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Discrete Schottky?

W

Watson A.Name

Jan 1, 1970
0
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?
 
K

Ken Smith

Jan 1, 1970
0
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.
 
R

Rene Tschaggelar

Jan 1, 1970
0
What speed did you have in mind ?
Perhaps with a FET ?

Rene
 
J

Jim Thompson

Jan 1, 1970
0
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?

(1) How "quick" does ANY circuit need to be when flashing an LED?

(2) A 2N2369A is gold-doped, thus does not need to be
Schottky-clamped to prevent charge storage.

(3) The 1N5711 data sheet shows only 2pF of capacitance, so it should
be OK.

...Jim Thompson
 
J

John Larkin

Jan 1, 1970
0
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?

How fast? How much current?

John
 
L

Larry Brasfield

Jan 1, 1970
0
....
(1) How "quick" does ANY circuit need to be when flashing an LED?

Maybe it is signaling to something not a person?
(2) A 2N2369A is gold-doped, thus does not need to be
Schottky-clamped to prevent charge storage.

Gold doped BJTs still exhibit charge storage. That doping
reduces carrier lifetime, so such transistors can switch off
in a few 10's of nS, but preventing saturation will still speed
up the turn-off.

If speed mattered, I would use a small MOSFET. They are
inexpensive and good for interface to logic.

....
 
J

Joerg

Jan 1, 1970
0
Hi Larry,
If speed mattered, I would use a small MOSFET. They are
inexpensive and good for interface to logic.

That depends. If the control signal originates from 3.3V logic or even
lower levels a FET becomes rather sluggish without some major level
shifting or toroid transformers. I have been in that pickle more than
once and reverted to bipolar. Besides the good old 2N2369 there are some
other fine RF transistors such as the BFS17A.

Regards, Joerg
 
C

clicliclic

Jan 1, 1970
0
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?

Have a look the Hewlett-Packard Application Note 1066: "Fiber-Optic
Solutions for 125 MBd Data Communication Applications at Copper Wire
Prices". They suggest direct drive by three 74HCTQ00 gates in
parallel. They easily achive rise/fall times below 10ns; the problem
then is the LED, not the driver.

Martin.
 
J

Jim Thompson

Jan 1, 1970
0
...

Maybe it is signaling to something not a person?


Gold doped BJTs still exhibit charge storage. That doping
reduces carrier lifetime, so such transistors can switch off
in a few 10's of nS, but preventing saturation will still speed
up the turn-off.

If speed mattered, I would use a small MOSFET. They are
inexpensive and good for interface to logic.

The 2N2369 at 10mA load turns off in 30ns. With the 1N5711's CJO=2pF
I doubt that you will get much improvement... might even be worse.

...Jim Thompson
 
N

Nico Coesel

Jan 1, 1970
0
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?

I usually put a diode parallel with the base resistor to discharge the
base capacitance more quickly. But like the others already asked,
what's the required speed? And why aren't you connecting the LED
directly to the open collector output?
 
R

Robert Baer

Jan 1, 1970
0
Jim said:
(1) How "quick" does ANY circuit need to be when flashing an LED?

(2) A 2N2369A is gold-doped, thus does not need to be
Schottky-clamped to prevent charge storage.

(3) The 1N5711 data sheet shows only 2pF of capacitance, so it should
be OK.

...Jim Thompson

I thought that the 2N2369A was fast because it was bribed with gold...
 
W

Winfield Hill

Jan 1, 1970
0
Jim Thompson wrote...
The 2N2369 at 10mA load turns off in 30ns.

One can actually do far better than that with a 2n2369.

I assume you get the 30ns number from the datasheet's MAX specs,
which are 13ns storage time and 18ns turn-off switching time at
Ic=10mA and Ib_on = 3mA (wow, way to high) and Ib_off = -1.5mA
(wow, way too low). Even under these less-than ideal conditions
a real transistor would be faster than the MAX specs. Moreover,
we'd be making a discrete circuit with our 2n2369, so we would
have capacitors available, which we'd use to increase the base
drive during transitions. This allows us to reduce steady ON
base drive, reducing OFF storage time, and to increase turn-off
base drive, speeding up switch-off time. In this fashion I have
been able to push 2n2369 style parts down to the 5ns territory.
With the 1N5711's CJO=2pF I doubt that you will get much
improvement... might even be worse.

2pF is the MAX value, for most of the swing the capacitance is
closer to 0.8pF (and smaller Schottky diodes are available for
super-fast switching). But even a 2pF 1n5711 would not be much
of a problem, 10mA applied to 2pF will slew 5 volts in just 1ns.
A Schottky diode is often worth its small extra capacitance.
 
F

Fred Bloggs

Jan 1, 1970
0
Ken said:
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.

There are plenty of drivers that avoid saturation and charge recovery
delays.
View in a fixed-width font such as Courier.
 
J

Jim Thompson

Jan 1, 1970
0
Jim Thompson wrote...

One can actually do far better than that with a 2n2369.

I assume you get the 30ns number from the datasheet's MAX specs,
which are 13ns storage time and 18ns turn-off switching time at
Ic=10mA and Ib_on = 3mA (wow, way to high) and Ib_off = -1.5mA
(wow, way too low). Even under these less-than ideal conditions
a real transistor would be faster than the MAX specs. Moreover,
we'd be making a discrete circuit with our 2n2369, so we would
have capacitors available, which we'd use to increase the base
drive during transitions. This allows us to reduce steady ON
base drive, reducing OFF storage time, and to increase turn-off
base drive, speeding up switch-off time. In this fashion I have
been able to push 2n2369 style parts down to the 5ns territory.


2pF is the MAX value, for most of the swing the capacitance is
closer to 0.8pF (and smaller Schottky diodes are available for
super-fast switching). But even a 2pF 1n5711 would not be much
of a problem, 10mA applied to 2pF will slew 5 volts in just 1ns.
A Schottky diode is often worth its small extra capacitance.

But that's a requirement of 10mA of DRIVE current.

...Jim Thompson
 
W

Winfield Hill

Jan 1, 1970
0
Jim Thompson wrote...
But that's a requirement of 10mA of DRIVE current.

Right, but provided by the speed-up cap or whatever, for a few ns.
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Fred Bloggs said:
Ken said:
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.

There are plenty of drivers that avoid saturation and charge recovery
delays.
View in a fixed-width font such as Courier.


.
.
.
. V
. dd
. |
. +-------+
. | |
. | c
. |\ |/ Voh-Vbe,on-Vled(I)
. ----| >----| I~ ------------------
. |/ |\ Re
. CMOS e
. buffer |
. / saturation not possible
. Re
. /
. \
. |
. ---
. \ /
. ---
. |
. gnd
.
.
.
.
.
.
.
. V+
. |
. ---
. \ /
. V ---
. dd |
. | c
. |\ |/ Voh-Vbe,on
. ----| >----| I~ ----------
. |/ |\ Re
. CMOS e
. buffer |
. / V+> Vo,h + Vled(I) + 0.7V
. Re
. /
. \
. | saturation not possible
. gnd
.

Thank you to all for all the advice. In the schem directly above, if the
Vdd is the same as V+, which is usually the case with only the single
supply, the conditions couldn't be met. So using the circuit in the
first schem above would be the wiser choice.

Here's how a guy in Prague, Czech Republic does it. He uses massively
paralleled 74HC04s to drive the HP spider LED.

http://ronja.twibright.com/transmitter/building.php (scroll down to near
bottom for the schem)And for the main project page:
http://ronja.twibright.com/main.php
 
J

James Arthur

Jan 1, 1970
0
On 12/4/04 The dark-vanquishing "Watson A.Name" wrote:

Thank you to all for all the advice. In the schem directly above, if the
Vdd is the same as V+, which is usually the case with only the single
supply, the conditions couldn't be met. So using the circuit in the
first schem above would be the wiser choice.

Here's how a guy in Prague, Czech Republic does it. He uses massively
paralleled 74HC04s to drive the HP spider LED.

http://ronja.twibright.com/transmitter/building.php (scroll down to near
bottom for the schem)And for the main project page:
http://ronja.twibright.com/main.php


Second the 2n2369a idea, which can be darn quick
when driven by a modern CMOS gate.

If you're set on push-pull drive as in the cool 'HC04 driver
you cited above, other CMOS families, such as 74AC, would
be much faster & stronger with fewer driver gates needed.

Need any free parts? -- they're yours for the asking.

Cheers,
James Arthur
 
N

Nicholas O. Lindan

Jan 1, 1970
0
James Arthur said:
Need any free parts? -- they're yours for the asking.

You _always_ pay, how is sometimes not apparent, but you
always pay.

Naked ye came, naked ye go. The book of accounts opens
with $0.00 and it closes with $0.00.
 
J

James Arthur

Jan 1, 1970
0
You _always_ pay, how is sometimes not apparent, but you
always pay.

Naked ye came, naked ye go. The book of accounts opens
with $0.00 and it closes with $0.00.


You may have overestimated those figures. Here in the U.S.
each newcomer inherits his/her portion of the national debt, whilst
those who exit early have generally failed to pay their share. Welchers.

But as to the spirit of your post...I've got nearly a hundred of
everything, you can't take 'em with you, and I'm local to Watson.

Payment? Yes, received in advance. (His posts, of course,
detailing his electronic adventures. I enjoy them.)

I enjoy swapping parts with my buddies at the local -- you
guessed it -- swapmeets. Today, for example, a pal boasted
he got a coffee can 2/3rds full of 2n2219's last week for $1. I
snagged a reel of 74HCT132Ds, some 50 & 100 gigohm
resistors, with AD549k fA op amps, some voltage references,
polypropylene caps, & a few other gizmos. Total cost < $10.

You can see how it's easy to load up on parts.

Not bought: a 5 1/2 digit Fluke DMM, offered for $25. I
hemmed over this one a while, but, already having three of like
capability, couldn't justify yet another.

Electronics is fun.

Cheers,
James
 
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