Current Commutated Chopper is a type of chopper in which commutation of thyristor used in the circuit take place due to oscillatory current through it. This article explains current commutated chopper along with its circuit diagram, working principle and various waveforms.

**Circuit Diagram:**

The circuit diagram of a current commutated chopper is shown below. In this diagram, T1 is the main thyristor. The commutation circuitry comprises of TA (the auxiliary thyristor), Capacitor C, Inductor L and Diode D1 & D2.

FD is the free-wheeling diode and R_{C} is the charging resistor. To make the analysis simple, following assumptions are made:

- Load current is constant
- SCRs and diodes are ideal switch
- Charging Resistor R
_{C}is very large and can be treated as open circuit during commutation. - Current i
_{c}, i_{T1}, i_{fd}and i_{o}are treated as positive when these are in the arrow directions marked. Similarly, voltages v_{c}, v_{T1}, v_{TA}and v_{o}are taken positive with the polarities as marked in the circuit diagram.

**Working Principle of Current Commutated Chopper:**

The working principle of current commutated chopper is based on current commutation technique of thyristor. The energy required for commutation comes from the charged capacitor C. This capacitor is first charged to source voltage V_{s} by charging resistor R_{C}. Once the capacitor C is charged to V_{s}, main thyristor T1 is turned ON so that load voltage v_{o} becomes equal to source voltage V_{s} and load current i_{o} = I_{o}. With the turning ON of T1, the commutation circuitry remains inactive. It comes into service when the auxiliary thyristor TA is fired.

The working principle of current commutated chopper can be divided into five modes just to have better understanding.

**Mode-1:**

At time t = 0, main SCR T1 is fired which results in load voltage and current to be equal to V_{s} and I_{o} respectively. This is because, the load is directly connected to source through T1. The equivalent circuit diagram for Mode-1 is shown below.

To start commutation of main thyristor T1, auxiliary thyristor TA is turned ON (say at t=t_{1}). This results in an oscillatory circuit comprising of C, L and TA. The voltage and current through this oscillatory circuit vary sinusoidally.

The magnitude of oscillatory current is given as

During the time interval (t_{2}-t_{1}), i_{c} and v_{c} vary sinusoidally through half a cycle (Refer waveform section). In this mode, the current through the oscillatory circuit is maximum when the voltage across the capacitor reduces to zero. At t_{2}, the current through the oscillatory circuit tend to reverse in the auxiliary thyristor TA and hence, it gets naturally commutated. At t_{2}, v_{c} = -V_{s} as shown in the waveform. This means, the lower plate of capacitor is positively charged while the upper plate is negatively charged. Note that, T1 remains unaffected during this mode and hence load current and voltage will remain to be I_{o} & V_{o} respectively.

**Mode-2:**

As TA is turned off at t_{2}, oscillatory current begins to flow through C, L, D2 and T1. Refer the equivalent circuit diagram for this mode shown below.

It should be noted that the current will flow through T1 not through D1. This is because the D1 is reversed biased by a small voltage drop across the conducting thyristor T1. Therefore, after t_{2}, i_{c} would pass through T1 not through D1.

In thyristor T1, i_{c} is in opposition to the load current io so that i_{T1} = (I_{o} – i_{c}). At some time t_{3} when i_{c} rises to Io, the current through the main thyristor will reduce to zero and hence it will be commutated i.e. turned off at t = t_{3}. *Since the oscillatory current through T1 turns it off, it is called current commutated chopper.*

During this mode, load voltage remains V_{s} through T1. For this mode, t_{2}<t<t_{3}.

**Mode-3:**

As T1 is turned off at t=t_{3}, ic becomes more that Io. After t_{3}, ic supplies load current Io and the excess current (i_{c} – I_{o}) is conducted through D1. This mode of operation is depicted in the figure below.

The voltage drop in D1 keeps T1 reversed biased for (t_{4}-t_{3}) = t_{c}; this is shown in the waveform of v_{T1}. At t_{4}, if v_{c} becomes more than V_{s}, FD comes into conduction otherwise Mode-4 would follow. During Mode-3, when i_{c} is at its peak value, voltage across capacitor becomes equal to zero. After this peak, capacitor voltage reverses and its upper and lower plate becomes positive & negative respectively.

**Mode-4:**

At t = t_{4}, capacitor current ic reduces to zero. This results in i_{D1} = 0 and diode D1 is turned off. After t_{4}, a constant load current I_{o} begins to flow through capacitor C, L and D2. Since current through capacitor is constant, it begins to charge linearly till voltage across it becomes equal to source voltage V_{s}. Note that, current is constant (equal to I_{o}) during this period i.e. (t_{5}-t_{4}).

The equivalent circuit diagram for Mode-4 operation of current commutated chopper is shown below.

As D1 is turned off at t_{4}, v_{T1} = v_{TA} = v_{c}; this is shown as ab in the waveform for v_{c}, v_{T1} and v_{TA}. Now the load voltage v_{o} = V_{s} – v_{c} = V_{s} – voltage ab at t_{4}. At t_{5}, v_{c} = V_{s}, therefore load voltage reduces to zero at this moment. During the time interval (t_{5}-t_{4}), v_{c} increases linearly, therefore load voltage v_{o} decreases linearly to zero during this time interval.

**Mode-5:**

At t_{5}, the capacitor is actually overcharged to a voltage somewhat more than the source voltage Vs. Therefore, free-wheeling diode FD gets forward biased and starts to conduct the load current Io at t_{5}. Load voltage reduces to zero at t_{5} as discussed in Mode-4. Mode-5 operation of current commutated chopper is shown below.

As i_{c} is not zero at t_{5}, the capacitor C is still connected to the load through V_{s}, C, L and D2. As a consequence, C is overcharged by the transfer of energy from L to C. At t_{6}, capacitor current becomes zero and the voltage across it becomes more than the source voltage.

During (t_{6}-t_{5}), capacitor current and current through free-wheeling diode feeds the load i.e. i_{c}+i_{fd} = I_{o}. From t_{5} onwards, i_{o} freewheels through FD. As i_{c} is zero and D2 is open circuited, C now discharges through R_{C} for the freewheeling interval of chopper. After t_{5}, v_{T1} remains constant at V_{s}, because V_{s} reaches T1 terminal through FD. AT t=T, the main SCR is again triggered and the Mode-1 to Mode-5 is repeated again.

**Waveforms of Current Commutated Chopper:**

Various waveforms for current commutated chopper are shown below.

**Advantages of Current Commutated Chopper:**

Following are the advantages of current commutated chopper:

- Commutation is reliable so long as the load current is less than the peak commutating current.
- Capacitor is always charged with correct polarity.
- Auxiliary thyristor is naturally commutated as its commutating current passes through zero value in the oscillatory circuit formed by L & C.