1. EE462L, Spring 2014 Diode Bridge Rectifier (DBR)

2. Rectifier Rectifiers convert ac into dc Some commercial rectifiers
(Used to charge batteries)

3. 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
Equivalent DC load resistance RL
Idc + 1 3
Iac +
≈ 28√2Vdc ≈ 40Vdc −
120V
Variac
120/25V
Transformer +
≈ 28Vac rms – 4 2 –

4. 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)

5. Example of Assumed State AnalysisRL + – +
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
Thus, our assumed state is correct

6. Example of Assumed State Analysis+
11V – +
11V – +
10V –
− 1V + RL
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

7. Assumed State Analysis+ 1 3
What are the states of the diodes – on or off? RL +
Vac – 4 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.

8. Assumed State Analysis, cont.+ − 1 + − RL + − +
Vac > 0 – 2
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”

9. AC and DC Waveforms for a Resistive Load+ V ac
> – 1 2 dc – V ac
< + 4 3 dc
Vdc
-40
-20 20 40
0.00
8.33
16.67
25.00
33.33
Milliseconds
Volts
Vac
-40
-20 20 40
0.00
8.33
16.67
25.00
33.33
Milliseconds
Volts
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!

10. EE362L_Diode_Bridge_Rectifier.xls F - Hz C uF
VAC P W 60
18000 28
200
Diode bridge conducting. AC system replenishing capacitor energy. 5 10 15 20 25 30 35 40 45
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 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.

11. DC-Side Voltage and Current for Two Different Load Levels
200W Load f T 1 =
Vdc
Idc
800W Load
Ripple voltage increases
Average current increases (current pulse gets taller and wider)

12. Approximate Formula for DC Ripple Voltage
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

13. Approximate Formula for DC Ripple Voltage, cont.
and Δt
T/2
For low ripple,

14. AC Current Waveform f T 1 =
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

15. “Vampire” Loads ? =
“Vampire” loads have high leakage currents and low power factor.
Your new lab safety tool:

16. Schematic

17. Thermistor Characteristics
For our thermistor, 1pu = 1kΩ

18. Measuring Diode Losses with an Oscilloscope1 4 3 2
Scope probe
Scope alligator clip
Estimate on oscilloscope the average value V
avg
of diode forward voltage drop over conduction
interval T
cond
Esti
mate on oscilloscope the average value I
avg of
ac current over conduction interval T
cond
i(t)
v(t) T
cond
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
• T
cond
. Each diode has one conduction interval per
60Hz period, so the average power absorbed by all four diodes is then
cond
avg Hz T I V P
240 4 60 =
Watts.

19. Forward Voltage on One Diode
Zero
Conducting
Zoom-In
Forward voltage on one diode
Forward voltage on one diode

20. 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