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Impossible FET or BJT design

L

legg

Jan 1, 1970
0
Graham,

Fred is using ascii characters to map out a circuit. Drawing the
proper symbol for the transistor is dificult. I can't tell from the
ascii representation of the transistor if it is a JFET or MOSFET.

The application involves switching power supplies to an IC. I don't
want to use op-amps.

It won't be a simple linear or buck, as there's no way to get -15V at
some current from -15V, if there's a pass element controlling the
current.

If you'd be satisfied with a few millivolts less than -15, then it's a
trivial linear task.

see a.b.s.e. transfer1

RL
 
T

Tom Bruhns

Jan 1, 1970
0
I'm just guessing here, 'cuz you haven't said what you're trying to do
with the 100uA output, but I'm going to guess that it's for an on/off
control pin on a switcher module or some such...at least I've been
faced with 'zactly that problem (and solved it with a very similar
circuit as what has been posted). Does the output have to stop at
-10Vmax and -15Vmin? Or is -10 the minimum allowable for that state?
Can it be -0.05V instead? If it's to a control pin, is the current
from that pin known to be in one direction? If all that's true, you
may be able to simplify further, and probably not need a connection to
the +5 at all: drive the base w/ a voltage divider between the input
and -15, and also replace the 20k w/ a short if the output is OK at
nom. 0V. Presumably, you don't need _exactly_ -15 and -10V outputs;
what is the allowed range, and what it the output current max and min
at each? You could probably use a (depletion-mode, of course)
P-channel JFET, source grounded, gate driven directly by the logic
input, drain driving that 20k-10k divider (or maybe even just a 10k
pulldown). Note that a 10k pulldown in the original circuit or with
the jfet will allow a 1 volt drop with 100uA: the output could be only
-14 in that case. Though a jfet would eliminate a couple resistors,
unfortunately, use of jfets is deprecated.

This is such an easy task that I'm surprised that you posted with
"impossible" in the subject...the table of desired values immediately
tells you you need an inversion, and you get that with a
common-emitter/common-source stage. Then it's just a matter of where
you tie the emitter/source, and how you drive the base/gate to get the
device to switch. Where you tie the emitter/source also gives a strong
hint which polarity device to use: is the e/s above or below the
output range you want to use? Things like, "I need to sink some
current, but don't have to source any," or "I need to source current
but don't have to sink any more than a couple microamps" can affect
how you do the job.

Cheers,
Tom
 
F

Fred Bloggs

Jan 1, 1970
0
Tom said:
I'm just guessing here, 'cuz you haven't said what you're trying to do
with the 100uA output, but I'm going to guess that it's for an on/off
control pin on a switcher module or some such...at least I've been
faced with 'zactly that problem (and solved it with a very similar
circuit as what has been posted). Does the output have to stop at
-10Vmax and -15Vmin? Or is -10 the minimum allowable for that state?
Can it be -0.05V instead? If it's to a control pin, is the current
from that pin known to be in one direction? If all that's true, you
may be able to simplify further, and probably not need a connection to
the +5 at all: drive the base w/ a voltage divider between the input
and -15, and also replace the 20k w/ a short if the output is OK at
nom. 0V. Presumably, you don't need _exactly_ -15 and -10V outputs;
what is the allowed range, and what it the output current max and min
at each? You could probably use a (depletion-mode, of course)
P-channel JFET, source grounded, gate driven directly by the logic
input, drain driving that 20k-10k divider (or maybe even just a 10k
pulldown). Note that a 10k pulldown in the original circuit or with
the jfet will allow a 1 volt drop with 100uA: the output could be only
-14 in that case. Though a jfet would eliminate a couple resistors,
unfortunately, use of jfets is deprecated.

People usually scream latch-up with the slightest little substrate
current, but your circuit would look something like this:
View in a fixed-width font such as
Courier.
 
J

joseph2k

Jan 1, 1970
0
Pooh said:
Why the need for fets ? This is a perfect application for an op-amp.

Graham

Not to be too picky, but how many (rail to rail) op-amps do you know of that
produce good outputs when driven to the rail.
 
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