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Numerical Problems:
1.
The trigger circuit of a thyristor has a source voltage of 15V and the load line has a slope of -
120V per ampere. The minimum gate current to turn on the SCR is 25mA. Compute
i.
Source resistance required in the gate circuit
ii.
The trigger voltage and trigger current for an average gate power dissipation of 0.4
watts
Solution:
i.
The slope of load line gives the required gate source resistance.
From the load line,
series resistance required in the gate circuit is 120Ω
ii.
Here V
g
I
g
= 0.4W
For
the gate circuit E
s
= R
s
I
g
+ V
g
15 = 120I
g
+0.4/I
g
120I
g
2
–
15 I
g
+ 0.4 = 0
Its solution gives Ig = 38.56mA or 86.44 mA
V
g
=
= 10.37V
V
g
=
= 4.627V
So choose the value for Ig which gives less voltage Ig = 86.44 mA and V
g
= 4.627V from
minimum gate current of 25mA.
2.
For an SCR the gate-cathode characteristic has a straight line slope of 130.
For trigger source
voltage of 15V and allowable gate power dissipation of 0.5 watts,
compute the gate source
resistance.
3.
SCRs with a rating of 1000V and 200A are available to be used in a string to handle 6kV and 1kA.
Calculate the number of series and parallel units required in case de-rating factor is 0.1 and 0.2
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4.
It is required to operate 250A SCR in parallel with 350A SCR with
their respective on state
voltage drops of 1.6V and 1.2V. Calculate the value of resistance to
be inserted in series with
each SCR so that the share the total load of 600A in proportion to their current ratings.
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