Power Electronics - Choppers

A chopper uses high speed to connect and disconnect from a source load. A fixed DC voltage is appped intermittently to the source load by continuously triggering the power switch ON/OFF. The period of time for which the power switch stays ON or OFF is referred to as the chopper’s ON and OFF state times, respectively.

Choppers are mostly appped in electric cars, conversion of wind and solar energy, and DC motor regulators.

Symbol of a Chopper

Classification of Choppers

Depending on the voltage output, choppers are classified as −

Step Up chopper (boost converter)

Step Down Chopper(Buck converter)

Step Up/Down Chopper (Buck-boost converter)

Step Up Chopper

The average voltage output (V_o) in a step up chopper is greater than the voltage input (V_s). The figure below shows a configuration of a step up chopper.

Current and Voltage Waveforms

V₀ (average voltage output) is positive when chopper is switched ON and negative when the chopper is OFF as shown in the waveform below.

Where

T_ON – time interval when chopper is ON

T_OFF – time interval when chopper is OFF

V_L – Load voltage

V_s – Source voltage

T – Chopping time period = T_ON + T_OFF

V_o is given by −

When the chopper (CH) is switched ON, the load is short circuited and, therefore, the voltage output for the period T_ON is zero. In addition, the inductor is charged during this time. This gives V_S = V_L

$Lfrac{di}{dt}=V_{S},$ $frac{Delta i}{T_{ON}}=frac{V_{S}}{L}$

Hence,$Delta i=frac{V_{S}}{L}T_{ON}$

Δi = is the inductor peak to peak current. When the chopper (CH) is OFF, discharge occurs through the inductor L. Therefore, the summation of the V_s and V_L is given as follows −

$V_{0}=V_{S}+V_{L},quad V_{L}=V_{0}-V_{S}$

But $Lfrac{di}{dt}=V_{0}-V_{S}$

Thus,$Lfrac{Delta i}{T_{OFF}}=V_{0}-V_{S}$

This gives,$Delta i=frac{V_{0}-V_{S}}{L}T_{OFF}$

Equating Δi from ON state to Δi from OFF state gives −

$frac{V_{S}}{L}T_{ON}=frac{V_{0}-V_{S}}{L}T_{OFF}$, $V_{S}left ( T_{ON}+T_{OFF} ight )=V_{0}T_{OFF}$

$V_{0}=frac{TV_{S}}{T_{OFF}}=frac{V_{S}}{frac{left ( T+T_{ON} ight )}{T}}$

This give the average voltage output as,

The above equation shows that V_o can be varied from V_S to infinity. It proves that the output voltage will always be more than the voltage input and hence, it boosts up or increases the voltage level.

Step Down Chopper

This is also known as a buck converter. In this chopper, the average voltage output V_O is less than the input voltage V_S. When the chopper is ON, V_O = V_S and when the chopper is off, V_O = 0

When the chopper is ON −

$V_{S}=left ( V_{L}+V_{0} ight ),quad V_{L}=V_{S}-V_{0},quad Lfrac{di}{dt}=V_{S}-V_{0},quad Lfrac{Delta i}{T_{ON}}=V_{s}+V_{0}$

Thus, peak-to-peak current load is given by,

$Delta i=frac{V_{s}-V_{0}}{L}T_{ON}$

Circuit Diagram

Where FD is free-wheel diode.

When the chopper is OFF, polarity reversal and discharging occurs at the inductor. The current passes through the free-wheel diode and the inductor to the load. This gives,

Rewritten as −$quad Lfrac{Delta i}{T_{OFF}}=V_{0}$

Equating equations (i) and (ii) gives;

$frac{V_{S}-V_{0}}{L}T_{ON}=frac{V_{0}}{L}T_{OFF}$

$frac{V_{S}-V_{0}}{V_{0}}=frac{T_{OFF}}{T_{ON}}$

$frac{V_{S}}{V_{0}}=frac{T_{ON}-T_{OFF}}{T_{ON}}$

The above equation gives;

Equation (i) gives −

$Delta i=frac{V_{S}-DV_{S}}{L}DT$, from $D=frac{T_{ON}}{T}$

$=frac{V_{S}-left ( 1-D ight )D}{Lf}$

$f=frac{1}{T}=$chopping frequency

Current and Voltage Waveforms

The current and voltage waveforms are given below −

For a step down chopper the voltage output is always less than the voltage input. This is shown by the waveform below.

Step Up/ Step Down Chopper

This is also known as a buck-boost converter. It makes it possible to increase or reduce the voltage input level. The diagram below shows a buck-boost chopper.

When the chopper is switched ON, the inductor L becomes charged by the source voltage V_s. Therefore, V_s = V_L.

Because −

$D=frac{T_{ON}}{T}$ and $f=frac{1}{T} .............................................. left ( iii ight )$

When the chopper is switched OFF, the inductor’s polarity reverses and this causes it to discharge through the diode and the load.

Hence,

$Lfrac{Delta i}{T_{OFF}}=-V_{0}$, thus $Delta i=-frac{V_{0}}{L}T_{OFF}................................left ( iv ight )$

Evaluating equation (iii) and (iv) gives −

$frac{DV_{S}}{Lf}=-frac{V_{0}}{L}T_{OFF}$, $DV_{S}=-DV_{S}=-V_{0}T_{OFF}f$

$DV_{S}=-V_{0}frac{T-T_{ON}}{T}=-V_{0}left ( 1-frac{T_{ON}}{T} ight )$, $V_{0}=-frac{DV_{S}}{1-D}$

Because $D=frac{T_{ON}}{T}=frac{T-T_{OFF}}{1-D}$

This gives,

$V_{0}=frac{DV_{S}}{1-D}$

D can be varied from 0 to 1. When, D = 0; V_O = 0

When D = 0.5, V_O = V_S

When, D = 1, V_O = ∞.

Hence, in the interval 0 ≤ D ≤ 0.5, output voltage varies in the range 0 ≤ V_O < V_S and we get step down or Buck operation. Whereas, in the interval 0.5 ≤ D ≤ 1, output voltage varies in the range V_S ≤ V_O ≤ ∞ and we get step up or Boost operation.