Whereisshutdownthatatthestartofanyturnoff-say,atA1-C1
musthavenochargeonit.Attheendoftheturnoff(atB),C1hasslowed
up
the
voltage
rise
time
but
has
accumulated
a
voltage
2Vdc
next
(neglectingforthemomenttheleakageinductancespike).
AtthestartofthenextturnoffatA2,C1mustagainhavenovoltage
acrossit.Thus,atsometimebetweenBandA2,C1mustbedischarge.It
isdischargedintervalCtoA2byresistorR1.WhenQ1turnsonatC,the
topendofC1goesimmediatelytogroundandC1dischargesthoughQ1
andR1.
Thus,onceC1hasbeenselectedlargeenoughtoyieldasufficiently
slowed
up
collector
voltage
raise
time,R1
is
chosen
to
discharge
C1
to
within5percentofitisfullchargeintheminimumton.Or
3R1C1=
t
on(min)
(11.1)
But
if
C1
accumulates
a
voltage
2V
dc
at
each
turnoff
,
it
stores
an
energyof0.5C1(2Vdc)2joules.IfitdissipatesthatenergyinR1during
eachontime,PowerdissipationinR1inwatts(forTinseconds)is
PD
R1
=0.5C1(2V
dc
)
2
/T
(11.2)
ItwillbeseeninthefollowingsectionthatpowerdissipationinR1
given
by
Eq.11.2
seriously
limits
how
much
collector
voltage
rise
time
canbeslowedupbyincreasingC1.