Power Electronics - DC to DC Converters Complete Study Notes

Choppers:

A chopper is a static circuit that converts fixed DC input voltage to a variable DC output voltage directly. Chopper circuits offer smooth control, high efficiency, fast response and regeneration.

The power semiconductor devices used for a chopper circuit can be

1. Force commutated Thyristor
2. Power BJT
3. Power MOSFET
4. GTO
5. IGBT

In chopper circuits, these power semiconductor devices are represented by a switch ‘SW’ or ‘CH’ with an arrow.

                  Figure : Representation of Chopper Switch ‘SW’

There are mainly two types of chopper:

1. Step-Down chopper or Buck converter
2. Step-Up chopper or Boost Converter
3. Step Up-Down chopper or Buck-Boost Chopper

Principle of Operation of Step-Down Chopper:

                       Figure : Step Down Chopper

During T_on period, chopper is ON and load voltage is equal to source voltage V_s. During T_off period, chopper is off, therefore load current flows through the freewheeling diode ‘FD’. Hence, load terminals are short circuited and load voltage becomes zero during turn off period T_off. During T_on, load current rises whereas during T_off load current decays.

                  Figure : Step Down Chopper (Different Waveforms)

Average load voltage V₀ is given by

Where,  T_on = on time; T_off = off-time

T = T_on + T_off = chopping period

This load voltage can be controlled by varying duty cycle α.

V₀ = f. T_on .V_s

NOTE: Variation of T_on means adjustment of pulse width, this is also called Pulse-Width-Modulation scheme.

Average output current,

Power Input = Power delivered

V_sI_s = V_oI_o

By putting the expression of output voltage:

Principle of Operation of Step-Up Chopper:

When the average output voltage V₀ greater than input voltage V_s, then the chopper is called step up chopper.

                   Figure : Step Up Chopper

Case I: When the chopper CH is ON, then the circuit reduces to

During turn on period, inductor will store energy.

Case II: When chopper CH is OFF, then the circuit reduces to

During turn off time, inductor current cannot die down instantaneously. Therefore, this current is forced to flow through the diode and load for a time T_off. As the current tends to decrease, polarity of the emf induced in L is reversed as shown above.

Therefore, voltage across the load is given by

It is clearly seen from the above equation that V₀ exceeds the source voltage V_s.

When CH is ON, current through inductor 'L' increase from I₁ to I₂ and when CH is off, current would fall from I₂ to I₁.

When CH is ON, source voltage is applied to L i.e. V_L = V_S.

When CH is off, KVL gives

V_L – V₀ + V_S = 0

V_L = (V₀ – V_S)

The energy input to inductor from the source, during the period T_on, is

W_on = (Voltage across L) x (Average current through L) x T_on

During the time T_off, when chopper is off, the energy released by inductor to the load is

W_off = (voltage across L) x (Average current through L) x T_off

Considering the system to be lossless, these two energies will be equal

1. V_S. T_on = (V₀ – V_S) × T_off

V₀T_off = V_S(T_on + T_off) = V_S .T

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