Current Source Inverter (CSI)

Current Source Inverters (CSI), and its waveforms, will be described, the device used here is a thyristor. The Auto-Sequential Commutated mode of operation for this Inverter (ASCI), using thyristors, will be discussed, with waveforms. Then, the circuit and operation of three-phase CSI, along with relevant waveforms, will be presented.

In the current source inverter, the current is nearly constant. The voltage changes here, as the load is changed. In an Induction motor, the developed torque changes with the change in the load torque, the speed is constant, with no acceleration/deceleration. The input current in the motor also changes, with the input voltage being constant. So, the CSI, where current, but not the voltage, is the main point of interest, is used to drive such motors, with the load torque changing.

1-Phase Current Source Inverter

Current Source Inverter (CSI) diagram

A constant current source is assumed here, which may be realized by using an inductance of suitable value, which must be high, in series with the current limited dc voltage source. The thyristor pairs, Th1 and Th3, and Th2 and Th4, are alternatively turned ON to obtain a nearly square wave current waveform. Two commutating capacitors C1 in the upper half, and C2 in the lower half, are used.

Four diodes, D1–D4 are connected in series with each thyristor to prevent the commutating capacitors from discharging into the load. The output frequency of the inverter is controlled in the usual way, i.e., by varying the half time period, (T/2), at which the thyristors in pair are triggered by pulses being fed to the respective gates by the control circuit, to turn them ON, as can be observed from the waveforms given below the inductance (L) is taken as the load, in this case, the reason(s) for which need not be stated, is well known.

waveform of current source inverter

Operating Principle of Current Source Inverter

The operation is explained by two modes.

Mode 1:

Starting from the instant,t = 0- , the thyristor pair, Th2 and Th4, is conducting (ON), and the current (I) flows through the path, Th2, D2 load (L), D4, Th4, and source, I. The commutating capacitors are initially charged equally with the polarity as given, i.e. vC1 = vC2 = - VC0. This means that both capacitors have a right-hand plate positive and left-hand plate negative. If two capacitors are not charged initially, they have to pre-charged.

Starting from the instant,t = 0 , the thyristor pair, Th2 and Th4, is conducting (ON), and the current (I) flows through the path, Th2, D2  load (L), D4, Th4, and source, I. The commutating capacitors are initially charged equally with the polarity as given, i.e. vC1 = vC2 = – VC0. This means that both capacitors have a right-hand plate positive and left-hand plate negative. If two capacitors are not charged initially, they have to pre-charged.

At time, t = 0, thyristor pai, Th1, and Th3, is triggered by pulses at the gates. The conducting thyristor pair, Th2 and Th4, is turned OFF by application of reverse capacitor voltages. Now, thyristor pair, Th1, and Th3, conducts current (I). The current path is through Th1 ,C1  D2,  L , D4, C2, Th4, and source, I. Both capacitors will now begin charging linearly from (- VC0) by the constant current, I. The diodes, D2 and  D4, remain reverse biased initially.

when it is forward biased, is obtained by going through the closed path, abcda as

when it is forward biased, is obtained by going through the closed path, abcda as

It may be noted the voltage across load inductance, L is zero (0), as the current, I is constant.

It may be noted the voltage across load inductance, L is zero (0), as the current, I is constant

As the capacitor gets charged, the voltage VD1 across D1 increases linearly. At some time, say t1, the reverse bias across D1 becomes zero (0), the diode, D1 starts conducting. An identical equation can be formed for diode, D3 also. Actually, both diodes, D1 & D3, start conducting at the same instant,t1. The time t1 for which the diodes, D1 & D3 remain reverse biased is obtained by equating vD1 = -VCO + ((2 *I*t1)/C) = 0

The time is given by t1 = (C/(2*I)) VCO

The value of vC

The value of vC

The value of t1 obtained earlier. This means that the voltages across C1 and C2, varies linearly from -VCO to zero in time t1. Mode 1 ends, when t = t1, and vC = 0. Note that t1 is the circuit turn-off time for the thyristors.

Mode 2:

The circuit for this mode is shown in Fig Diodes, D2, and D4, which are already conducting, but at, t = t1 diodes, D1, and D3, get forward biased, and start conducting. Thus, at the end of time t1, all four diodes, D1- D4 conduct.

The circuit for this mode is shown in Fig Diodes, D2, and D4, which are already conducting, but at, t = t1 diodes, D1, and D3, get forward biased, and start conducting. Thus, at the end of time t1, all four diodes, D1– D4 conduct. As a result, the commutating capacitors now get connected in parallel with the load (L). For simplicity in analysis, the circuit is redrawn as

the commutating capacitors now get connected in parallel with the load (L). For simplicity in analysis, the circuit is redrawn as

The voltage balance equation is

The voltage balance equation of current source inverter

The total commutation interval is

The total commutation interval in current source inverter

At the end of the process, constant current flows in the path, Th1, D1, L, D3, Th3, and source, I. This continues till the next commutation process is initiated by the triggering of the thyristor pair, Th2 and Th4.

The process (mode 1) starts with the triggering of the thyristor pair, Th1, and Th3. Earlier, the thyristor pair, Th2 and Th4 were conducting. With the two commutating capacitors charged earlier with the polarity, the conducting thyristor pair, Th2 and Th4 turns off by the application of reverse voltage. Then, the voltages across the capacitors decrease to zero at a time, (end of mode I), as constant (source) current, I flow in the opposite direction.

Now Mode 2 starts as the diodes, D1 & D3, get forward biased, and start conducting. So, all four diodes D1– D4, conduct, and the load inductance, L are now connected in parallel with the two commutating capacitors. The current in the load reverses to the value –I, after time t2, (end of mode 2), and the two capacitors also are charged to the same voltage in the reverse direction, the magnitude remaining same, as it was before the start of the process of commutation (t = 0). It may be noted that the constant current, I flows in the direction as shown, a part of which flows in the two capacitors.

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