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J

John O'Flaherty

Jan 1, 1970
0
Everything except the pinout, which could have been improved had pins 3
& 4 been swapped.

As it is now, you could put it in backwards without causing any
problem.
 
S

Spehro Pefhany

Jan 1, 1970
0
Yes, I have.. and still..


Okay, say I have 5V on the base, and 4.4V across the relay coil, and
(say) 10mA is flowing out of the emitter. Steady state. Now I suddenly
drop the base voltage to 0V. Current continues to flow out the
emitter, but the emitter voltage drops to -0.6V (because of coil
inductance). It starts out at 10mA or so, and drops towards zero from
there.



Best regards,
Spehro Pefhany
 
R

Robert Latest

Jan 1, 1970
0
John said:
Well, I have a 3.3 volt cpu and 5-volt relays here! I'm using some
cute little Fujitsu dpdt telecom relays. They pull in and drop out in
under a millisecond.

Yeah, mechanical relays are pretty darn good even for currents waaay below
spec. I've grown fond of the Pansonic TN2 series. Which ones are you using?

robert
 
L

legg

Jan 1, 1970
0
On Tue, 13 Nov 2007 22:20:07 -0500, Spehro Pefhany
Okay, say I have 5V on the base, and 4.4V across the relay coil, and
(say) 10mA is flowing out of the emitter. Steady state. Now I suddenly
drop the base voltage to 0V. Current continues to flow out the
emitter, but the emitter voltage drops to -0.6V (because of coil
inductance). It starts out at 10mA or so, and drops towards zero from
there.

.....So long as the base driver supports positive current in that
quadrant. Digital or overdriven linear parts are usually cut off,
making the only source of base current the parasitic diodes to
substrate.

.......You might measure a larger negative voltage on the emitter than
is anticipated here.

RL
 
J

John Larkin

Jan 1, 1970
0
On Tue, 13 Nov 2007 22:20:07 -0500, Spehro Pefhany


....So long as the base driver supports positive current in that
quadrant. Digital or overdriven linear parts are usually cut off,
making the only source of base current the parasitic diodes to
substrate.

......You might measure a larger negative voltage on the emitter than
is anticipated here.

RL

Any regular cmos part will go low pretty hard, whether sourcing or
sinking current, it still looks resistive. Even a bipolar TTL will
stop at ground in this situation.

John
 
H

Hot Jock

Jan 1, 1970
0
As it is now, you could put it in backwards without causing any
problem.

Yes, that's the upside, but PCB layout could have been simpler.
 
S

Spehro Pefhany

Jan 1, 1970
0
Yes, that's the upside, but PCB layout could have been simpler.

Probably there are two dies inside as John L. suggested, so there
wasn't much option.

Best regards,
Spehro Pefhany
 
J

Jamie

Jan 1, 1970
0
Spehro said:
Okay, say I have 5V on the base, and 4.4V across the relay coil, and
(say) 10mA is flowing out of the emitter. Steady state. Now I suddenly
drop the base voltage to 0V. Current continues to flow out the
emitter, but the emitter voltage drops to -0.6V (because of coil
inductance). It starts out at 10mA or so, and drops towards zero from
there.



Best regards,
Spehro Pefhany

Have it your way.
You didn't mention if you were dropping the base to common or just
floating it?
So, I'll assume after your description that you are pulling the base
low to common. That being the case, with the return potential from the
coil and expecting the emitter to handle that is risky.
I've seen circuits like that many times where it simply destroyed the
transistor gain but past the simple diode test.

and if your not pulling the base to common I wouldn't dare to think
as to where that voltage is going.


I'll stick with my wheeling diode protection, thank you.
 
J

John Larkin

Jan 1, 1970
0
So, I'll assume after your description that you are pulling the base
low to common. That being the case, with the return potential from the
coil and expecting the emitter to handle that is risky.

Why? It's still acting like a normal emitter follower.
I've seen circuits like that many times where it simply destroyed the
transistor gain but past the simple diode test.

What would kill the gain? Nothing's zenering here.

Speff's circuit is fine, if you don't mind losing 0.6 volts of coil
drive. The relay is off if the port is hi-Z before being configured,
it drives fine, it clamps the flyback fine.

John
 
J

Jamie

Jan 1, 1970
0
John said:
Why? It's still acting like a normal emitter follower.




What would kill the gain? Nothing's zenering here.

Speff's circuit is fine, if you don't mind losing 0.6 volts of coil
drive. The relay is off if the port is hi-Z before being configured,
it drives fine, it clamps the flyback fine.

John
Maybe.

I'll stick with experience and what's obvious to me. Maybe it's
flawed how ever, it's been working for me so far.

look at these articles. They have Emitter relay driver examples.
You'll notice they insist on diodes.
I wonder why.

http://www.uoguelph.ca/~antoon/tutorial/xtor/xtor3/xtor3.html
 
J

John Larkin

Jan 1, 1970
0
Maybe.

I'll stick with experience and what's obvious to me. Maybe it's
flawed how ever, it's been working for me so far.

A diode won't hurt; it's just unnecessary. You could use seven diodes,
too.
look at these articles. They have Emitter relay driver examples.
You'll notice they insist on diodes.
I wonder why.

It's obvious; the guy doesn't understand things.

John
 
S

Spehro Pefhany

Jan 1, 1970
0
Because they're idiots. Figs 4 and 5 are dangerous nonsense, for a
number of reasons.

John

UofG a.k.a. "Moo U"


Best regards,
Spehro Pefhany
 
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