Survey

Document related concepts

Analog-to-digital converter wikipedia, lookup

Nanofluidic circuitry wikipedia, lookup

Transistor–transistor logic wikipedia, lookup

Josephson voltage standard wikipedia, lookup

Integrating ADC wikipedia, lookup

Valve RF amplifier wikipedia, lookup

Wilson current mirror wikipedia, lookup

Operational amplifier wikipedia, lookup

Schmitt trigger wikipedia, lookup

Resistive opto-isolator wikipedia, lookup

Power MOSFET wikipedia, lookup

Voltage regulator wikipedia, lookup

Power electronics wikipedia, lookup

Surge protector wikipedia, lookup

Current source wikipedia, lookup

Switched-mode power supply wikipedia, lookup

Current mirror wikipedia, lookup

Transcript

Diode Rectifiers Chapter 5 5-1 Diode Rectifier Block Diagram • Inexpensive rectifiers with diodes to convert the input ac into dc in an uncontrolled manner • Power flow is only from the ac side to the dc side • Large capacitor is used as a filter to achieve ripple-free dc output 5-2 A Simple Diode Circuit with a Pure Resistive Load • The load voltage vd and the current i have an average component 5-3 A Simple Circuit with R-L Load Subsequent to t=0, diode becomes forward biased and current begins to flow Current in the circuit is governed by di v s Ri L dt Until t1, Current builds up and inductor stored energy increases as vL (=vs-vR) is positive Beyond t1, vL becomes negative, and current begins to decrease Beyond t2, current continues to flow for a while even after the input voltage has gone negative. This is because of stored inductor energy 5-4 A Simple Circuit (Load has a dc voltage) • Load consists of an inductor and a dc voltage • Current begins to flow when the input voltage exceeds the dc voltage at t1 • The current reaches its peak at t2 • Current continues to flow for a while even after the input voltage has gone below the dc voltage and decays to zero at t3 5-5 Single-Phase Diode Rectifier Bridge • Large filter capacitor at the dc side • Utility supply modeled as a sinusoidal voltage source vs in series with its internal impedance (Ls = Zs) • To improve line current waveform, an inductor may be added in series on the ac side, which will increase Ls more 5-6 Diode-Rectifier Bridge Analysis Top group Bottom group • • • • Simplification: Ls=0 and replace dc side by R or a constant dc current source It is very unlikely that a purely resistive load will be supplied Constant dc current is equivalent to a large inductor in series at the dc output Current flows continuously through one diode of the top group and one diode of the bottom group. 5-7 Redrawing Diode-Rectifier Bridge Top group Bottom group • Current flows continuously through one diode of the top group and one diode of the bottom group. • In the top group, cathodes of the two diodes are at the same potential. Therefore, the diode with its anode at the highest potential will conduct id • When vs goes negative, id instantaneously commutes to D3 as D1 is reversed biased. • Similar argument could be applied to the bottom group. 5-8 Items that are worth noting: • Waveforms with a purely resistive load and a purely dc current at the output • When vs is positive, D1 and D2 conduct vd=vs and id=is When vs is negative, D3 and D4 conduct vd= -vs and id= -is • In both cases, the dc-side voltage and current waveforms are the same • The average value of the dc output voltage Vdo 1 T2 2Vs sin t dt T 2 0 2 2Vs 0.9Vs 5-9 Diode-Rectifier Bridge Analysis with AC-Side Inductance • Effect of Ls on the circuit operation (especially on the current commutation) • The dc-side can be represented by a constant dc current Id • Current commutation time: because of Ls, current change from negative to positive will not be instantaneous, requires some time for transition. The time interval required for such a transition is called the current commutation time. 5-10 Understanding Current Commutation • Let us consider a hypothetical circuit with two diodes and a sinusoidal voltage source. The output is represented by a constant dc current. • For comparison purpose, 5-11(b) shows vs, vd and is waveforms with Ls=0 5-11 Understanding Current Commutation (cont.) • Prior to time time t=0, vs is negative, D1 is open, Id circulates through D2 with vd =0 and is =0 • At t=0, vs starts to be positive, D1 is conducting, is builds up as long as 0 < is < Id. • D2 conducts as long as is < Id. • The current is through the inductor starts with a value of zero at the beginning of the commutation interval and ends up with a value of Id at the end. 5-12 Current Commutation Waveforms 5-13 Calculation of the Commutation Interval m dis V L 2V s sin t Ls dt 0 t m 2V s sin t d t Ls dis m Id 0 0 2V s sin t d t Ls dis Ls I d m Am 2V s sin t d t 2V s 1 cos m Ls I d 0 L s I d m cos 1 2V s 1 • If Ls=0, current commutation is instantaneous with m=0. • For a given frequency , the commutation interval m increases with Ls and Id and decreases with increasing voltage Vs. 5-14 Reduction in the Average Output Voltage Due to m With Ls =0, the average value of the output voltage, L 0 Vdos 1 2 2 2 V sin t d t V s 0.45V s s 2 0 2 With a finite Ls, the average value of the output voltage, L 0 Vdos 1 2V s sin t d t 2 m 1 1 m 2V s sin t d t 2V s sin t d t 2 0 2 0 L s I d 0.45V s 2 Reduction in the average value of the output voltage, L 0 Vdo Vdos L 0 Vdos L s I d 2 5-15 Current Commutation in Single-Phase Full-Bridge Rectifier • Prior to time t=0, vs is –ve, diodes 3 and 4 are conducting, Is=-Id • At t=0, vs starts to be +ve, diodes 1 and 2 become forward biased • All four diodes conduct during the commutation period, so Vd =0 5-16 Rectifier with a dc-Side Voltage • This time it is assumed that the dcside voltage is constant • Constant dc side voltage is equivalent to a large value of C • When vs exceeds Vd at qb, D1 and D2 begin to conduct. • The current reaches its peak at qp, beyond which vL is negative. • The current becomes zero at qf when the volt-sec areas A and B become equal and negative of each other. 5-17 Average Value of the DC Current When the current is flowing, the inductor voltage is given by did v L Ls 2V s sin t Vd dt Ls did 2V i d q q q qb qb q s sin t Vd d t 1 2V s sin q Vd dq Ls qb The average value Id of the dc current, qf i d q dq Id qb 5-18 Single-Phase Voltage Doubler Rectifiers • Applications where input line voltage is insufficient in magnitude to meet the dc output voltage requirement. • Can be used to avoid a voltage step-up device • Switch at 230V position and input 230V: acts like a full-bridge rectifier • Switch at 115V position and line input 115V: C1 is charged to 115V peak in +ve half and C2 to 115V peak in the –ve half, like 230V operation 5-19 Three-Phase, Full-Bridge Rectifier • In industrial applications, where three-phase power is available, it is preferable to use three-phase rectifier circuits because of … – Higher power handling capability – Lower ripple content in the waveforms 5-20 Three-Phase, Full-Bridge Rectifier: Redrawn • • Current flows through one diode from the top group and one from the bottom Diode with highest anode potential from the top and diode with lowest cathode potential from the bottom will conduct 5-21 Three-Phase, Full-Bridge Rectifier Waveforms • Six-pulse rectifier: Vd (=VPn-VNn) waveform consists of six segments per cycle • Each diode conducts for 120o • Conduct sequence: 1-2-3…. • Average dc output voltage, Vdo 1 6 2V cos t d t 3 6 LL t 6 6 3 2V LL 1.35V LL 5-22 Summary • Line-frequency diode rectifiers converts line-frequency ac into dc in an uncontrolled manner • To improve line current waveform, an inductor is added in series on the ac side • Various diodes rectifier circuits have been discussed • A large filter capacitor is used to achieve ripple-free dc output voltage • Voltage doubler rectifiers are used to double the dc output voltage magnitude • Three-phase rectifiers are preferable in most respects over the singlephase ones • Rectifiers inject large amounts of harmonic currents into the utility systems – remedies would have to be implemented 5-23 4 Three-phase diode rectifier. The diode rectifier shown in the figure below, supplies a DC machine, which has a constant load torque T = 100 Nm. The flux is held constant and Ka· = 1. This gives an armature current Ia = 100 A. The armature inductance of the machine, La, is so large that the armature current may be considered to be constant. The line voltage of the grid, VLL, is equal to 230V. Assume ideal grid, Ls=0. (a) (b) (c) (d) Sketch the armature voltage vd(t) and the line currents ir(t). Calculate the average dc voltage, Vd. Calculate the rms current in phase r, Ir,rms. List the advantages for a three-phase rectifier compared to a single-phase rectifier? 5-24 Solution: Example 4 5-25