1. 1 EE462L, Fall 2011 Diode Bridge Rectifier (DBR)

2. 2 120/25V Transformer 120V Variac Important − never connect a DBR directly to 120V ac or directly to a variac. Use a 120/25V transformer. Otherwise, you may overvoltage the electrolytic capacitor Diode Bridge Rectifier (DBR) + – + ≈ 28V ac rms – 1 4 3 2 Equivalent DC load resistance R L + ≈ 28√2V dc ≈ 40Vdc − I ac I dc Be extra careful that you observe the polarity markings on the electrolytic capacitor

3. 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. 4 Example of Assumed State Analysis + V ac – Consider the V ac > 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 (V ac = I R L ) is satisfied RLRL Thus, our assumed state is correct +–+–

5. 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” RLRL + 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. 6 Assumed State Analysis Consider the V ac > 0 case + – + V ac – 1 4 3 2 RLRL 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 R L. 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 –V ac terminal. I cannot flow into diode #4, so diode #4 must be “off.” If current is flowing, then diode #1 must be “on.”

7. 7 Assumed State Analysis, cont. + V ac > 0 – 1 2 RLRL A check of voltages confirms that diode #4 is indeed reverse biased as we have assumed We see that KVL (V ac = I R L ) 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 V ac < 0, where it can be seen that diodes #3 and #4 are “on,” and diodes #1 and #2 are “off”

8. 8 AC and DC Waveforms for a Resistive Load Vac -40 -20 0 20 40 0.008.3316.6725.0033.33 Milliseconds Volts With a resistive load, the ac and dc current waveforms have the same waveshapes as Vac and Vdc shown above + V ac > 0 – 1 2 + V dc – – V ac <0 + 4 3 + V dc – Vdc -40 -20 0 20 40 0.008.3316.6725.0033.33 Milliseconds Volts Note – DC does not mean constant!

9. 9 EE362L_Diode_Bridge_Rectifier.xls F- Hz C- uF VAC P- W 60 18000 28 200 0 5 10 15 20 25 30 35 40 45 0.002.785.568.3311.1113.8916.67 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. 10 DC-Side Voltage and Current for Two Different Load Levels 800W Load Ripple voltage increases Average current increases (current pulse gets taller and wider)

11. 11 Approximate Formula for DC Ripple Voltage 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. 12 Approximate Formula for DC Ripple Voltage, cont. For low ripple, and ΔtΔt T/2

13. 13 AC Current Waveform f T 1 =

14. 14 Schematic

15. 15 Mounting the Toggle Switch Space left between hex nut and body of switch

16. 16 Careful!

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

18. 18 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 R therm 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. Measuring the temperature on the backside of a solar panel

19. 19 Measuring Diode Losses with an Oscilloscope T cond avg Hz condavg TIV T TIV P240 4 60  Watts. 1 4 3 2 Scope probe Scope alligator clip Estimate on oscilloscope the average value I avg of ac current over conduction interval T cond Estimate on oscilloscope the average value V avg of diode forward voltage drop over conduction interval 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 i(t) v(t)

20. 20 Forward Voltage on One Diode Zero Conducting Zoom-In Zero Forward voltage on one diode

21. 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