🚀 go-pugleaf

jon_sree@world.std.com (Jon Sreekanth) writes:

>My question is : when I turn off the transistor, what is the leakage
>current drawn from the + 100V supply ?

>I have a Sprague data book, which says, for MPSA42, Icbo is 100nA
>at Vcb of 200V. Is Icbo the same thing as the leakage Ic in the
>above circuit ? Sprague conveniently ignored to show their measurement
>circuit, or define their terms. 

>(To turn off the transistor, I have the choice of applying either 0V
>or -12V through the 1K resistor. -12V will break down the base-emitter
>junction, but the 1K prevents damage. I'll do whichever gets me
>lower leakage current from the 100V Supply) 

So, the leakage depends on how you turn the transistor off.  Icbo is the
leakage from collector to base; if the base were left _open_, you could
nominally multiply that leakage by the 'beta' of the transistor (grounded-
emitter current gain) to get the open-base, grounded-emitter leakage.  The
leakage will be relatively independent of the collector voltage, up to
breakdown.  If you arrange to suck all the base current out before it
gets to the base-emitter junction and starts getting amplified, you can
keep the leakage quite low.  I would _NOT_ recommend letting that 1k
resistor limit the reverse-bias base current, with -12 on the left end!!
Put a diode across the base-emitter junction of the transistor, and 
turning off to -12 would be fine (diode to keep base from going more than
~.7V negative).  Or, if you really hold the left end of the 1k to 0
volts, you will get most of the base leakage current through it and the 
collector leakage will be essentially Icbo.  Still, do be sure the
Vceo is at least 100 volts...(I don't have a book in front of me).