10−3
1 × 10−3
= 0.05 ln
− 0.05 ln
I
10 × Is2
s2 −3
9 × 10
10 × Is2
= 0.05 ln
×
1 × 10−3
Is2
−3
9 × 10 × 10
= 0.05 ln
1 × 10−3
= 0.22 V
Figure 3.2:
4
CHAPTER 3. DIODES, HOME WORK SOLUTIONS
Now, we change I2 to make V = 50 mV, while keeping I1 and iD1 the same, we then write
V as in the last equation, but using iD1 and iD2 and noting that iD1 = I2 and iD2 = I1 - I2 :
= VD2 − VD1
10 × iD2
= 0.05 ln
iD1
10 × (I1 − I2 )
= 0.05 ln
I2
10 × (10 × 10−3 − I2 )
= 0.05 ln
I2
10 × (10 × 10−3 − I2 )
0.05 = 0.05 ln
I2
1
I2 e = 10 × (0.01 − I2 )
I2 = 7.86 mA
V
3.3. PROBLEM 3.45
3.3
5
Problem 3.45
A p+ − n diode is one in which the doping concentration in the p region is much greater than
that in the n region. In such a diode, the forward current is mostly due to hole injection
across the junction. Show that
Dp
I ≈ Ip = Aqn2i
eV /VT − 1
L p ND
For the specific case in which ND = 5 × 1016 /cm3 , Dp = 10 cm2 /s, τp = 0.1 µs, and
A = 104 µm2 , find Is and the voltage V obtained when I = 0.1 mA. Assume operation at
300 K where ni = 1.5 × 1010 /cm3 . Also calculate the excess minority-carrier charge and the
value of the diffusion capacitance at I = 0.1 mA.
Solution
The hole and electron currents add together in a forward biased diode. Theses currents are
given by:
Dp
eV /VT − 1
Ip = Aqn2i
L p ND
Dn
In = Aqn2i
eV /VT − 1
L n NA
The total current I is then:
I = Ip + In
= Aqn2i
Dp
Dn
eV /VT − 1 + Aqn2i
eV /VT − 1
L p ND
L n NA
In a p+ − n junction NA ND , so Ip In , we then have:
I ≈ Ip = Aqn2i
Since Lp =
Is
Dn
eV /VT − 1
L n NA
p
τp Dp , Is is then given by:
Dp
2
= Aqni
L p ND
!
D
pp
= Aqn2i
N D τ p Dp
4
−12
= 10 × 10
−19
× 1.6 × 10
= 7.2 × 10−16 A
10
6 2
× (1.5 × 10 × 10 )
10 × 10−4
√
5 × 1016 × 106 0.1 × 10−6 × 10 × 10−4
6
CHAPTER 3. DIODES, HOME WORK SOLUTIONS
The current I can be written as, noting that V VT :
I = Is eV /VT − 1
0.1 × 10−3
V
≈ Is eV /VT
= 7.2 × 10−16× eV /0.025 10−4
= 0.025 × ln
7.2 × 10−16
= 0.641 V
The excess minority charge is:
Q =
=
≈
=
=
=
Qp + Qn
τp Ip + τn In
τp Ip
0.1 × 10−6 × 0.1 × 10−3
10−11 C
10 pC
The diffusion capacitance is given by:
Cd =
≈
=
=
=
=
τT I
VT
τp I
VT
0.1 × 10−6 × 0.1 × 10−3
0.025
−10
4 × 10
400 × 10−12
400 pF
3.4. PROBLEM 3.62
3.4
7
Problem 3.62
For the circuit in Figure (3.3) utilize Thevenin’s theorem to simplify the circuits and find the
values of of the labeled currents and voltages. Assume that the diodes can be represented
by the constant-voltage drop model (VD = 0.7 V).
+15V
+15V
10kΩ
+10V
10kΩ
I
10kΩ
I
V
V
20kΩ
20kΩ
10kΩ
10kΩ
Figure 3.3:
Solution
In the circuit on the left side of Figure (3.3), the voltage divider composed of 10 kΩ, 20
kΩ and 15 V source can be replaced, using Thevenin’s theorem, by a voltage source V =
Vs × 20/(10 + 20) = 15 × 20/30 = 10V and a resistor that is the parallel equivalent of the two
resistors, i.e. 6.67 kΩ. The circuit on the right side of Figure (3.3) has two voltage dividers
that can be replaced with their Thevenin equivalents. The simplified circuits are shown in
8
CHAPTER 3. DIODES, HOME WORK SOLUTIONS
6.67 kΩ
I
5 kΩ
I
5 kΩ
V
VD
V
10 V
7.5 V
5V
20kΩ
Figure 3.4:
Figure (3.4). The diode in the circuit on the left side of Figure (3.4) is forward biased, the
current I that is conducted by the diode and passes through the 20 kΩ resistor is given by:
10 − 0.7
6.67 + 20
9.3
=
26.67
= 0.35 mA
I =
The voltage V is the voltage across the 20 kΩ resistor:
V = 0.35 × 20 = 7 V
The diode in the circuit on the right side of Figure (3.4) is reverse biased, so the current
conducted by the diode is zero. The voltage across the diode is then the difference between
the voltage sources, i.e. V = 5 − 7.5 = −2.5 V
3.5. PROBLEM 3.71
3.5
9
Problem 3.71
For the circuit of Figure (3.5), replace the diode by its small-signal resistance, and thus
sketch the circuit for calculating the transfer function v◦ /vs assuming vs is a sinusoid of
small amplitude (less than 10 mV) and frequency ω. Find an expression for f3db in terms of
the bias current I. If I is to vary over the range 10 µA to 1 mA, find the value of C required
to ensure that f3db is at most 100 Hz. What is the range of f3db obtained? Assume n = 2
and Rs = 1kΩ.
I
Rs
C
vo
vs
Figure 3.5:
Solution
Replacing the diode with it small-signal resistance, rd = nVT /I is shown in Figure (3.6).
This is a high pass filter. The input voltage vs is given by:
vs = i(Rs + rd + XC )
1
= i Rs + rd +
jωC
10
CHAPTER 3. DIODES, HOME WORK SOLUTIONS
C
Rs
vo
rd
vs
Figure 3.6:
where XC is the reactance of the capacitor. The output voltage vo is the voltage across rd ,
i.e. vo = ird . The transfer function vo /vs of this filter is then:
vo
ird
= vs
i Rs + rd +
=
1
jωC
jωCrd
1 + jωC(Rs + rd )
The break point of a filter is the frequency at which the output signal drops by 3db from
the maximum. For a high pass filter the f3db is given by:
f3db =
1
2πCR
1
2πC(Rs + rd )
1
=
2πC(Rs + nVI T )
I
=
2πC(IRs + nVT )
=
for I = 10 µA :
3.5. PROBLEM 3.71
11
f3db
10−5
=
2πC(10−5 × 1 × 103 + 2 × 0.025)
10−5
=
2πC × 0.06
for I = 1 mA :
f3db
10−3
=
2πC × 1.05
For the same value of C the maximum f3db occurs at I = 1 mA, we can then find the value
of C that makes f3db ≤ 100Hz from:
10−3
2πC × 1.05
10−3
C ≥
2π × 105
= 1.52 µF
100 ≥
Using this value of C we can calculate the range of f3db when the current varies between
10 µA and 1 mA to be 17.5 Hz to 99.7 Hz. The maximum value of f3db is less that 100 Hz
due to the rounding off of the value of C.
12
3.6
CHAPTER 3. DIODES, HOME WORK SOLUTIONS
Problem 3.88
It is required to design a full-wave rectifier circuit using the circuit of Figure (3.7) to provide
an average output voltage of:
(a) 10 V,
(b) 100 V.
In each case find the required turns ratio of the transformer. Assuming that a conducting
diode has a voltage drop of 0.7 V. The ac line voltage is 120 V rms.
D1
vs
R
Vo
ac Line
Voltage
vs
D2
Figure 3.7:
Solution
When the peak of the input signal vs vD we can use the result of Problem 3.91 (solution is
available in the solution for extra problems that is posted on the course web page), namely:
vo ≈
For v o = 10 V we get:
2
vs − vd
π
3.6. PROBLEM 3.88
13
π
(v o + vD )
2
π
=
(10 + 0.7)
2
= 16.81 V
√
The peak value of the line voltage is 120 2. The turns ratio Np /Ns = vp /vs , where p refers
to the transformer’s primary and s refers to it secondary, we then have:
√
Np
120 2
=
Ns
16.81
= 10.1
Np = 10.1 × Ns
for each half of the secondary
= 5.05 × Ns
for the entire center tapped secondary
vs =
For v o = 100 V we get:
vs =
=
Np
=
Ns
=
Np =
=
π
(100 + 0.7)
2
158.2
√V
120 2
158.2
1.073
1.073 × Ns
0.536 × Ns
for each half of the secondary
for the entire center tapped secondary