Questions about current and BS2
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I'm just starting out, and I'm worried about frying my BS2. Please
forgive naive questions. :-)
I have an unregulated 6.3v power supply, based on a center + side tap
of a transformer marked (6.3v-0-6.3v 1.2A). It gone through a full
bridge rectifier and across a big cap (ripple removal by brute
force). I want to use this to power the BS2 (using its internal power
regulator). First question: is the internal regulator going to be
able to handle a 1.2A supply, or will something get cooked?
Second, I want to take outputs from the BS2 and drive a ULN2803A
Darlington Array, so I can operate relays. The ULN2803A would be
powered by the unregulated 6.3v supply, and the BS2 would drive it
through a 1K resistor, and and it would be pulled low with a 10K
resistor. I assume there's no risk to the BS2 in this, but I want to
make sure. (Is the 1K resistor needed?)
Third, and this is the one that worries me, I want to take an output
from the ULN2803A, and use it as an input to the BS2. The ULN2803A
would be operating at 6.3v or so and can provide 500ma. The BS2
safely sinks < 20ma and presumably expects 5v. I assume I pull the
BS2 input pin low with a 10K resistor and then connect the output of
the darlington to the BS2 through a 1K (or more?) resistor, to limit
current to the BS2... But I'm worried about presenting an unregulated
6v to the BS2. What's the *right* way to do this? I'm trying to avoid
regulating the 6.3 down to 5v, because I want to drive relays with
it, and I'd rather the regulator didn't have to handle any serious
loads.
This might all be paranoia, and I'm sure the BS2 is tougher than I
think, but I want to make sure. I've smelled smoke before. Thanks.
forgive naive questions. :-)
I have an unregulated 6.3v power supply, based on a center + side tap
of a transformer marked (6.3v-0-6.3v 1.2A). It gone through a full
bridge rectifier and across a big cap (ripple removal by brute
force). I want to use this to power the BS2 (using its internal power
regulator). First question: is the internal regulator going to be
able to handle a 1.2A supply, or will something get cooked?
Second, I want to take outputs from the BS2 and drive a ULN2803A
Darlington Array, so I can operate relays. The ULN2803A would be
powered by the unregulated 6.3v supply, and the BS2 would drive it
through a 1K resistor, and and it would be pulled low with a 10K
resistor. I assume there's no risk to the BS2 in this, but I want to
make sure. (Is the 1K resistor needed?)
Third, and this is the one that worries me, I want to take an output
from the ULN2803A, and use it as an input to the BS2. The ULN2803A
would be operating at 6.3v or so and can provide 500ma. The BS2
safely sinks < 20ma and presumably expects 5v. I assume I pull the
BS2 input pin low with a 10K resistor and then connect the output of
the darlington to the BS2 through a 1K (or more?) resistor, to limit
current to the BS2... But I'm worried about presenting an unregulated
6v to the BS2. What's the *right* way to do this? I'm trying to avoid
regulating the 6.3 down to 5v, because I want to drive relays with
it, and I'd rather the regulator didn't have to handle any serious
loads.
This might all be paranoia, and I'm sure the BS2 is tougher than I
think, but I want to make sure. I've smelled smoke before. Thanks.
Comments
for a microprocessor, but you can still easily
kill input/output pins with some of what you're trying.
The good news is -- a few serial resistors in the
right places can completely remove the risk.
1. If your transformer/rectifier/cap is really putting out
6.3 volts, that should be enough for the BS2 regulator.
You don't need to worry about current in this case, since
the BS2 will only 'pull' the amount of current it needs.
The regulator will only 'cook' if you put too much
voltage across it.
2. This sounds OK. The only problem here could be
turning off the darlington. Someone with more
darlington experience can help here. Yes, the
1K resistor is needed. If driving relays, you'll
also need reversed biased protection diodes
across the coils. When the coil releases, the
diodes give the current from the collapsing
magnetic field someplace to go. Note many relays
have these internally.
3. If you put a 20 Kohm resistor in series with the
BS2 input pin, the BS2 internal protection diodes
should 'clip' the signal to something the BS2 can
handle. The 20 Kohm resistor is needed so when the
protection diodes clamp the voltage, the resulting
current doesn't fry the protection diodes.
The BS2 can safely receive +- 12 volt signals on
its input pins, when the 20 Kohm series resistor
is used. What the BS2 sees 'inside' is +5 to 0.
I don't think your suggested input series resistor of
1K is enough -- go to 20 K for safety.
--- In basicstamps@yahoogroups.com, "scottamayo" <scott@m...> wrote:
> I'm just starting out, and I'm worried about frying my BS2. Please
> forgive naive questions. :-)
>
> I have an unregulated 6.3v power supply, based on a center + side
tap
> of a transformer marked (6.3v-0-6.3v 1.2A). It gone through a full
> bridge rectifier and across a big cap (ripple removal by brute
> force). I want to use this to power the BS2 (using its internal
power
> regulator). First question: is the internal regulator going to be
> able to handle a 1.2A supply, or will something get cooked?
>
> Second, I want to take outputs from the BS2 and drive a ULN2803A
> Darlington Array, so I can operate relays. The ULN2803A would be
> powered by the unregulated 6.3v supply, and the BS2 would drive it
> through a 1K resistor, and and it would be pulled low with a 10K
> resistor. I assume there's no risk to the BS2 in this, but I want
to
> make sure. (Is the 1K resistor needed?)
>
> Third, and this is the one that worries me, I want to take an
output
> from the ULN2803A, and use it as an input to the BS2. The ULN2803A
> would be operating at 6.3v or so and can provide 500ma. The BS2
> safely sinks < 20ma and presumably expects 5v. I assume I pull the
> BS2 input pin low with a 10K resistor and then connect the output
of
> the darlington to the BS2 through a 1K (or more?) resistor, to
limit
> current to the BS2... But I'm worried about presenting an
unregulated
> 6v to the BS2. What's the *right* way to do this? I'm trying to
avoid
> regulating the 6.3 down to 5v, because I want to drive relays with
> it, and I'd rather the regulator didn't have to handle any serious
> loads.
>
> This might all be paranoia, and I'm sure the BS2 is tougher than I
> think, but I want to make sure. I've smelled smoke before. Thanks.
allan.lane@h... writes:
> > Second, I want to take outputs from the BS2 and drive a ULN2803A
> > Darlington Array, so I can operate relays. The ULN2803A would be
> > powered by the unregulated 6.3v supply, and the BS2 would drive it
> > through a 1K resistor, and and it would be pulled low with a 10K
> > resistor. I assume there's no risk to the BS2 in this, but I want
> to
> > make sure. (Is the 1K resistor needed?)
>
A 1K resistor is not necessary. The 2803 has a 2.7K in series with each
output. The 2803 aslo has clamp diodes so those will not be needed externally.
The voltage for the 2803 has to be the same as the voltage for what you are
operating. If you are using 12V relays then the 2803 must have 12 volts. If
you
are using 5 V relays then the 2803 must have 5 volts.
A high from the Stamp turns the Darlington on -output collector goes low.
Why in the world would you want a 2803 output fed back to the Stamp. If the
output of the Darlington is low, you will feed back a logic low if there is a
load on the output. If the 2803 output is high and the Stamp input is low,
that would be the same as turning on the 2803, = PLUS - the Stamp would have to
sink all the current the 2803 load is drawing, which would probably fry your
Stamp.
Sid Weaver
W4EKQ
Port Richey, FL
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