Lecture Notes: 2304154 Physics and Electronics
Lecture 5 (2nd Half), Year: 2007
Physics Department, Faculty of Science, Chulalongkorn University
25/10/2007
Contents
1 Ideal-Diode Model
1
2 Rectifier Circuits
2.1 Half-Wave Rectifier Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Full-Wave Rectifier Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
6
3 Linear Small-Signal Equivalent Circuits
7
4 Problems
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1
Ideal-Diode Model
Assumed States for Ideal-Diode
iD
forward: iD = +, vD = 0
Diode on
reverse: iD = 0, vD = −
vD
Diode off
A Procedure for Analyzing Idel Diode Circuits
1. Assume a state for each diode, either on (a short circuit) or off (an open circuit).
2. Determine the current through the diode (assumed to be on), and the voltage across
the diode (assumed to be off).
3. In case that current = + (forward bias) and voltage = − (reverse bias), finish this
analysis. Otherwise, return to step 1 and choose a different combination of diode
states.
Example 1.
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Figure 1 Circuit diagram for example 1.
Figure 2 Equivalent circuit assuming D1 off and D2 on for example 1.
Figure 3 Equivalent circuit assuming D1 on and D2 off for example 1.
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Figure 4 Circuit diagram for half–wave rectifier with resistive load.
Assume D1 off and D2 on: Loop on the left–hand–size is opened loop. Thus, at
closed loop;
iD2 = 0.5 mA
10 V = vD1 + [(6 kΩ)(0.5mA)]
vD1 = 10 V − [(6 kΩ)(0.5mA)]
= +7 V
This assumption is wrong.
Assume D1 on and D2 off: Loop on the right–hand–size is opened loop. Thus, at
closed loop;
iD1 = 1.0 mA
3 V = [(6 kΩ)(1mA)] + vD2
vD2 = 3 V − [(6 kΩ)(1mA)]
= −3 V
This assumption is right.
2
Rectifier Circuits
Definition of Rectifier Circuits
Rectifier circuits: The circuits used to convert ac power to dc power.
2.1
Half-Wave Rectifier Circuits
Output Voltage of Half-Wave Rectifier (in Figure 4 and 5)
Only the positive half–cycles of the source voltage appear across the load.
Battery-Charging Circuit (in Figure 6)
When the ac source voltage is less than the battery voltage, the diode is reverse
biased. Hence, the current flows only in the direction that charges the battery.
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Figure 5 Source and output voltage versus time for half–wave rectifier.
Figure 6 Half–wave rectifier used to charge a battery.
Figure 7 Circuit diagram of the half–wave rectifier with smothing capacitor.
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C is charged.
Since vs < vL ; C is discharged.
Figure 8 Waveform of the half–wave rectifier with smothing capacitor.
Figure 9 Define of ripple.
Figure 10 Define of ripple and dc voltage.
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Figure 11 Circuit diagram of 2 diodes full–wave rectifier.
Figure 12 Voltage diagram of 2 diodes full–wave rectifier.
HWR with Smoothing Capacitor (in Figure 7 and 8)
Q∼
= IL T ; T = period of the ac voltage
Q = Vr C
C = IL T /Vr
VL = VDC ∼
= Vm − Vr /2
Peak Inverse Voltage (PIV)
PIV: The maximum value of reverse voltage across the diode.
For HWR: PIV = Vm
For HWR with smoothing capacitor: PIV ∼
= 2Vm (the maximum source voltage +
voltage stored on the capacitor.)
2.2
Full-Wave Rectifier Circuits
2 Diodes Circuit
These 2 source voltages are out of phase (from transformer).
PIV for each diode = Vm
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Figure 13 Circuit diagram of diodes bridge full–wave rectifier.
Figure 14 Comparing ripple voltage between half–wave and full–wave rectifier with smoothing capacitor.
Diode Bridge Circuit
This circuit uses only 1 source voltage.
PIV for each diode = Vm
FWR with a capacitive filter
C=
3
IL T
2Vr
Linear Small-Signal Equivalent Circuits
Dynamic Resistance
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From Ohm Law
V = IR
R = (I/V )−1
−1
diD
(rd )Q =
dvD Q
From Shockley equation
VDQ
1
diD
= Is
exp
dvD Q
nVT
nVT
VDQ
∼
= Is exp nV
T
At VDQ VT , IDQ
nVT
IDQ
1
rd ∝
IDQ
rd =
Given Vac = VDQ + vD and Iac = IDQ + iD
iD ∼
= vD /rd
Ac Voltage of Fixed Amplitude in Diode
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Figure 15 For Problem 1
4
Problems
1. Find the values of I and V for the circuits of Figure 15, assuming that the diodes are
ideal.
2. Find the values of I and V for the circuits of Figure 16, assuming that the diodes are
ideal.
3. Consider the battery–charging circuit with Vm = 20 V, R = 10Ω, and VB = 14 V. a)
Find the peak current assuming an ideal diode. b) Find the percentage of each cycle
for which the diode is in the on state.
4. A power–supply circuit is needed to deliver 0.1 A and 15 V (average) to a load. The
ac source has a frequency of 60 Hz. Assume that the circuit of Figure 7 is to be
used. The peak–to–peak ripple voltage is to be 0.4 V. Instead of assuming an ideal
diode, allow 0.7 V for forward diode drop. Find the peak ac voltage Vm needed and
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Figure 16 For Problem 2
the approximate value of the smoothing capacitor. (Hint: To achieve an average load
voltage of 15 V with a ripple of 0.4 V, design for a peak load voltage of 15.2 V.)
5. Repeat Problem 4 using the circuit of Figure 14 with the smoothing capacitor in
parallel with the load RL .
6. A 20–V rms 60–Hz ac source is in series with an ideal diode and a 100–Ω resistance.
Determine the peak current and peak inverse voltage (PIV) for the diode.
7. Most dc voltmeters produce a reading equal to the average value of the voltage measured. The mathematical definition of the average value of a periodic waveform is:
Vavg
1
=
T
Z
T
v(t)dt
0
in which T is the period of the voltage v(t) applied to the meter.
(a) What does a dc voltmeter read if the applied voltage is v(t) = Vm sin(ωt)?
(b) What does the meter read if the applied voltage is a half–wave rectified version
of the sinewave?
(c) What does the meter read if the applied voltage is a full–wave rectified version
of the sinewave?
8. A certain diode has IDQ = 4 mA and iD (t) = 0.5 sin(200πt) mA. Find an expression
for iac (t), and sketch it to scale versus time.
9. A breakdown diode has
iD =
−10−6
; for -5 V < vD < 0
(1 + vD /5)3
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where iD is in amperes. Plot iD versus vD in the reverse–bias region. Find the
dynamic resistance of this diode at IDQ = −1 mA and at IDQ = −10 mA.
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