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'[PIC]: Tutorial of sorts - analog design with tran'
2000\06\20@055356 by Russell McMahon

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Just a flag to the pure PIC people.
(No PICs in this case but ...)

The challenge thread

    Re: [EE]: Analog Design Challenge - explain this

and my "answer"

   Re: [EE]: The Answer :-) - Analog Design Challenge - explain this

cover the use of transistors in unusual circumstances which may reach out
and bite you some-day (as it did to me in a  commercial design).
If you are interested in analog design with transistors you may want to read
the thread.
In summary -

i - A transistor with opposite to normal voltage applied to C-E and base
open will breakdown at some voltage - typically around 9 volts or so

ii - Adding a base to emitter resistor greatly reduces this voltage and
softens the knee.

iii - Adding instead a collector - base resistor does similar but with
somewhat less severity.

iv - IT MUST NOT BE ASSUMED THAT A REVERSE BIASED BIPOLAR TRANSISTOR WILL
NOT CONDUCT THROUGH A REVERSE BIASED JUNCTION !!!!!!!!!!!!!

v - Adding a series diode in series with the transistor's collector is a
good idea when reverse biased C-E junction will occur if well defined
operation is desired,


RM

2000\06\20@131018 by Dan Michaels

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Russell McMahon wrote:
.........
>
>i - A transistor with opposite to normal voltage applied to C-E and base
>open will breakdown at some voltage - typically around 9 volts or so
>

I wonder this isn't acually referring to the "base-emitter breakdown
voltage", which is fairly low, eg, < 10v, for general-purpose transistors.
In your scenario above, the C-B junction diode will be forward-biased
and the B-E reverse-biased, so the applied reverse C-E voltage is mainly
seen at the B-E junction.

I would not have thought this would be a problem with your ckt,
unless Vin = 10V, and the battery totally discharged. However, as I
suggested previously, a diode from C to RL would fix this.
============

Also, I wonder - if Q1 actually provides as much charge current to
the battery as you say, about 150 mA while in inverse mode, then
maybe the problem is too much power dissipation.

Vec * Ic = ~4v * 150 mA = ~0.6W.

Certainly, a lot for a little plastic BJT. Possibly, BJTs
operating in pathological inverse mode don't like to see so much
power dissipation, since the dissipation will now be in the
[labelled] emitter region [generally tiny area] instead of the
[labelled] collector region [large area].

- DanM

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