Transcript Slide 1
EE462L, Fall 2011 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 120Vac or directly to a variac. Use a 120/25V transformer. Otherwise, you may overvoltage the electrolytic capacitor Idc + 1 3 Iac 120V Variac 120/25V Transformer + ≈ 28Vac rms – Equivalent DC load resistance RL 4 2 + ≈ 28√2Vdc ≈ 40Vdc − – 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) 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 + 11V – + 10V – − 1V + + Auctioneering circuit RL 11V – • 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 5 Assumed State Analysis + 1 + Vac – 3 RL 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 Volts 40 0.00 0 0.00 -20 -20 -40 -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 EE362L_Diode_Bridge_Rectifier.xls F - Hz 60 C - uF 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 200W Load T = 800W Load Ripple voltage increases 1 f Idc Average current increases (current pulse gets taller and wider) 10 Approximate Formula for DC Ripple Voltage 1 1 2 2 CV peak CVmin = Pt 2 2 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 2 2 V peak Vmin = 2 Pt C (V peak Vmin )(V peak Vmin ) = (V peak Vmin ) = 2 Pt C 2Pt C (V peak Vmin ) 11 Approximate Formula for DC Ripple Voltage, cont. (V peak Vmin ) = 2Pt C (V peak Vmin ) T/2 For low ripple, V peak Vmin 2Vpeak , and t T 2 Δt 1 T= f P (V peak Vmin ) = V peakto peak ripple 2 fCV peak 12 AC Current Waveform T= 1 f 13 Schematic Notch or + sign 120V Variac Iac 120/25V Transformer + Vac – Red wire + Idc + 3 1 16-18mF 4 2 Iac – 0.01Ω input current sensing resistor 2kΩ, 2W discharge resistor 3.3kΩ, + 1W V dc LED – 0.01Ω output current sensing resistor 14 Mounting the Toggle Switch Space left between hex nut and body of switch 15 Careful! 16 Thermistor Characteristics Thermistor Resistance 3.5 3 Ohms - pu 2.5 2 1.5 1 0.5 0 0 10 20 30 40 50 60 70 80 90 100 110 Temp - deg C For our thermistor, 1pu = 1kΩ 17 Measuring the temperature on the backside of a solar panel Rtherm • Thermistor in series with 470Ω resistor 2.5V • Series combination energized by 2.5Vdc To data logger 470Ω • The voltage across the 470Ω resistor then changes with temperature as shown below Excel curve fit. Coefficients entered into data logger. Solar Panel Temp y = 15.557x3 - 51.869x2 + 95.653x - 25.875 80 70 Thermistor Resistance As the thermistor gets hotter, more of the 2.5V appears across the 470Ω resistor 60 3 Ohms - pu 2.5 2 1.5 1 0.5 0 0 10 20 30 40 50 60 Temp - deg C 70 80 90 100 110 Temp - Degrees C 3.5 50 40 30 20 10 0 -10 -20 0 0.5 1 Voltage 1.5 2 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 60Hz period, so the average power absorbed by all four diodes is then 4Vavg I avg Tcond = 240Vavg I avg Tcond Watts. Pavg = T60 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 60Hz The shape is nearly triangular, so the average value is approximately one-half the peak 21