**Choppers **

A chopper is basically a dc to dc converter whose main function/usage is to create adjustable dc voltage from fixed dc voltage sources through the use of semiconductors.

**Types of choppers **

The main classification of the types of choppers is given in another post. Take a look –** TYPES OF CHOPPER CIRCUITS**

There are two types of choppers – AC and DC.

**AC Link Chopper**

In the case of an ac link chopper, first dc is converted to ac with the help of an inverter. After that, AC is stepped-up or stepped-down by a transformer, which is then converted back to dc by a diode rectifier. Ac link chopper is costly, bulky and less efficient as the conversion is done in two stages.

**DC Chopper **

A DC chopper is a static device that converts fixed dc input voltage to a variable dc output voltage directly. A chopper can be said as dc equivalent of an ac transformer as they behave in an identical manner. This kind of choppers are more efficient as they involve one stage conversion. Just like a transformer, a chopper can be used to step up or step down the fixed dc output voltage. Choppers are used in many applications all over the world inside various electronic equipments. A chopper system has a high efficiency, fast response and a smooth control.

**Principle of Chopper Operation **

A chopper can be said as a high speed on/off semiconductor switch. Source to load connection and disconnection from load to source happens in a rapid speed. Consider the figure, here a chopped load voltage can be obtained from a constant dc supply of voltage, which has a magnitude V_{s.} Chopper is the one represented by “SW” inside a dotted square which can be turned on or off as desired.

**Output Voltage and Current Waveforms **

Let us now take a look of the output current and voltage wave forms of a chopper. During the time period T_{on }the chopper is turned on and the load voltage is equal to source voltage V_{s}. During the interval T_{off }the chopper is off and the load current will be flowing though the freewheeling diode FD . The load terminals are short circuited by FD and the load voltage is therefore zero during T_{off}. Thus, a chopped dc voltage is produced at the load terminals. We can see from the graph that the load current is continuous. During the time period T_{on}, load current rises but during T_{off } load current decays .

Average load Voltage is given by

V_{0 } = T_{on}/ (T_{on} +T_{off}) * Vs = (T_{on}/T) V = A V_{s}………………(1.0)

T_{on} : on -time

T_{off} : off- time

T = T_{on} +T_{off}= chopping period

A = T_{on} /T = duty cycle

So we know that the load voltage can be controlled by varying the duty cycle A. equation 1.0 shows that the load voltage is independent of load current it can be also written as

V_{0} = *f*. T_{on }.V_{s}

*f*= 1/T = chopping frequency

**Step – up Choppers **

In the case of the chopper circuit (*Refer figure named – “chopper circuit”*) shown in beginning of this article, V_{0 } or the average output voltage is less than the input voltage V_{s }so this type of chopper is called a step down chopper. For a step-up chopper we can obtain an average output voltage V_{0 }greater than input voltage. Figure (a) shows the elementary form of a step-up chopper.

**Working Principle of a Step-up Chopper**

In step-up chopper a large inductor, L is in series with the source voltage V_{s}. This forms a closed path as shown in the figure (b). During the time period T_{on} the chopper is on the inductor stores energy. When the chopper is turned off the current is forced to flow through the diode and load for a time T_{off} and as the inductor current cannot die suddenly. When the current decreases the polarity of the emf induced in L is reversed. Fig (c). As a result the total voltage available across the load is given by the equation V_{0 }= V_{s }+ L (*di/dt*) . The voltage V_{0 }exceeds the source voltage and hence the circuit acts as a step-up chopper and the energy which is stored in L is released to the load.

**Voltage and current waveforms **

When the chopper is turned ON the current through the inductance L will increase from I_{1} to I_{2. }As the chopper is on the source voltage is applied to L that is v_{L = }V_{S } .

When the chopper is OFF, the KVL for the figure (c) can be written as

v_{L }– V_{0}+V_{s} =0 or v_{L} =V_{0 }-V_{s } where v_{L } is the voltage across L. Variation of source voltage v_{S ,} source current I_{S } , load voltage v_{0 } and load current i_{O } is sketched in the fig (d) . Let us assume that the variation of output current is linear, the energy input to inductor from the source, during the time period T_{on }, is

W* _{in}*= V

_{s }(I

_{1}+I

_{2}/2) T

_{on }

During the time T* _{off }* the chopper is off, so the energy released by the inductor to the load is

W* _{off}* = (V

_{0}-V

_{s})(I

_{1}+I

_{2}/2).T

_{off }Let us assume that the system is lossless, then the two energies say W* _{in }*and W

*are equal.*

_{off }So equating these two we will get

V_{s }(I_{1}+I_{2}/2) T_{on = }(V_{0}-V_{s})(I_{1}+I_{2}/2).T_{off }

V_{s }T_{on = }(V_{0}-V_{s}) T_{off }

V_{0}T* _{off = }*V

_{s}(T

*T*

_{off + }_{on}) = V

_{s }.T

V_{0} = V_{S} (T/T_{off}) = V_{S } (T/T-T_{on}) =V_{S} (1/(1-A) ………….(2.0)

From the equation 2.0 we can see that the average voltage across the load can be stepped up by varying the duty cycle. If the chopper in the figure (a) is always off, A=0 and V_{0}= V_{s}. If the chopper is always on, A =1 and V_{0} = infinity as we can see from the graph. In practical applications the chopper is turned on and off so that the required step-up average output voltage, more source voltage is obtained.

Figure shows variation of load voltage V_{0} with duty cycle .

**Application of Step-up Chopper**

Figure shows regenerative braking of dc motor.

The principle of step-up chopper can be used for the regenerative braking of DC motors. The armature voltage E_{a } is analogy to the V_{S } and voltage V_{0} is the dc source voltage. When the chopper is on the inductor L stores the energy and when it is off the inductor release the energy. If E_{a }/ (1-A) exceeds V_{0} , the dc machine will work as a dc generator and the armature current will flow in a direction opposite to the motoring mode. As the power now is flowing from dc machine to the source V_{0} it will cause regenerative breaking of the dc motor. Even at decreasing motor speeds, regenerative breaking can be provided as the motor armature E_{a} is directly proportional to the field flux and motor speed.

how do we get vs in dc chopper example?. is the vs a constant value

DC-DC convertors operating under discontinuous mode, when compared to their conduction mode will have

(a) high gain (vo/vin)

(b) low gain (vo/vin)

(c) gain depending on duty ratio

(d) none

please reply

It is assumed that energy supplied to inductor is equal to energy supplied by chopper when the chopper is in off state. How can we say that they are equal? Maybe before all the energy is supplied by the inductor, the chopper turns on again.

i need the how 12v dc will convert into the 400v dc either its possible with chopper circuit plz explain

can we use the chopper circulatory to boost the 12v dc into the 400v dc plz explain with circuit giagram

Is this same same as clipper and clamper circuits ?

It can be considered as an analogy to clipper and clamper circuit, the basic difference here is these are controlled by external signals and depend on the triggering time which is not in the case of clipper clamper circuit which is diode based.

But why at all we should use this power supply when new technology SMPS supplies are available in ample range.

The actual purpose of using this type of power supply is not explained here.

much more knowlge about electronics thankyou