Principle Of Step-up Chopper L + D + V Chopper C L O A D V O 1
Step-up chopper is used to obtain a load voltage higher than the input voltage V. The values of L and C are chosen depending upon the requirement of output voltage and current. When the chopper is, the inductor L is connected across the supply. The inductor current rises and the inductor stores energy during the time of the chopper, t. 2
When the chopper is off, the inductor current is forced to flow through the diode D and load for a period, t OFF. The current tends to decrease resulting in reversing the polarity of induced EMF in L. Therefore voltage across load is given by d V V L i. e., V V O O dt 3
A large capacitor C connected across the load, will provide a continuous output voltage. Diode D prevents any current flow from capacitor to the source. Step up choppers are used for regenerative braking of dc motors. 4
Expression For Output Voltage Assume the average inductor current to be during and OFF time of Chopper. When Chopper is Voltage across inductor L V Therefore energy stored in inductor Where = V.. t t period of chopper. 5
When Chopper is OFF (energy is supplied by inductor to load) Voltage across Energy supplied by inductor where t OFF L V V O L V V t OFF period of Chopper. Neglecting losses, energy stored in inductor L = energy supplied by inductor L O OFF 6
Vt V V t V O Where O OFF V t OFF VO V T t t t OFF T T = Chopping period or period of switching. 7
T t t V O OFF 1 V t 1 T 1 VO V 1 d t Where d duty cyle T 8
For variation of duty cycle ' d ' in the range of 0 d 1 the output voltage will vary in the range V V O V O 9
Performance Parameters The thyristor requires a certain minimum time to turn and turn OFF. Duty cycle d can be varied only between a min. & max. value, limiting the min. and max. value of the output voltage. Ripple in the load current depends inversely on the chopping frequency, f. To reduce the load ripple current, frequency should be as high as possible. 10
Problem A Chopper circuit is operating on TRC at a frequency of 2 khz on a 460 V supply. f the load voltage is 350 volts, calculate the conduction period of the thyristor in each cycle. 11
V 460 V, V = 350 V, f = 2 khz dc Chopping period Output voltage T T V dc 1 f 1 210 t T 3 V 0.5 msec 12
Conduction period of thyristor t T V V dc t t 3 0.510 350 460 0.38 msec 13
Problem nput to the step up chopper is 200 V. The output required is 600 V. f the conducting time of thyristor is 200 sec. Compute Chopping frequency, f the pulse width is halved for constant frequency of operation, find the new output voltage. 14
V 200 V, t 200 s, V 600V dc Solving for V V dc T t 600 200 T T T T 200 10 6 T 300s 15
Chopping frequency f f 1 T 1 30010 6 Pulse width is halved 3.33KHz 6 20010 t 100 2 s 16
Frequency is constant f 3.33KHz 1 T 300s f T Output voltage = V T t 6 30010 200 300 Volts 6 300 100 10 17
Problem A dc chopper has a resistive load of 20 and input voltage V S = 220V. When chopper is, its voltage drop is 1.5 volts and chopping frequency is 10 khz. f the duty cycle is 80%, determine the average output voltage and the chopper on time. 18
V 220 V, R 20, f 10 khz d V S ch t 0.80 T = Voltage drop across chopper = 1.5 volts Average output voltage t V V V T V dc S ch dc 0.80 220 1.5 174.8 Volts 19
Chopper time, Chopping period, 1 T f 1 T 3 1010 Chopper time, t t t dt 0.800.110 t dt 3 0.1 10 secs 100 μsecs 3 3 0.08 10 80 μsecs 20
Problem n a dc chopper, the average load current is 30 Amps, chopping frequency is 250 Hz, supply voltage is 110 volts. Calculate the and OFF periods of the chopper if the load resistance is 2 ohms. 21
30 Amps, f 250 Hz, V 110 V, R 2 dc 3 Chopping period, T 410 4 msecs 1 1 f 250 Vdc dc & Vdc dv R dv dc R dcr 30 2 d 0.545 V 110 22
Chopper period, t Chopper OFF period, OFF OFF OFF dt t T t t t 3 0.545 4 10 2.18 msecs 410 2.1810 3 3 3 1.82 10 1.82 msec 23
A dc chopper in figure has a resistive load of R = 10 and input voltage of V = 200 V. When chopper is, its voltage drop is 2 V and the chopping frequency is 1 khz. f the duty cycle is 60%, determine Average output voltage RMS value of output voltage Effective input resistance of chopper Chopper efficiency. 24
Chopper i 0 + V R v 0 V 200 V, R 10, Chopper voltage drop V 2V d 0.60, f 1 khz. ch 25
Average output voltage V d V V dc ch V dc 0.60 200 2 118.8 Volts RMS value of output voltage O V d V V ch V O 0.6 200 2 153.37 Volts 26
Effective input resistance of chopper is R R Output power is i dc i V S dc Vdc 118.8 11.88 Amps R 10 V V 200 16.83 11.88 S V dc 0 0 V V 2 dt 2 dt 1 v0 1 ch PO dt dt T R T R 27
P P nput power, P O O i P O 1 T 1 T d V 0 0 2 0.6 200 2 dt dt V R 10 O ch Vi dt V V 2 V R ch 2352.24 watts dt 28
P P O O Chopper efficiency, P dv V R 0.6200 200 2 O P i 100 V 10 ch 2376 watts 2352.24 100 99% 2376 29
Problem A chopper is supplying an inductive load with a free-wheeling diode. The load inductance is 5 H and resistance is 10.. The input voltage to the chopper is 200 volts and the chopper is operating at a frequency of 1000 Hz. f the /OFF time ratio is 2:3. Calculate Maximum and minimum values of load current in one cycle of chopper operation. Average load current 30
L 5 H, R 10, f 1000 Hz, V 200 V, t : t 2 : 3 Chopping period, T t t t 1 1 1 msecs f 1000 OFF 2 3 2 3 t OFF OFF 31
T t t 5 T t 3 OFF 2 T t t OFF 3 OFF t OFF 3 T 5 OFF T 3 1 10 3 0.6 msec 5 32
Duty cycle, t T t t max OFF 1 0.6 10 0.4 msec 3 3 t 0.410 d 0.4 3 T 110 Maximum value of load current is given by drt L V 1 e E RT R R L 1 e 33
Since there is no voltage source in the load circuit, E = 0 max V R 1 e 1 e drt L RT L 200 1 e 1 e 0.4101 10 5 max 10 1 10 10 5 3 3 34
0.810 3 3 max 2 10 max Minimum value of load current with E = 0 is given by min 1 e 20 1 e 8.0047A V R e e drt L RT L 1 1 35
min 10 10 1 10 5 max 3 0.4101 10 200 e 5 1 3 e 1 Average load current dc dc 2 min 7.995 A 8.0047 7.995 8 A 2 36
Problem A chopper feeding on RL load is shown in figure, with V = 200 V, R = 5, L = 5 mh, = 1 khz, d = 0.5 and E = 0 V. Calculate Maximum and minimum values of load current. Average value of load current. RMS load current. Effective input resistance as seen by source. RMS chopper current. f 37
V = 200 V, R = 5, L = 5 mh, f = 1kHz, d = 0.5, E = 0 Chopping period is T 1 1 3 f 110 3 1 10 secs Chopper i 0 + R FWD L v 0 E 38
39 3 3 3 3 max 0.5 5 1 10 5 10 max 5 1 10 5 10 0.5 max 1 Maximum value of load current is given by 1 1 200 1 0 5 1 1 40 24.9 A 1 drt L RT L V e E R R e e e e e
Minimum value of load current is given by min drt L V e 1 E RT R R L e 1 0.551 10 3 3 510 200 e 1 min 5 1 10 5 min 1 e 510 3 3 1 0.5 e 1 40 15.1 A e 1 0 40
Average value of load current is 1 2 dc 2 for linear variation of currents 24.9 15.1 dc 20 A 2 RMS load current is given by 1 2 2 2 max min min min max min ORMS 3 41
O RMS 1 2 2 2 24.9 15.1 15.1 15.1 24.9 15.1 3 96.04 228.01 147.98 20.2 A O RMS 3 RMS chopper current is given by ch d 0.5 20.2 14.28 A O RMS 1 2 42
Effective input resistance is S R = Average source current i S S S dc 0.520 10 A Therefore effective input resistance is R i V d V 200 20 10 S 43