SCRs are connected in parallel to fulfill the ___________ demand
Number of devices connected in parallel can carry huge amounts of current.
The term used to measure the degree of utilization of SCRs connected in series & parallel is
String Efficiency = Rating of the whole string/(rating of one SCR x number of SCRs)
To have maximum possible string efficiency
Having similar ratings does not mean they have similar charc.
For a string voltage of 3300 V, let there be six series connected SCRs each of voltage 600V. Then the string efficiency is
String efficiency = 3300/(6 x 600) = 98.54.
The measure of reliability of string is given by the factor
DRF is derating factor given by the above expression.
When an extra SCR is connected in series with a string
DRF = 1 – String efficiency
String Efficiency = Rating of the whole string/(rating of one SCR x number of SCRs)
Extra SCR will reduce the string efficiency which in turn increase the DRF.
The most practical way of obtaining a uniform distribution of series connected SCRs is to
For uniform distribution of voltage across series connected SCRs, a resistor of value R in parallel with each series connected SCR.
3 SCRs are connected in series. The string efficiency is 91%. SCRs 1, 2 & 3 have leakage currents 4 mA, 8 mA & 12 mA. Which SCR will block more voltage?
The SCR with lower leakage current block more voltage.
Two parallel connect SCRs have same voltage drop (Vt) having rated current = 2I_{1}. SCR1 carries a current of I_{1}=2.6 A whereas SCR2 carries a current of I_{2}=1.4 A. Find the string efficiency.
The total current would be I1+I2 & rated current is 2I_{1}
String efficiency = (I_{1}+I_{2})/2I_{1}.
SCRs with a rating of 1000 V & 200 A are available to be used in a string to handle 6 KV & 1 KV. Calculate the number of series & parallel units required in case the derating factor is 0.1. (Round off the fraction to the greatest & nearest integer)
DRF = 1S.E
Therefore
0.1 = (16000/1000Ns) = (11000/200Np)
Ns = 6.6 = 7(say)
Np = 5.5 = 6(say)
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