SCR in Series

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 Series

If the input voltage is higher than the voltage rating of a single SCR, two or more SCR’s can be connected in series their forward blocking capability, as in any other device, the characteristics of two SCR’s (even two SCR’s of the same make and rating) are different. Two SCR’s in series divide the voltage between them in inverse proportion to their leakage currents. This leads to an unequal distribution of voltage across the two series SCR’s. Figure 1 shows the leakage current of two identical SCR’s, SCR1 and SCR2.

Each having a forward blocking voltage VBO. When two such SCR’s are connected in series, the same current flows through the two devices; however, the voltage across SCR1. Figure 1: The sharing of voltages between two series-connected SCR’s, SCR1 and SCR2

(V1) is higher than that across SCR2 (V2) because the leakage current for SCR1 is smaller than that of SCR2 for the same voltage. Therefore, the two SCR’s do not share the supply voltage equally. The maximum voltage that the SCR’s can block is only V1 + V2, not 2VBO. To use the full forward blocking capabilities of each SCR1 the forward blocking voltage must be equally distributed.

A nearly equal distribution of voltages during blocking is easily accomplished by connecting voltage equalizing resistor R1 and R2 (Figure 2(a)) in parallel with each SCR such that each parallel combination has the same resistance. However, when several SCR’s are connected in series, this method becomes uneconomical. A second approach, which permits a reasonably uniform distribution of voltages, is to use the same value resistance in parallel with each SCR. This allows a different but fixed voltage to appear across each SCR. In this arrangement (Figure 2(b)), the SCR with the lower leakage current will have a greater portion of the blocking voltage than the SCR with the higher leakage current. Figure 2: (a) Resistive Equalization (b) Inductive Equalization

Let us assume that the leakage current of SCR1 (ISCR1) is greater than the leakage current of SCR2 (ISCR2). SCR2 will be required to have the larger voltage (V2).

V2 = I2R

The voltage across the series combination is

Vm = V2 + I1R

Applying Kirchhoff’s current law (KCL) to the middle node

ISCR1 + I1 = ISCR2 + I2

ISCR1 + ISCR2 = I1 - I2 = ΔI

Or

I1 = I2ΔI

Vm = V2 + (I2 – ΔI) R

= V2 + I2R – ΔIR

= V2 + V2 – ΔIR

= 2 V2 – ΔIR

R = (2 V2 – Vm) /ΔI

Unequal voltage distribution among SCR’s in series also occurs during turn-ons and off. One SCR may turn ON or turn OFF before the other. The OFF SCR will then be subjected to the full source voltage. Shunt capacitors are very effective in equalizing voltages during switching. The capacitor also forces voltage sharing during sudden changes in supply voltage. A resistor is added in series with the capacitor to limit the current and di/dt (due to discharge of the capacitor through SCR) during turn ON. This CS and RS combination. Shown in Figure 3, is essentially a snubber circuit. A diode (D) connected across RS shunts it for forward voltages. Figure 3: RC equalization for SCR’s Connected in series

Example 1

The voltages across two SCR’s connected in series are 200V and 180V. calculate the value of the required equalizing resistor if the SCR’s have a maximum difference of 1mA in latching current. Also find the power dissipated by the blocking resistor.

Solution:

R = (2V2 –  Vm) / ΔI

= (2(200) – 380) / 1(10-3) = 20 kΩ

PR = (V21 / R) / (V22 / R)

= (1802 + 2002) / (20 (103) = 3.62 W