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EE462L, Fall 2011 Diode Bridge Rectifier (DBR). Diode Bridge Rectifier (DBR). Be extra careful that you observe the polarity markings on the electrolytic capacitor. - PowerPoint PPT Presentation
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1
EE462L, Fall 2011Diode Bridge Rectifier (DBR)
2
120/25VTransformer
120VVariac
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
Diode Bridge Rectifier(DBR)
+
–
+≈ 28Vac rms
–
1
4
3
2
Equivalent DC load resistance RL
+≈ 28√2Vdc ≈ 40Vdc−
Iac
Idc
Be extra careful that you observe the polarity markings on the electrolytic capacitor
3
Variac, Transformer, DBR Hookup
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)
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).
4
Example of Assumed State Analysis
+
Vac–
•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
RL
•Thus, our assumed state is correct
+
–
5
Example of Assumed State Analysis
+10V
–
• 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”
RL
+
11V
–
+11V
–
•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
− 1V +
•The bottom diode is reverse biased, and thus confirmed to be “off”
•Current cannot flow backward through the bottom diode, so it must be “off”
•Thus our assumed state is correct
Auctioneering circuit
6
Assumed State Analysis
• Consider the Vac > 0 case
+
–
+Vac–
1
4
3
2RL
What are the states of the diodes – on or off?
• We make an intelligent guess that I is flowing out of the source + node.
• 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.
• I cannot flow into diode #4, so diode #4 must be “off.” If current is flowing, then diode #1 must be “on.”
7
Assumed State Analysis, cont.
+Vac > 0
–
1
2RL
• A check of voltages confirms that diode #4 is indeed reverse biased as we have assumed
• We see that KVL (Vac = I • RL ) is satisfied
• Thus, our assumed states are correct
+−
+
−
• A check of voltages confirms that diode #3 is indeed reverse biased as we have assumed
+−
• 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
Vac
-40
-20
0
20
40
0.00 8.33 16.67 25.00 33.33
Milliseconds
Vol
ts
With a resistive load, the ac and dc current waveforms have the same waveshapes as Vac and Vdc shown above
+ Vac > 0
–
1
2
+ Vdc
– – Vac < 0
+
4
3 + Vdc
–
Vdc
-40
-20
0
20
40
0.00 8.33 16.67 25.00 33.33
MillisecondsV
olts
Note – DC does not mean constant!
9
EE362L_Diode_Bridge_Rectifier.xls
F - Hz C - uF VAC P - W 60 18000 28 200
05
101520
2530
35
4045
0.00 2.78 5.56 8.33 11.11 13.89 16.67
Milliseconds
Volts Vsource
Vcap
Peak-to-peak ripple voltage
C charges C discharges to load
Diode bridge conducting. AC system replenishing capacitor energy.
Diode bridge off. Capacitor discharging into load.
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
800W Load
200W Load
fT
1
Vdc
Idc
Ripple voltage increases
Average current increases (current pulse gets taller and wider)
11
Approximate Formula for DC Ripple Voltage
CtPVVpeak
22min
2
CtPVVVV peakpeak
2))(( minmin
)(2)(
minmin VVC
tPVVpeak
peak
tPCVCVpeak 2min
221
21
Energy given up by capacitor as its voltage drops from Vpeak to Vmin
Energy consumed by constant load power P during the same time interval
12
Approximate Formula for DC Ripple Voltage, cont.
)(2)(
minmin VVC
tPVVpeak
peak
fT 1
peakripplepeaktopeakpeak fCV
PVVV2
)( min
, 2min peakpeak VVV For low ripple,
2Tt and
Δt
T/2
13
AC Current Waveform
fT 1=
14
Schematic
+
–
+ Vac
–
1
4
3
2
16-18mF Iac 2kΩ, 2W
discharge resistor 0.01Ω input current
sensing resistor
Iac
120/25V Transformer
120V Variac Idc
+ Vdc
–
Red wire
0.01Ω output current sensing resistor
LED
3.3kΩ, 1W
Notch or + sign
+
15
Mounting the Toggle Switch
Space left between hex nut and body of switch
16
Careful!
17
Thermistor Characteristics
Thermistor Resistance
0
0.5
1
1.5
2
2.5
3
3.5
0 10 20 30 40 50 60 70 80 90 100 110
Temp - deg C
Ohm
s - p
u
For our thermistor, 1pu = 1kΩ
18
Solar Panel Temp
y = 15.557x3 - 51.869x2 + 95.653x - 25.875
-20
-10
0
10
20
30
40
50
60
70
80
0 0.5 1 1.5 2
Voltage
Tem
p - D
egre
es C
• Thermistor in series with 470Ω resistor
• Series combination energized by 2.5Vdc
• The voltage across the 470Ω resistor then changes with temperature as shown below
470Ω
Rtherm
2.5V
To data logger
As the thermistor gets hotter, more of the 2.5V appears across the 470Ω resistor
Excel curve fit. Coefficients entered into data logger.
Thermistor Resistance
0
0.5
1
1.5
2
2.5
3
3.5
0 10 20 30 40 50 60 70 80 90 100 110
Temp - deg C
Ohm
s - p
u
Measuring the temperature on the backside of a solar panel
19
Measuring Diode Losses with an Oscilloscope
Tcond
condavgavg
Hz
condavgavgavg TIV
T
TIVP 240
4
60 Watts.
1
4
3
2
Scope probe
Scope alligator clip
Estimate on oscilloscope the average value Iavg of ac current over conduction interval Tcond
Estimate on oscilloscope the average value Vavg of diode forward voltage drop over conduction interval 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 • Tcond . Each diode has one conduction interval per 60Hz period, so the average power absorbed by all four diodes is then
i(t)
v(t)
20
Forward Voltage on One Diode
Zero
Conducting
Zoom-In
Zero
Forward voltage on one diode
Forward voltage on one diode
21
AC Current Waveform
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
View this by connecting the oscilloscope probe directly across the barrel of the 0.01Ω current-sensing resistor