1. Simulate the circuit long enough (about 5 time constants) to capture enough information about the. i s R. S v C. Figure 1: Problem PS3.

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1 Problem PS3.1 Consider the circuit shown in Figure 1 where the switch is closed at t =. Prior to closing the switch, the capacitor voltage is zero, (i.e. v C ( ) = ). This is the circuit that was analyzed in Problem PS1.1. We will now analyze the behavior of this system using a simulator of your choice. For the simulation, assume that = 12V, R = 1kΩ, C = 1µF. 1. Simulate the circuit long enough (about 5 time constants) to capture enough information about the waveforms. PSfrag replacements 2. Produce hardcopies of waveforms for capacitor voltage v C (t) and current i C (t). i s R S v C i C C Figure 1: Problem PS3.1 Solution First, here s the netlist for the circuit. ps3_1.cir - Switched RC Circuit Vs 1 12 ; dc voltage source * S and Vsw model a switch that turns on at t= S sx ; voltage controlled switch Vsw 5 pulse( 1 1u 1u 1 1) ; control voltage for switch S R 2 3 1k ; 1 kohm resistance C 3 1u (ic=) ; 1 uf capacitor with initial voltage=v * switch model (turn-on voltage, turn-off voltage, on resistance, off resistance).model sx vswitch (von=.9, voff=.1, ron=1m, roff=1meg).tran 1u 1m 1u uic.probe.end Running this circuit in PSpice produces the waveforms shown in Figure 2. Based on our solutions from Problem Set 1, the capacitor voltage and current are given by v C (t) = (1 e t/rc ) u(t) = 12(1 e 1t ) u(t) V i C (t) = C dv C(t) = dt R et/rc u(t) = 12 e 1t u(t) ma where time constant of the circuit is τ = RC = 1ms and u(t) is the unit step function. Examining Figure 2 shows that these are precisely the waveforms that were computed. The waveforms shown in Figure 2 reach their steady state values in about five time constants (5τ = 5ms) which is what we would have expected. V. Caliskan 1 September 27, 22

2 Date/Time run: 9/16/2 2:33:38 16mA ps3_1.cir - Switched RC Circuit (D) ps3_1.dat Temperature: 27. 1mA capacitor current A 16V I(C) 1V capacitor voltage SEL>> V s 2ms 4ms 6ms 8ms 1ms V(3) Time Vahe Caliskan, ScD MIT Lab for Electromagnetic and Electronic Systems Figure 2: Simulation results for Problem PS3.1. Problem PS3.2 Consider the dc-excited LC circuit shown in Figure 3 where all of the elements are ideal. The switch is closed at t =. Prior to closing the switch, the capacitor voltage and inductor current are both zero, (i.e. v C ( ) =, i L ( ) = ). This is the circuit that was analyzed in Problem PS1.3. We will now analyze PSfrag replacements the behavior of this system using a simulator of your choice. For the simulation, assume that = 24V, L = 1µH, C = 1µF. R 1. Simulate the circuit long enough to capture enough information about the waveforms. 2. Produce hardcopies of waveforms for the capacitor voltage v C (t) and inductor current i L (t). i s L i c S i L v C C Figure 3: Problem PS3.2 Solution First, here s the netlist for the circuit. ps3_2.cir - dc-excited LC circuit Vs 1 24 ; dc voltage source * S and Vsw implement a switch that turns on at t= S sx Vsw 5 pulse( 1 1u 1u 1 1) L 2 3 1u (ic=) ; 1 uh inductor with zero initial current V. Caliskan 2 September 27, 22

3 C 3 1u (ic=) ; 1 uf capacitor with zero initial voltage * switch model (turn-on voltage, turn-off voltage, on resistance, off resistance).model sx vswitch (von=.9, voff=.1, ron=1m, roff=1meg).tran 1u 1m 1u uic.probe.end Running this circuit in PSpice produces the waveforms shown in Figure 4. Based on our solutions from Problem Set 1, the capacitor voltage and current are given by i L (t) = Z sin(ω t) (1) = 24 sin(1t) A (2) v C (t) = (1 cos(ω t)) (3) = 24 (1 cos(1t)) V (4) where Z = L/C = 1Ω and ω = 1/ LC = 1rad/s. The period of the oscillations is simply /ω = 628µs. Examining Figure 4 verifies that our solutions are correct. Date/Time run: 9/16/2 22:15:21 5V ps3_2.cir - dc-excited LC circuit (E) ps3_2.dat Temperature: 27. capacitor voltage V 3A V(3) inductor current A SEL>> -3A s.2ms.4ms.6ms.8ms 1.ms I(L) Time Vahe Caliskan, ScD MIT Lab for Electromagnetic and Electronic Systems Figure 4: Simulation results for Problem PS3.2. V. Caliskan 3 September 27, 22

4 Problem PS3.3 Consider the dc-excited LCD (inductor-capacitor-diode) circuits shown in Figure 5. The switch is closed at PSfragt replacements =. Prior to closing the switch, the capacitor PSfrag replacements voltage and inductor current are both zero, (i.e. v C ( ) =, i L ( ) = ). This is the circuit that was analyzed in Problem PS2.1. We will now analyze the behavior of this systemr using a simulator of your choice. For ther simulation, assume that = 24V, L = 1µH, C = 1µF. The netlists for the circuits are given below. 1. Simulate the circuit long enough to capture enough information about the waveforms. 2. Produce hardcopies of waveforms for the capacitor voltage v C (t) and inductor current i L (t). i c i s S D L i L v C C i c i s S i L L v C C D Solution for the series diode circuit First, here s the netlist for the circuit. Figure 5: Problem PS3.3 ps3_3a.cir - dc-excited LC circuit with series diode Vs 1 24 ; dc voltage source * S and Vsw implement a switch that turns on at t= S sx Vsw 5 pulse( 1 1u 1u 1 1) D 2 3 dx ; almost-ideal diode L 3 4 1u (ic=) ; 1 uh inductor with zero initial current C 4 1u (ic=) ; 1 uf capacitor with zero initial voltage * diode model.model dx d (rs=.1m, n=1u) * switch model (turn-on voltage, turn-off voltage, on resistance, off resistance).model sx vswitch (von=.9, voff=.1, ron=1m, roff=1meg).tran 1u 1m 1u uic.probe.end Running this circuit in PSpice produces the waveforms shown in Figure 6. Based on our solutions from Problem Set 2, the capacitor voltage and current are given by i L (t) = Z sin(ω t) (5) = 24 sin(1t) A (6) v C (t) = (1 cos(ω t)) (7) = 24 (1 cos(1t)) V (8) where Z = L/C = 1Ω and ω = 1/ LC = 1rad/s. The period of the oscillations is simply /ω = 628µs. But as shown in the Problem Set 2 solutions, these voltages and currents only last for /ω = 314µs. After this time the voltage stays at 2 = 48V and the current stays at. V. Caliskan 4 September 27, 22

5 ps3_3a.cir - dc-excited LC circuit with series diode Date/Time run: 9/16/2 22:17:13 (F) ps3_3a.dat Temperature: 27. 5V capacitor voltage SEL>> V 3A V(4) 2A inductor current 1A -A s.2ms.4ms.6ms.8ms 1.ms I(L) Time Vahe Caliskan, ScD MIT Lab for Electromagnetic and Electronic Systems Figure 6: Simulation results for Problem PS3.3 series diode circuit. Solution for the parallel diode circuit First, here s the netlist for the circuit. ps3_3b.cir - dc-excited LC circuit with parallel diode Vs 1 24 ; dc voltage source * S and Vsw implement a switch that turns on at t= S sx Vsw 5 pulse( 1 1u 1u 1 1) L 2 3 1u (ic=) ; 1 uh inductor with zero initial current C 3 1u (ic=) ; 1 uf capacitor with zero initial voltage D 3 dx ; almost-ideal diode * diode model.model dx d (rs=.1m, n=1u) * switch model (turn-on voltage, turn-off voltage, on resistance, off resistance).model sx vswitch (von=.9, voff=.1, ron=1m, roff=1meg).tran 1u 1m 1u uic.probe.end Running this circuit in PSpice produces the waveforms shown in Figure 7. Based on our solutions from Problem Set 2, the diode never turns on and the results are identical to the system shown in PS3.2. V. Caliskan 5 September 27, 22

6 ps3_3b.cir - dc-excited LC circuit with parallel diode Date/Time run: 9/16/2 22:18:49 (G) ps3_3b.dat Temperature: 27. 5V capacitor voltage SEL>> V 3A V(3) inductor current A -3A s.2ms.4ms.6ms.8ms 1.ms I(L) Time Vahe Caliskan, ScD MIT Lab for Electromagnetic and Electronic Systems PSfrag replacements Figure 7: Simulation results for Problem PS3.3 parallel diode circuit. PSfrag replacements Problem PS3.4 Consider the half-wave rectifier circuits shown in Figure 8. The ac voltage source is given by sin(ωt). These are the circuits C that were analyzed in Problems PS2.2 andc PS2.3. We will now analyze the behavior of these rectifiers using a simulator of your choice. For the simulation, assume that = 17V, ω = (6)rad/s, R = 1.2Ω. For the inductance use L = 1mH for the single diode rectifier and L = 5mH for the two diode rectifier. Thei s netlists for the circuits are given below. i s v C v C 1. Simulate i c the circuit long enough to capture enoughi c information about the waveforms. 2. Produce hardcopies of waveforms for the inductor current i L (t) and filter drive voltage v d (t). S S D L D 1 L D v v i L s d R v s D v d 2 D 1 i L R D 2 Solution for half-wave rectifier with inductive load First, here s the netlist for the circuit. Figure 8: Problem PS3.4 ps3_4a.cir - half-wave rectifier with inductive load Vs 1 sin( 17 6 ) ; ac voltage source 17*sin(2*pi*6*t) D 1 2 dx ; almost-ideal diode L 2 3 1m (ic=) ; 1 mh inductor with zero initial current R ; 1.2 Ohm load resistor * diode model V. Caliskan 6 September 27, 22

7 .model dx d (rs=.1m, n=1u) * switch model (turn-on voltage, turn-off voltage, on resistance, off resistance).model sx vswitch (von=.9, voff=.1, ron=1m, roff=1meg).tran 1u 5m 1u uic.probe.end The results of the simulation are shown in Figure 9. These waveforms are consistent with our predictions from Problem Set 2 Solutions. Note that since the inductor current is reset to zero in every cycle, there is no startup transient. ps3_4a.cir - half-wave rectifier with inductive load Date/Time run: 9/16/2 23:12:53 (H) ps3_4a.dat Temperature: 27. 2V filter drive voltage 1V V -5V 15A V(2) 1A inductor current 5A SEL>> -A s 1ms 2ms 3ms 4ms 5ms I(L) Time Vahe Caliskan, ScD MIT Lab for Electromagnetic and Electronic Systems Figure 9: Simulation results for Problem PS3.4 half-wave rectifier with inductive load. Solution for half-wave rectifier with inductive load and circulation diode First, here s the netlist for the circuit. ps3_4b.cir - half-wave rectifier with inductive load and circulation diode D2 Vs 1 sin( 17 6 ) ; ac voltage source 17*sin(2*pi*6*t) D1 1 2 dx ; almost-ideal diode D2 2 dx ; almost-ideal circulating diode L 2 3 5m (ic=) ; 5 mh inductor with zero initial current R ; 1.2 Ohm load resistor * diode model.model dx d (rs=.1m, n=1u) * switch model (turn-on voltage, turn-off voltage, on resistance, off resistance).model sx vswitch (von=.9, voff=.1, ron=1m, roff=1meg).tran 1u 1m 1u uic.probe.end V. Caliskan 7 September 27, 22

8 The results of the simulation are shown in Figure 1. These waveforms are consistent with our predictions from Problem Set 2 Solutions. Since we have circulation diode the inductor current is not reset to zero in every cycle, so we do see a startup transient. In other words it takes time for the inductor current to build up to its average value. The results that we gave in Problem Set 2 Solution assumed that the startup transient had passed and we were already in periodic steady state. ps3_4b.cir - half-wave rectifier with inductive load and circulation diode D2 Date/Time run: 9/16/2 23:16:51 Temperature: 27. (I) ps3_4b.dat 2V filter drive voltage 1V SEL>> V 5A V(2) inductor current A s 2ms 4ms 6ms 8ms 1ms I(L) Time Vahe Caliskan, ScD MIT Lab for Electromagnetic and Electronic Systems Figure 1: Simulation results for Problem PS3.4 half-wave rectifier with inductive load and circulation diode. V. Caliskan 8 September 27, 22

9 Problem PS3.5 Consider the half-wave and full-wave rectifier circuits shown in Figure 11. Each circuit has a commutating inductance L c on the ac side. The inductive lowpass filter that is on the output of each of the rectifiers has been approximated by a constant dc current source of value. Assume that all of the diodes in the rectifiers are ideal. For each of these circuits do the following: 1. Write down all possible circuit structures (based on the diode states) and determine which of these structures are valid (i.e., do not violate any basic laws). 2. Write down all of the constraints on circuit variables for the valid subcircuits. 3. Calculate and sketch the source current i s (t) and rectifier output voltage v d (t). 4. Calculate and sketch v d, the average value of v d (t), as a function of the dc load current. 5. What would be the average value of the rectifier output voltage v d if the commutating inductance L c = (i.e., a short circuit)? 6. Calculate the power factor of the source for the case when L c =. PSfrag replacements v s D 1 L c D 1 L c i s vd v i s D s v d 2 D 2 rag replacements R D 3 D 4 v C Figure 11: Problem PS3.5 Solution for half-wave rectifier with constant-current load 1. With the presence of two diodes, there are four possible states that the circuit can possess. These states are the following: (1) D 1 on, D 2 off; (2) D 1 on, D 2 on; (3) D 1 off, D 2 on; (4) D 1 off, D 2 off. All of these states except (4) is valid. State (4) is invalid because it violates Kirchoff s current law (KCL). The three valid subcircuits for the half-wave rectifier are shown in Figure 12. Note that the presence of the commutating inductance results in subcircuit (2) being a valid structure. If L c was not present on the ac side, subcircuit (2) would violate Kirchoff s voltage law (KVL). D 1 D 2 i s L c i D1 L c i D1 L c vd1 v Lc i s i s v s v D2 v d v s i v d v s v d D2 i D2 (1) D 1 on, D 2 off (2) D 1 on, D 2 on (3) D 1 off, D 2 on Figure 12: Subcircuits for the half-wave rectifier with commutation inductance L c constant-current load. V. Caliskan 9 September 27, 22

10 2. For the subcircuits corresponding to states (1), (2) and (3) in Figure 12, one can write some inequalities involving the diode voltages and currents. Subcircuit (1) with D 1 on and D 2 off: In this subcircuit i s = i D1 = > and v D2 <. When v s becomes negative, v D2 = v s will become positive and D 2 will become forward-biased and we will move to subcircuit (2). Subcircuit (2) with D 1 on and D 2 on: In this subcircuit i D1 i D2 =, i D1 >, i D2 >, v Lc = v s. We will enter this circuit mode when v s crosses and will stay in this circuit mode until the load current transitions from one diode to other. Once the transition has occurred, one of the diodes will carry all of the current while the other will have zero current and will turn off. The turning off of the diode will move us to either to subcircuit (1) or (3) depending of the direction of the current diversion. Subcircuit (3) with D 1 off and D 2 on: In this subcircuit v D1 < and i D2 = >. When v s becomes positive, v D1 = v s will become positive and D 1 will become forward-biased and we will move to subcircuit (2). 3. The sequence of the subcircuits is In subcircuit (2), the voltage v s is directly applied the commutating inductance L c as shown in Figure 12. Since a sinusoidal voltage is directly applied across L c, the current i s through the inductor will also be a sinusoid. The current i s will either increase or decrease depending on the polarity of v s. Taking into account the sequence of the subcircuits and the expected sinusoidal transitions, the typical waveforms for the output voltage v d and the diode current waveforms are shown in Figure 13. Upon closer examination of Figure 13, we see that it takes u radians for the diode currents to transition from to or from to. To determine the angle u, we can take a look at any of the transitions. For example, let us take a look at the interval < ωt < u. At ωt =, i D1 = i s = and i D2 =. Applying the initial condition on the inductor current i s (ωt = ) = to subcircuit (2) gives: i s (ωt) = i s () = ωt ωt v Lc (ωt)d(ωt) ωl c sin(ωt)d(ωt) = ωl c (1 cos(ωt)) This expression is valid over the angular interval < ωt < u. We further know from examining Figure 13 that i s = i D1 evaluated at ωt = u should give. Evaluating the above expression at ωt = u and solving for u yields: i s (ωt = u) = = ωl c (1 cos( u)) = (1 cos(u)) ωl ( c u = cos 1 1 ωl ) c The angle u is usually referred to as the commutation angle. 4. As shown in Figure 13, when diode D 2 conducts, the output voltage v d is zero. In the absence of the commutating inductance (i.e. L c = ), the voltage v d was a half-rectified sinusoid. With the commutating V. Caliskan 1 September 27, 22

11 Problem PSfrag replacements Set 3 Solutions Z Vs Z v d (t) /2 3/2 5/2 3 7/2 4 ωt i s = i D1 u u i L (t) I L i L (ωt = ) v s (t) i D2 u /2 u 3/2 u 5/2 3 3u 7/2 4 ωt /2 3/2 5/2 3 7/2 4 ωt u u u 3u Figure 13: Output voltage v d and diode currents i D1 and i D2 for the half-wave rectifier with constant-current load. inductance, the voltage v d is zero for an additional interval u when the current is transitioning from D 2 to D 1. The results in an overall reduction of the average output voltage v d. The average value of v d can be calculated as v d = 1 v d (ωt)d(ωt) = = (1 cos(u)) = = V ( s 1 ωl ) c 2 u ( 2 ωl c sin(ωt)d(ωt) ) Note that the term / is the average value of the output voltage in the absence of the commutation inductance. However, for a given value of the commutation inductance, as the load current is increased, average output value v d decreases. The plot of the average output voltage versus output current is shown in Figure Average value of the output voltage in the absence of the commutating inductance is /. This was already shown in the Lecture 7 Notes. V. Caliskan 11 September 27, 22

12 v d PSfrag replacements 2 ωl c Figure 14: Average output voltage v d as a function of load current for a half-wave rectifier. 6. The power factor for the case with zero commutation inductance can be found by examining the waveforms shown in Figure 3 of Lecture 7 Notes. Or we can take our results from above and set the commutation angle u =. k p = p(t) = v s(t)i s (t) = v d(t) = v d (t) S,rms I s,rms,rms I s,rms,rms I s,rms So to figure out the power factor all we need to do is to figure out v d (t),rms and I s,rms. These are all pretty easy to calculate. v d = 1 1 I s,rms =,rms = 2 v d (ωt)d(ωt) = i 2 1 sd(ωt) = sin(ωt)d(ωt) = I 2 d d(ωt) = 2 Putting these values into the power factor expression yields k p = Vs 2 2 = V. Caliskan 12 September 27, 22

13 Solution for full-wave rectifier with constant-current load 1. With the presence of four diodes, there are sixteen possible subcircuits that the circuit can possess. These states and their validity are given in the following table: rag replacements Subcircuit D 1 D 2 D 3 D 4 valid? (1) off off off off, violates KCL, no path for to flow (2) off off off on, violates KCL, no path for to flow (3) off off on off, violates KCL, no path for to flow (4) off off on on, violates KCL, no path for to flow (5) off on off off, violates KCL, no path for to flow (6) off on off on, if D 2 and D 4 are on, both D 1 and D 3 must be on (7) off on on off (8) off on on on, if D 2 and D 4 are on, both D 1 and D 3 must be on (9) on off off off, violates KCL, no path for to flow (1) on off off on (11) on off on off, if D 1 and D 3 are on, both D 2 and D 4 must be on (12) on off on on, if D 1 and D 3 are on, both D 2 and D 4 must be on (13) on on off off, violates KCL, no path for to flow (14) on on off on, if D 2 and D 4 are on, both D 1 and D 3 must be on (15) on on on off, if D 1 and D 3 are on, both D 2 and D 4 must be on (16) on on on on (commutation (7) (1)) The only valid subcircuits are (7), (1) and (16). These subcircuits for the full-wave rectifier are shown in Figure 15. L c D 1 D 2 D 1 D 2 D 1 D 2 L c v s i s v d i s i I s d v s v d I v d d v s I d D 3 D 4 D 3 D 4 D 3 D 4 L c (7) (1) (16) Figure 15: Valid subcircuits for the full-wave rectifier. 2. For the subcircuits corresponding to subcircuits (7), (1) and (16) in Figure 15, one can write some inequalities involving the diode voltages and currents. Subcircuit (7) with D 2, D 3 on and D 1, D 4 off: This subcircuit is valid for v s < during which i s =. Subcircuit (1) with D 1, D 4 on and D 2, D 3 off: This subcircuit is valid for v s > during which i s =. Subcircuit (16) with D 1, D 2, D 3 & D 4 on: This subcircuit is valid during the zero crossings of the source voltage v s (t) when the source current i s (t) transitions between and in both directions. 3. The sequence of the subcircuits is In subcircuit (16), the voltage v s is directly applied the commutating inductance L c. Since a sinusoidal voltage is directly applied across L c, the current i s through the inductor will also be a sinusoid. The current i s will either increase or decrease V. Caliskan 13 September 27, 22

14 depending on the polarity of v s. Taking into account the sequence of the subcircuits and the expected sinusoidal transitions, the typical waveforms for the output voltage v d and the diode current waveforms are shown in Figure 16. Upon closer examination of Figure 16, we see that it takes u radians for the source currents (and also the diode currents) to transition from to or from to. To determine the angle u, we can take consider any of the transitions. For example, let us take a look at the interval < ωt < u when we are in subcircuit (1). At ωt =, i D1 = i D4 = i s = and i D2 = i D3 =. Applying the initial condition on the inductor current i s (ωt = ) = to subcircuit (1) gives: i s (ωt) = i s () = ωt ωt v Lc (ωt)d(ωt) ωl c sin(ωt)d(ωt) = ωl c (1 cos(ωt)) This expression is valid over the angular interval < ωt < u. We further know from examining Figure 16 that i s = i D1 = i D4 evaluated at ωt = u should give. Evaluating the above expression at ωt = u and solving for u yields: i s (ωt = u) = = ωl c (1 cos( u)) = (1 cos(u)) ωl ( c u = cos 1 1 2ωL ) c The angle u is usually referred to as the commutation angle. 4. As shown in Figure 16, when all diodes conduct during commutation, the output voltage v d is zero. In the absence of the commutating inductance (i.e. L c = ), the voltage v d was a half-rectified sinusoid. With the commutating inductance, the voltage v d is zero for an additional interval u when the current is transitioning to/from and. This results in an overall reduction of the average output voltage v d. The average value of v d can be calculated as v d = 1 v d (ωt)d(ωt) = = (1 cos(u)) = = 2V ( s 1 ωl ) c u ( 2 2ωL c sin(ωt)d(ωt) ) u sin(ωt)d(ωt) Note that the term 2 / is the average value of the output voltage in the absence of the commutation inductance. However, for a given value of the commutation inductance, as the load current is increased, average output value v d decreases. The plot of the average output voltage versus output current is shown in Figure 17. V. Caliskan 14 September 27, 22

15 PSfrag replacements Z Vs Z ECE 445 Analysis and Design of Power Electronic Circuits v d (t) i L (t) I L i L (ωt = ) v s (t) /2 3/2 5/2 3 7/2 4 ωt i s = i D1 i D2 i s (t) u u u u /2 u 3/2 u 5/2 3 3u 7/2 4 ωt Figure 16: Output voltage v d and the source current i s (t) for the full-wave rectifier with constant-current load. 5. Average value of the output voltage in the absence of the commutating inductance is 2 /. This was already shown in the Lecture 8 Notes. 6. Let us try to apply the power factor calculations to the full-bridge rectifier with a constant-current load as shown in Figure 18 where the source voltage v s (t) = sin(ωt). Also illustrated in the figure are the source current i s (t) and the rectifier output voltage v d (t). We now have all of the information necessary to calculate the power factor of the source. First let us calculate the rms values of the source variables. The rms value of the source voltage v s (t) and source current i s (t) are given by The average power supplied by the source is given by,rms = 2 and I s,rms = (9) p(t) = v s (t)i s (t) (1) Since the rectifier is lossless, all of the power supplied by the source is delivered to the load; therefore, average power can also be calculated as 1 T 2 T/2 2 p(t) = v d (t) = v d (t) = v d (t)dt = sin(ωt)dt = (11) T T Now the power factor can be calculated by substituting (9) and (11) into the expression for power factor k p = p(t) p(t) = = I 2 d S,rms I = (12) s,rms 2 V. Caliskan 15 September 27, 22

16 v d PSfrag replacements ωl c Figure 17: Average output voltage v d as a function of load current for a full-wave rectifier. rag replacements D 1 D 2 D 3 D 4 L c v s i s v d v d i s 3 4 ωt 3 4 ωt Figure 18: Full-bridge rectifier with a constant-current load. V. Caliskan 16 September 27, 22