ECE 1750 Week 3 (part 1) Rectifiers 1
Rectifier Rectifiers convert ac into dc Some commercial rectifiers: (Used to charge batteries like those on the right)
Example of Assumed State Analysis V ac R L Consider the V ac > 0 case We make an intelligent t guess that t 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 Thus, our assumed state is correct 3
Example of Assumed State Analysis 1V 11V 10V R L 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 h the bottom diode, d 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 4
Assumed State Analysis V ac 1 3 4 R L What are the states of the diodes on or off? Consider the V ac > 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 R L. I comes to the junction of diodes # and #4. We have already determined that diode #4 is off. If current is flowing, then diode # must be on, and I continues to the V ac terminal. 5
Assumed State Analysis, cont. V ac > 0 1 R L 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 (V ac = I R L ) is satisfied Thus, our assumed states are correct 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 # are off 6
AC and DC Waveforms for a Resistive st e Load V ac >0 1 3 V dc 4 V ac < 0 V dc Vac Vdc 40 40 0 0 ts Vol 0 0.00 8.33 16.67 5.00 33.33-0 Vo olts 0 0.00 8.33 16.67 5.00 33.33-0 -40 Milliseconds -40 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! 7
Half-wave rectifier 1 T V p Vdc vrdt sin( ) d T 0 0 V p t V p V dc t V ac R L 8
Full wave rectifier Waveforms v 1 T V p Vdc vrdt sin( ) d T 0 0 V p vr V p v V dc t t t R tifi d d Rectifier nd order Filter
Diode Bridge Rectifier (DBR) v Vp 10 V dc 170 V v Vdc 10 V dc 170Vdc t DC link t I ac 10V ac rms 1 3 4 First order low pass filter 10 V dc 170Vdc 10
DC-Side Voltage and Current for Two Different Load Levels Vdc 00W Load 800W Load T 1 f Ripple voltage increases Idc Average current increases (current pulse gets taller and wider) I ac 1 I dc = I ac 3 10V ac rms 4 11
Approximate ppo Formula uafor DC Ripple ppevoltage otage With a first order low pass filter 1 1 CV peak CV min Pt 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 V V V peak min V peak V min Pt C ( V peak Vmin )( V peak Vmin ) Pt C ( V peak V min ) C( V P t V peak min ) 1
( V Approximate Formula for DC Ripple Voltage, cont. peak V min ) Pt C ( V Vmin ) peak min T/ For low ripple, V peak V min V V peak, and t T t T 1 f (V V Vmin ) V peak peak to peak ripple P fcv peak A second order low-pass filter realized by adding an inductor in the dclink allows to reduce the required capacitance for a given ripple goal. 13
AC Current Waveform 1 T = f 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 1 T ptdt () Average Power 0.. T pf Total Used (Apparent) Power VRMS IRMS V I I p. f. V I I 60 Hz, RMS 60 Hz, RMS 60 Hz, RMS 60 Hz, RMS RMS RMS 14
Vampire Loads =? Vampire loads have high leakage currents and low power factor. Your new lab safety tool: 15
Estimated Diode Conduction Losses Estimate the average value I avg of the i(t) v(t) ac current over the conduction interval T cond Estimate the average value Vavg of diode forward voltage drop over one conduction interval ltcond 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 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 avgtcond P 40V I T Watts. avg T 60 Hz avg avg cond 16
Forward Voltage on One Diode Zero Conducting Forward voltage on one diode Zero Zoom-In Forward voltage on one diode 17
AC CCurrent Waveform eo 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 18