Chopper With D.C Motor As Load
This topic shows the operation of the Step-Down chopper with D.C motor coupled to the mechanical load. The operation of it is classified into two categories i.e on time and off time. And the equation of the output current waveform is derived for both on time and off time of the d.c chopper.
Summary
This topic shows the operation of the Step-Down chopper with D.C motor coupled to the mechanical load. The operation of it is classified into two categories i.e on time and off time. And the equation of the output current waveform is derived for both on time and off time of the d.c chopper.
Things to Remember
1) The operation of chopper with D.C motor as Load can be categorized into two types based on on time and off time of thyristor.
2) During on time, the gate signal is provided to the gate terminal of the thyristor and it remains in conduction on forward biased state.Input voltage appears across the output of the motor.
3) During off time, thyristor is turned off by force commutation using external circuitry. And it acts as the open switch and input voltage isn’t supplied to the motor.
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Chopper With D.C Motor As Load
Chopper With D.C Motor As Load

Let Vdc = Input Voltage,
Ra = Armature Resistance,
L = Inductance of Armature Winding,
E = Back emf generated by the motor,
Th1 = Switching Elements
D1 = Freewheeling diode.
The operation of it can be classified into two categories:
a)During On time : The gate signal is provided to the gate terminal of the thyristor and it remains in conduction in the forward biased state.Input voltage appears across the output of the motor.
b)During off time : Thyristor is turned off by force commutation using external circuitry. And it acts as the open switch and input voltage isn’t supplied to the motor.
Mathematical Calculation:
1)During On Time

The equation can be written as:
$$\eqalign{
& Vdc - i\,Ra - {{Ldi} \over {dt}} - E = 0 \cr
& {{Vdc - E} \over {Ra}} = i\, + {L \over {Ra}}{{\,di} \over {\,dt}} \cr
& Suppose\;y = {{Vdc - E} \over {Ra}} \cr
& y - i = {L \over {Ra}}{{\,di} \over {\,dt}} \cr
& or,Ra\,dt = {{Ldi} \over {y - i}} \cr
& Integrating\;this\;equation,we\;get \cr
& {{Ra} \over L}t = \; - \ln (y - i)\; + \;k \cr
& From\;the\;figure,\;we\;can\;observe\;that\;at\;time\;t = 0,\;the\;current\;is\;Ia. \cr
& Therefore,\;k = \ln (y - IA.) \cr
& Subsituting\;k\;in\;above\;equation,\;we\;get \cr
& {{Ra} \over L}t = \; - \ln (y - i)\; + \;\ln (y - Ia) \cr
& {{Ra} \over L}t = \ln ({{y - Ia} \over {y - i}}) \cr
& solving\;it,we\;get \cr
& i = \;IA\;{e^{ - {{Ra} \over L}t}} + {{Vdc - E} \over {Ra}}(1 - {e^{ - {{Ra} \over L}t}}) - - - - - - - - - - - - (i) \cr} $$
2) During off Time
The load current for mode 2 can be found by solving the equation
$$\eqalign{
& i\,Ra + {{Ldi} \over {dt}} - E = 0(with\;initial\;condition\;Ib\,) \cr
& The\;value\;of\;IB\;can\;be\;{\mathop{\rm determined}\nolimits} \;by\;subsituting\;t\; = \;t2\;in\;eqn(i), \cr
& we\;get\;the\;equation\;when\;the\;switch\;is\;open: \cr
& i = \;IB{e^{ - {{Ra} \over L}t}}\; - {E \over R}(1 - {e^{ - {{Ra} \over L}t}}) \cr
& \cr} $$
Hence, in this way the equation for charging and discharging current can be determined. And the peak-to-peak ripple current can be determined as Idiff = IB - Ia .
References
1) Department of Radioelectronics. Power System b. n.d.—. Power System Blockset. n.d. 15 05 2016. <http://radio.feld.cvut.cz/matlab/toolbox/powersys/stud_24a.gif>
2).B Classes. CHOPERS(DC-DC Converters). n.d. 10 05 2016. <http://www.bclasses.in/images/material/eee/powerelectronics/newdcchoppers/46.jpg>
Lesson
DC Chopper
Subject
Electrical Engineering
Grade
Engineering
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