SCR In Parallel
The maximum power that can be controlled by a single SCR is determined by its rated forward current and rated forward blocking voltage. To maximize one of these two ratings, the other must be reduced. Although SCR’s are currently available with very high voltage ratings, in many applications, such as transmission lines, the required voltage rating exceeds the voltage that can be provided by a single SCR. Then it is necessary to connect two or more SCR’s in series. Similarly, for very high current applications, SCR’s must be connected in parallel. For high-voltage, high-current applications, series-parallel combinations of SCR’s are used.
SCR's in Parallel
When the load current exceeds the rating of a single SCR, SCR’s are connected in parallel to increase their common current capability. If SCR’s are not perfectly matched, this results in an unequal sharing of current between them. Figure 1 shows the V-I characteristics of two SCR’s, SCR1 and SCR2. The ratings of the SCR’s are the same. When these SCR’s are connected in parallel, they will have equal voltage drops VSCR across them. However, due to their mismatch in characteristics. SCR2 is carrying the rated current (I2), while SCR1 is carrying a current I1, which is much less than its rated value. The total rated current of the parallel connection is only I1 + I2 instead of 2l2.
(a) two SCRs in parallel
(b) on-state characteristics
Matched-pair SCR’s are generally available for parallel connection, but they are very expensive. With unmatched SCR’s equal current sharing is enforced by adding a low-value resistor or inductor in series with each SCR. Forced current sharing using equal-value resistors is shown in Figure 2. The basic requirement is to make current I1 close to I2; a maximum difference of 20% is accept across SCR2, the value of R can be obtained from.
I1R + V1 = I2R + V2
R = (V1 – V2) / (I2 – I1)
Equalization using resistor is inefficient due to the extra power loss in the resistor. Moreover, resistors do not compensate for unequal SCR turn ON or turn OFF times. One of the SCR’s may turn ON or turn OFF before the other. In either case the ON SCR must carry the full load current momentarily until both are switched, and it can easily be damaged due to overloading. Figure 3 shows a center tapped reactor in which the SCR carrying the greater current will induce a voltage proportional to the imbalance in current and with the polarity shown. Voltage in reactor L1 opposes the flow of current, and voltage in L2 causes an increase of current flow through the SCR that originally carried the lower current. As a result, a balanced current distribution is achieved. The reactors, although expensive, are more efficient.
Since the main problem in high current applications is excessive junction temperature, SCR’s used in parallel must be derated by at least 15%. For example, two 50 A SCR’s connected in parallel may carry only 85 (47.5 + 47.5) A. in addition, the SCR’s are normally mounted on a common heat sink to equalize temperatures.
For the circuit in Figure 2, find the value of the current-sharing resistors if VS = 1000 V, RL = 1Ω, and the rated current of each SCR is 700A. also find the power loss in the resistors.
Solution:
Choose
V1 = 1.55V
V2 = V1 = 1.45V
IL = 1000/1 = 1000A
Let one thyristor carry
I1 = 400A
Therefore
I2 = 1000 – 400 = 600A
R = (1.55 – 1.45) / 200 = 0.5 mΩ
PLOSS = (600)2(0.0005) + (400)2(0.0005) = 260 W
Example 2
In the circuit in Figure 2, the source voltage is 500V and the load resistance is 5Ω. Two SCR’s with the characteristics shown in Figure 1(b). each rated at 70A, are connected in parallel to share the load current. Find the value of the resistor that will give the proper current sharing. Also find the voltage drop across the parallel combination and the power dissipated by the current sharing resistors.
Solution:
The load current is
IL = 500 / 5 = 100 A
Let us choose a current difference of 15% between the two SCR’s, so
I2 = 50 + 15 = 65 A
I1 = 50 – 15 = 35 A
From the characteristic curve
V1 = 1.6 V
V2 = 1.5 V
Therefore
R = (V1 – V2) / (I2 – I1) = (1.6 – 1.5) / (65 – 35) = 3.3 mΩ
The voltage drops across the two resistors are
I1R = 35 (0.0033) = 0.12 V
I2R = 65 (0.0033) = 0.22 V
The voltage drop across the parallel branch is
V1 + I1R = V2 + I2R
=1.6 + 0.12 = 1.5 + 0.22 = 1.72V
The power loss in the resistor is
I21 R = 352 (0.0033) = 4.1 W
I22 R = 652 (0.0033) = 14.1 W