EE462L, Spring 2014
Diode Bridge Rectifier (DBR)
1
Rectifier
• Rectifiers convert ac into dc
• Some commercial rectifiers
(Used to charge batteries)
2
Diode Bridge Rectifier
(DBR)
Be extra careful that you
observe the polarity markings
on the electrolytic capacitor
Important − never connect a DBR directly to
120Vac or directly to a variac. Use a 120/25V
transformer. Otherwise, you may
overvoltage the electrolytic capacitor
Idc
+
1
3
Iac
120V
Variac
120/25V
Transformer
+
≈ 28Vac rms
–
Equivalent DC load
resistance RL
4
2
+
≈ 28√2Vdc ≈ 40Vdc
−
–
3
Variac, Transformer, DBR Hookup
The variac is a one-winding
transformer, with a variable output tap.
The output voltage reference is the
same as the input voltage reference
(i.e., the output voltage is not isolated).
The 120/25V transformer has separate input
and output windings, so the input voltage
reference is not passed through to the output
(i.e., the output voltage is isolated)
4
Example of Assumed State Analysis
+
Vac
–
+
RL
–
• Consider the Vac > 0 case
• We make an intelligent guess that I is
flowing out of the source + node.
• If current is flowing, then the diode must be “on”
• We see that KVL (Vac = I • RL ) is satisfied
• Thus, our assumed state is correct
5
Example of Assumed State Analysis
+
11V
–
+
10V
–
− 1V +
+
RL 11V
–
• We make an intelligent guess that I is flowing out of the 11V source
• If current is flowing, then the top diode must be “on”
• Current cannot flow backward through the bottom diode, so it must
be “off”
• The bottom node of the load resistor is connected to the source
reference, so there is a current path back to the 11V source
• KVL dictates that the load resistor has 11V across it
• The bottom diode is reverse biased, and thus confirmed to be “off”
• Thus our assumed state is correct
6
Assumed State Analysis
+
1
+
Vac
–
3
RL
4
What are the states of
the diodes – on or off?
2
–
• Consider the Vac > 0 case
• We make an intelligent guess that I is flowing out of the source + node.
• I cannot flow into diode #4, so diode #4 must be “off.” If current is flowing,
then diode #1 must be “on.”
• I cannot flow into diode #3, so diode #3 must be “off.” I flows through RL.
• I comes to the junction of diodes #2 and #4. We have already determined
that diode #4 is “off.” If current is flowing, then diode #2 must be “on,” and
I continues to the –Vac terminal.
7
Assumed State Analysis, cont.
+
1
+
Vac > 0
–
−
+
−
2
+
RL
−
• A check of voltages confirms that diode #4 is indeed reverse biased as we
have assumed
• A check of voltages confirms that diode #3 is indeed reverse biased as we
have assumed
• We see that KVL (Vac = I • RL ) is satisfied
• Thus, our assumed states are correct
• The same process can be repeated for Vac < 0, where it can be seen that
diodes #3 and #4 are “on,” and diodes #1 and #2 are “off”
8
AC and DC Waveforms for a Resistive Load
1
+
Vac > 0
–
3
+
Vdc
–
2
–
Vac < 0
+
4
Vdc
Vac
20
20
0
8.33
16.67
25.00
33.33
Volts
40
Volts
40
0.00
0
0.00
-20
-20
-40
-40
Milliseconds
+
Vdc
–
8.33
16.67
25.00
33.33
Milliseconds
With a resistive load, the ac and dc current waveforms
have the same waveshapes as Vac and Vdc shown above
Note – DC does not mean constant!
9
EE362L_Diode_Bridge_Rectifier.xls
F - Hz
60
C - uF
18000
C charges
VAC
28
P-W
200
Diode bridge
conducting. AC system
replenishing capacitor
energy.
C discharges to load
45
Peak-to-peak
ripple voltage
40
35
Volts
30
25
Vsource
20
Vcap
Diode bridge off.
Capacitor discharging
into load.
15
10
5
0
0.00
2.78
5.56
8.33
11.11
13.89
16.67
Milliseconds
From the power grid point of view, this load is not a “good
citizen.” It draws power in big gulps.
10
DC-Side Voltage and Current for Two Different
Load Levels
Vdc
200W Load
T =
800W Load
Ripple voltage
increases
1
f
Idc
Average current increases (current
pulse gets taller and wider)
11
Approximate Formula for DC Ripple Voltage
1
1
2
2
CV peak
 CVmin
= Pt
2
2
Energy consumed by constant load
power P during the same time interval
Energy given up by capacitor as its
voltage drops from Vpeak to Vmin
2
2
V peak
 Vmin
=
2 Pt
C
(V peak  Vmin )(V peak  Vmin ) =
(V peak  Vmin ) =
2 Pt
C
2 Pt
C (V peak  Vmin )
12
Approximate Formula for DC Ripple Voltage,
cont.
(V peak  Vmin ) =
2 Pt
C (V peak  Vmin )
T/2
For low ripple, V peak
 Vmin  2V peak ,
and t 
T
2
Δt
1
T=
f
P
(V peak  Vmin ) = V peak to  peak ripple 
2 fCVpeak
13
AC Current Waveform
T=
1
f
• The ac current waveform has significant harmonic content.
• High harmonic components circulating in the electric grid may create
quality and technical problems (higher losses in cables and transformers).
• Harmonic content is measurements: total harmonic distortion (THD) and
power factor
1 T
0 p(t )dt
Average Power
T
p. f . =
=
Total Used (Apparent) Power
VRMS I RMS
p. f . =
V60 Hz , RMS I 60 Hz , RMS
V60 Hz , RMS I RMS
=
I 60 Hz , RMS
I RMS
14
“Vampire” Loads
=
?
• “Vampire” loads have high leakage currents and low power
factor.
Your new lab safety tool:
15
Schematic
Notch or +
sign
120V Variac
Iac
120/25V
Transformer
+
Vac
–
Red wire
+
Idc
+
3
1
16-18mF
4
2
Iac –
0.01Ω input current
sensing resistor
2kΩ,
2W
discharge
resistor
3.3kΩ, +
1W V
dc
LED
–
0.01Ω output current
sensing resistor
16
Thermistor Characteristics
Thermistor Resistance
3.5
3
Ohms - pu
2.5
2
1.5
1
0.5
0
0
10
20
30
40
50
60
70
80
90
100
110
Temp - deg C
For our thermistor, 1pu = 1kΩ
17
Measuring Diode Losses with an Oscilloscope
Scope alligator clip
Scope probe
1
4
i(t)
v(t)
3
2
Estimate on oscilloscope the average value I avg of
ac current over conduction interval Tcond
Estimate on oscilloscope the average value Vavg
of diode forward voltage drop over conduction
interval T cond
Tcond
Since the forward voltage on the diode is approximately constant during the conduction
interval, the energy absorbed by the diode during the conduction interval is
approximately V avg • I avg • T cond . Each diode has one conduction interval per
60Hz period, so the average power absorbed by all four diodes is then
4Vavg I avg Tcond
= 240Vavg I avg Tcond Watts.
Pavg =
T60 Hz
18
Forward Voltage on One Diode
Zero
Conducting
Forward voltage on
one diode
Zoom-In
Zero
Forward voltage on
one diode
19
AC Current Waveform
View this by connecting the oscilloscope probe directly
across the barrel of the 0.01Ω current-sensing resistor
One pulse like this passes
through each diode, once
per cycle of 60Hz
The shape is nearly
triangular, so the average
value is approximately
one-half the peak
20