EE462L, Spring 2014 Diode Bridge Rectifier (DBR)

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

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 –

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)

Example of Assumed State Analysis RL + – + 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

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

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.

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”

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!

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.

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)

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

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

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

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

Schematic

Thermistor Characteristics For our thermistor, 1pu = 1kΩ

Measuring Diode Losses with an Oscilloscope 1 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.

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

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