Download 7: Rectifiers

• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Lecture 7
Rectifiers
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Half-wave rectifier
Full-wave rectifier
Output filtering
Impact of load resistance
Robert R. McLeod, University of Colorado
http://en.wikipedia.org/wiki/Rectifier
80
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Half-wave rectifier
Vsource > 0
Vsource < 0
0.65 V
+
Vsource = 3 sin (ωt )
Diode
forward
bias
voltage
Vsource
Vload
Diode
is open
circuit
1 V/div
200 µs/div
Robert R. McLeod, University of Colorado
81
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Half-wave rectifier with
two diodes
Vsource > 0
Vsource < 0
0.65 V
+
-
+
Vsource = 3 sin (2π 100t )
2x
diode
forward
bias
voltage
Vsource
Vload
Diodes
are open
circuit
1 V/div
200 µs/div
Robert R. McLeod, University of Colorado
82
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Full-wave rectifier
Vsource > 0
Vsource < 0
0.65 V
+
-
+
0.65 V
+
-
-
+
0.65 V
Vsource = 3sin (2π 100t )
2x
diode
forward
bias
voltage
Vsource
Vload
1 V/div
200 µs/div
Robert R. McLeod, University of Colorado
83
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Smoothing capacitor
No load
Vsource = 3sin (2π 100t )
Vsource
Vload
1 V/div
200 µs/div
Robert R. McLeod, University of Colorado
84
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Smoothing capacitor
Large resistance load
R1 = 1 MW
Vsource = 3sin (2π 100t )
RC = 1 s
Vsource
Vload
1 V/div
200 µs/div
Robert R. McLeod, University of Colorado
85
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Smoothing capacitor
Medium resistance load
R1 = 1 KW
Vsource = 3sin (2π 100t )
RC = 1 ms
Vsource
Vload
1 V/div
200 µs/div
Robert R. McLeod, University of Colorado
86
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Smoothing capacitor
Small resistance load
R1 = 100 W
Vsource = 3sin (2π 100t )
RC = 0.1 ms
Vsource
Vload
1 V/div
200 µs/div
Robert R. McLeod, University of Colorado
87
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 7.1
Q: The circuit shows a transformer with a center-tapped secondary.
Assume the turns ratio is 1:1. If the input on the left is driven with
a sine-wave, the arrangement of diodes would
A: Provide half-wave rectification with a peak voltage
approximately equal to the primary voltage.
B: Provide full-wave rectification with a peak voltage
approximately equal to the primary voltage.
C: Provide half-wave rectification with a peak voltage
approximately half that of the primary voltage.
D: Provide full-wave rectification with a peak voltage
approximately half that of the primary voltage.
The center tap is always halfway between
the top and bottom secondary outputs.
When the top line is at a higher voltage
than the center, D1 is ON and D2 is OFF,
so that R will see +V/2 (minus diode
forward bias drop). When the top line is
at a lower voltage than the center, D1 is
OFF and D2 is ON and the resistors sees
the same signal as before.
Robert R. McLeod, University of Colorado
http://en.wikipedia.org/wiki/Rectifier
88
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 7.2
Q: You have the choice of using a half-wave or a full-wave
rectifier to create a power supply. Ignoring diode forward bias loss,
what is the power penalty of using the half-wave rectifier? That is,
what is the power delivered from the half-wave rectifier relative to
the full-wave rectifier, all else being equal?
50%. The same signal is delivered for half
the period. For the other half the period,
the half-wave rectifier delivers zero power.
So the power is on half the time and the
penalty is ½.
You might be tempted to answer 25%,
since power is related to voltage squared.
This would be correct answer is the
amplitude of the voltage were halved, but
not if the time is halved, as is the case here.
Robert R. McLeod, University of Colorado
89
• Lecture 7: Rectifiers
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 7.3
Q: The load (shown as an unlabeled resistor on the right) of a
power supply without a regulator is specified to range from 1
KOhm to 10 MOhm. If the input frequency is 60 Hz, what
capacitance do you want to use in order to keep the output voltage
approximately constant?
We want a large enough capacitance that
the RC time constant is approximately
equal to the period of the waveform. As
shown in the notes, when the frequency
was 100 Hz (so period is 10 ms), a 1 ms RC
time constant let a fairly large ripple
appear. Thus a reasonable guess is that
we should set the RC time constant equal
to the period.
The small resistance is the limiting case.
Setting RC = 1/60 with R = 1000, we find
we would like a 17 microfarad capacitor.
This is why we use electrolytic capacitors
in this application – we want very large
capacitance.
Robert R. McLeod, University of Colorado
90