Re: Discrete Schottky?
From: Ken Smith (kensmith_at_green.rahul.net)
Date: 12/03/04
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Date: Fri, 3 Dec 2004 16:01:10 +0000 (UTC)
In article <18ebe27.0412030735.60f39e82@posting.google.com>,
Watson A.Name <veryfree123@hotmail.com> wrote:
>IIRC, with the LSTTL chips, the schottky TTLs use a schottky diode
>between base and collector to prevent the collector from going below a
>certain voltage into saturation, thus speeding up the turn-on. I've
>never seen a discrete circuit using this, but is it a workable
>solution to put a 1N5711 schottky signal diode between the base and
>collector of a 2N2369A or 2N3904 for example, to prevent it from
>saturating, and help speed up its switching? I was thinking of
>driving an IR LED with a signal that's fairly quick, and I don't think
>that an open collector TTL chip would have enough current. But then
>maybe parallel open collectors? Or what?
In discrete land, the voltage swing is often too large for just a
schottky. In those cases a "Baker clamp"[1] is used. I have used the
schottky clamp method at 10V and had it work.
[1] Baker clamp is like this
Simple version:
D1
-------+------->!--------+----
! !
! D2 !/
------->!------!
!\ e
!
It takes away base drive when the collector gets down to one diode drop.
D2 is often the E-B junction of another transistor. This circuit works ok
but the speed and capacitance of D1 is an issue during turn off.
All that said, you want to use a MOSFET to drive the IR LED. They switch
quite fast with no extra work. The Supertex TN0604 driven by HC logic
like this:
+12V
+-------+
! !
\ \
/R1 /R2
! \
V !
--- !
! !
!!----+-------
-----!!
!!--
!
GND
R1 sets the diodes current.
R2 speeds up the discharging of capacitances during turn off. It needs to
be about 100-1K depending on how fast you need.
-- -- kensmith@rahul.net forging knowledge
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