jon_sree@world.std.com (Jon Sreekanth) (02/14/91)
Hopefully this is a simple question, but I forgot my transistor theory
since it's been so long.
This is the circuit :
<--- Ic
__________ Vcc
|
> 300 ohm
<
|
|
1K ohm / C
|/ NPN
--- ^^^^^--- |\ MPSA42
\
input | E
|
____
--
Vcc is about 100V; the NPN transistor MPSA42 is in a conventional
common emitter circuit, collector resistance = 300 ohm, input
resistor (base resistor) is 1K ohm, emitter is grounded.
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)
Thanks much,
/ Jon Sreekanth
Assabet Valley Microsystems Fax and PC products
346 Lincoln St #722, Marlboro, MA 01752 508-562-0722
jon_sree@world.std.comgrege@gold.gvg.tek.com (Greg Ebert) (02/14/91)
[circuit deleted]
Icbo is collector-base leakage current with the emitter open. It is usually
specified for a 'large' collector-base voltage. It is also exponentially
related to temperature and approximately doubles every 8 degrees C (Ouch!).
Icbo is troublesome in the sense that when you connect the emitter (ie, use
the transistor in a real circuit), it will produce a collector-emitter current
approximately equal to (beta*Icbo). As collector current increases, so does
device heating, and thus Icbo, thus Ic... It is possible to construct a
circuit which will have sufficient positive thermal feedback to self-destruct.
But fortunately, you can shunt-away some/most Icb(o) by driving Vbe with
a voltage source, instead of leaving the base open. As suggested in the
original article, you can apply reverse Vbe bias without damaging the device
if you follow the spec. I did this with a 850V/75Amp (yes my friends, thats
AMPS, not mA or uA) device and its still quite healthy.
- - -
Though possibly off-subject, unclamped inductive loads (or improperly clamped)
are the most prevalent transistor-zappers in motor/inverter circuits. Have
respect for 1/2LI^2 ;if you don't provide a path for it to dissipate, it will
transmute your expensive device into a fuse.
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#######robf@mcs213k.cs.umr.edu (Rob Fugina) (02/15/91)
In article <1973@gold.gvg.tek.com> grege@gold.gvg.tek.com (Greg Ebert) writes: >device heating, and thus Icbo, thus Ic... It is possible to construct a >circuit which will have sufficient positive thermal feedback to self-destruct. > Would it be better to say that it's possible to construct a circuit which WON'T have sufficient positive thermal feedback to self-destruct? Me robf@cs.umr.edu
tomb@hplsla.HP.COM (Tom Bruhns) (02/15/91)
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).
jon_sree@world.std.com (Jon Sreekanth) (02/16/91)
In article <5170103@hplsla.HP.COM> tomb@hplsla.HP.COM (Tom Bruhns) writes: 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 ? ~.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). Thanks. That seems to settle it. I'll put the circuit together in the next few days. Vcbo and Vceo are quoted as 300V. Regards, / Jon Sreekanth Assabet Valley Microsystems Fax and PC products 346 Lincoln St #722, Marlboro, MA 01752 508-562-0722 jon_sree@world.std.com
rsd@sei.cmu.edu (Richard S D'Ippolito) (02/19/91)
In article <JON_SREE.91Feb13151509@world.std.com> Jon Sreekanth writes: >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. I hate to say this, but you need a better understanding of what you are reading -- Icbo IS a standard term whose definition you should know before reading a databook. Also, all of the measurement conditions (from which you can devine the circuit) are also given, namely, Vcb = 200V. It's like complaining that a recipe doesn't define 'tsp' -- if you don't know, you shouldn't be reading the recipe! Rich