ELC 4345 Fall 2013 Diode Bridge Rectifier DBR

ELC 4345, Fall 2013 Diode Bridge Rectifier (DBR) 1

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

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 120/25 V 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) 3

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 4

Example of Assumed State Analysis + 11 V – + 10 V – − 1 V + + RL 11 V Auctioneering circuit – • We make an intelligent guess that I is flowing out of the 11 V 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 11 V source • KVL dictates that the load resistor has 11 V across it • The bottom diode is reverse biased, and thus confirmed to be “off” • Thus our assumed state is correct 5

Assumed State Analysis + 1 3 RL + Vac – 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. 6

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” 7

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 0. 00 -20 -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! 8

EE 362 L_Diode_Bridge_Rectifier. xls F - Hz 60 C - u. F 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. 9

DC-Side Voltage and Current for Two Different Load Levels Vdc 200 W Load T = 800 W Load Ripple voltage increases 1 f Idc Average current increases (current pulse gets taller and wider) 10

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 11

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

AC Current Waveform T= 1 f 13

Schematic 14

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

Careful! 16

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

Rtherm Measuring the temperature on the backside of a solar panel • Thermistor in series with 470Ω resistor 2. 5 V • Series combination energized by 2. 5 Vdc 470Ω To data logger • The voltage across the 470Ω resistor then changes with temperature as shown below Excel curve fit. Coefficients entered into data logger. As thermistor gets hotter, more of the 2. 5 V appears across the 470Ω resistor 18

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 60 Hz period, so the average power absorbed by all four diodes is then 4 Vavg I avg Tcond = 240 V avg I avg Tcond Watts. Pavg = T 60 Hz 19

Forward Voltage on One Diode Zero Conducting Forward voltage on one diode Zoom-In Zero 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 60 Hz The shape is nearly triangular, so the average value is approximately one -half the peak 21