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'Current in 16f84'
1999\10\17@111327 by camerlin

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face
Hey everyone,

I'm using a 16F84 in a keyless entry project.  In this project a need to
unlock an electric door strike from port a.  All of the door strikes
that I have found run off of either 24V@120mA,  12V@280mA, or 5V at 1A
(or close to those currents).  The 16F84 will only source 50mA.  Does
anyone have any suggestion to make this work.  Should I amplify the
current and use an optocoupler or something else.
Thanks
Chris Camerlin

1999\10\17@121244 by John C. Frenzel

picon face
Chris,
Use the PIC to drive a small mech. relay such as a RS 275-232.   They have a
couple others to choose from.  Look for a 5V coil rating and a contact
rating that is greater than what you need.   I generally go 2 X on both the
volt and amp rating.
John

{Quote hidden}

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1999\10\17@125925 by Byron A Jeff

face picon face
>
> Hey everyone,
>
> I'm using a 16F84 in a keyless entry project.  In this project a need to
> unlock an electric door strike from port a.  All of the door strikes
> that I have found run off of either 24V@120mA,  12V@280mA, or 5V at 1A
> (or close to those currents).  The 16F84 will only source 50mA.  Does
> anyone have any suggestion to make this work.  Should I amplify the
> current and use an optocoupler or something else.

My standard way of dring stuff like this is using relays that are
connected to the PIC via an optocoupler.

BAJ

1999\10\17@134806 by Sean H. Breheny

face picon face
Perhaps it's just me,but this doesn't seem like it would need an
optocoupler. If you want to use a relay, just put a 1N914 or 1N4148 diode
across the coil to prevent voltage spikes and it should be fine. Just be
sure that your relay coil doesn't need more current than the pic pin can
source.

I'd be tempted to solve the whole problem with a power darlington (like one
of the TIP series). Just put the strike between 5V and the darlington's
collector. Base to the PIC via a 4.7k resistor, emitter to ground. Add an
additional 1.5k resistor from base to ground,to prevent self turn-on.

In either case, put a 1N4004 or similar diode across the strike (with
cathode toward positive - reverse biased) to prevent voltage spikes. A
capacitor such as a 0.1 uF ceramic disc would also be good across the
strike. I take it that the strike contains a motor?

BTW, do you need to reverse the strike (to unlock)? If so, an H-bridge
would be your best bet.

Sean

At 12:57 PM 10/17/99 -0400, you wrote:
{Quote hidden}

| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
spam_OUTshb7TakeThisOuTspamcornell.edu ICQ #: 3329174

1999\10\17@145212 by Byron A Jeff

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>
> Perhaps it's just me,but this doesn't seem like it would need an
> optocoupler. If you want to use a relay, just put a 1N914 or 1N4148 diode
> across the coil to prevent voltage spikes and it should be fine. Just be
> sure that your relay coil doesn't need more current than the pic pin can
> source.

The problem is that almost always the coil requires more current than a
PIC can source. Reed relays like that Radio Shack 275-0232 part only
requires 20ma though. It would handle the job.

I usually use the optocoupler not only for isolation but to bump up the
current.

>
> I'd be tempted to solve the whole problem with a power darlington (like one
> of the TIP series). Just put the strike between 5V and the darlington's
> collector. Base to the PIC via a 4.7k resistor, emitter to ground. Add an
> additional 1.5k resistor from base to ground,to prevent self turn-on.

Now this sounds like a good idea.

>
> In either case, put a 1N4004 or similar diode across the strike (with
> cathode toward positive - reverse biased) to prevent voltage spikes. A
> capacitor such as a 0.1 uF ceramic disc would also be good across the
> strike. I take it that the strike contains a motor?

Probably an electromagnet.

BAJ

1999\10\17@164113 by Kev Howard

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Chris,

I have sent a couple of diagrams for interfacing to you direct ( I dont want
to fill everyones mailbox!)

I have used both methods for many years for interfacing cmos to the world.--
simple but reliable.

Hope this helps

Kev

1999\10\17@175903 by paulb

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Here goes the "Optocoupler FAQ" again!

Sean H. Breheny wrote:

> Perhaps it's just me, but this doesn't seem like it would need an
> optocoupler.

 Shouldn't, and it is a wasted component if the PIC will be powered
from the same source as the electromagnet anyway.  Optocouplers are for
use in circuits where the supplies are isolated; that is *neither*
terminal of *either* supply is common to the other.

> I'd be tempted to solve the whole problem with a power darlington
> (like one of the TIP series).

 Or a logic FET.

> Add an additional 1.5k resistor from base to ground,to prevent self
> turn-on.

 That may apply to the FET, but where would you find enough leakage
current in your circuit, over the PCB or in the PIC to turn on a
darlington?  Of course once the PIC is initialised, it provides this
function as the output pulls down (when you tell it to!).

> I take it that the strike contains a motor?

 Not a door strike.  You're thinking of the *real* security products;
electric *bolts*.  The ones that cost money!

Byron A Jeff wrote:

> I usually use the optocoupler not only for isolation but to bump up
> the current.

 Only if you use one (darlington) with significantly greater than unity
gain.  The basic ones have a gain range between much less than unity and
somewhat greater.

 Look, getting back to the isolation/ supplies thing, the critical part
of this application is the layout of the wiring.  It must be arranged so
that the current/ voltage impulse cannot be coupled to the PIC
circuitry.  You certainly do not want the strike wiring using the same
cable or in close proximity with input wiring to the PIC if at all
possible.

 It will be assumed that the diode mentioned will be placed across the
coil wires somewhere between the driving transistor/ supply and the coil
itself.  At the point of switchoff of the transistor, the current in the
coil does not change but the voltage on the transistor collector moves
suddenly from its saturation voltage (1V for a Darlington) to one diode
drop above the supply.  Such a transient could couple capacitively to a
high-impedance PIC input so must be physically separate on the PCB.

 While the coil current changes only slowly due to the diode and its
inductance, the current in the supply and the darlington stops abruptly.
The traces carrying this current therefore must also be kept separate
from the PIC circuit including its power supply.  The PIC's regulator
should have separate wires returning to the main supply capacitor
(presumed that the supply for the strike is smoothed but not regulated)
to achieve this.

 Finally, the line from the PIC to the darlington should probably have
the base resistor at the darlington end and this line may be run in
company with the PIC (regulator) supply lines but not the strike supply
lines.

 It may be that the supply bridge and capacitor are on the same PCB
with the Darlington, the PIC regulator and the PIC.  If so, these
principles apply to the PCB design and the order of arrangement of
components should be just as listed.
--
 Cheers,
       Paul B.

1999\10\17@182227 by Sean H. Breheny

face picon face
At 07:58 AM 10/18/99 +1000, you wrote:
>> Add an additional 1.5k resistor from base to ground,to prevent self
>> turn-on.
>
>  That may apply to the FET, but where would you find enough leakage
>current in your circuit, over the PCB or in the PIC to turn on a
>darlington?  Of course once the PIC is initialised, it provides this
>function as the output pulls down (when you tell it to!).
>

No, not enough PCB leakage, BUT if the darlington gets hot, couldn't the
input BJT turn itself on  due to CB junction leakage, unless we have such a
resistor at the input? Yes, if the PIC is indeed driving the base low, this
isn't an issue. BUT if the PIC were to tristate the pin,it would be another
story. I just thought it was cheap protection.

Sean


|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
.....shb7KILLspamspam@spam@cornell.edu ICQ #: 3329174

1999\10\17@183049 by gwaiche

picon face
Hi!

For this kind of thing, I use the IRLZ24N Mosfet.
It as a TTL levels gate and can handle several amps.
More, it has a free wheel diode built in to drive
coils safely.

Gael


Chris Camerlin wrote:
{Quote hidden}

1999\10\17@183713 by paulb

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Sean H. Breheny wrote:

> No, not enough PCB leakage, BUT if the darlington gets hot, couldn't
> the input BJT turn itself on  due to CB junction leakage, unless we
> have such a resistor at the input?

 Wow!  I didn't realise that happened with Silicon!
--
 Cheers,
       Paul B.

1999\10\17@185204 by Sean H. Breheny

face picon face
Hi Paul,

Well, actually, I have never tried it with a darlington, but it certainly
did happen to me with my NPN power transistors in my power supply when they
got hot, until I placed a resistor from base to emitter (it was actually
100 ohms).

Are you saying that it probably won't happen with darlingtons? Perhaps you
are right,but if so,I don't understand why not.

Sean


At 08:37 AM 10/18/99 +1000, you wrote:
{Quote hidden}

| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
shb7spamKILLspamcornell.edu ICQ #: 3329174

1999\10\17@223842 by Richard Prosser

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Some darlingtons actually include a built-in resistor from base-emitter to
counteract this. You're generally safer though to always place a  resisistor
b-e if only to assist in those cases where something goes wrong & the
transistor gets leaky. This is one of the things us "old hands" look for
when checking the designs of new engineers. Several times I've had to repair
equipment where no resistor was fitted and the problem was a slightly leaky
transistor - which would've been OK with even a 1Meg pull down.
With a darlington, any leakage will be more significant due to the higher
gain.

Richard P

> {Original Message removed}

1999\10\18@004135 by paulb

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Gael Waiche wrote:

> More, it has a free wheel diode built in to drive coils safely.

 So it has four terminals then?

Sean H. Breheny wrote:

> Well, actually, I have never tried it with a darlington, but it
> certainly did happen to me with my NPN power transistors in my power
> supply when they got hot, until I placed a resistor from base to
> emitter (it was actually 100 ohms).

> Are you saying that it probably won't happen with darlingtons?

 Nope, I'm not saying; I would generally include it.  But in this
application the PIC pulldown should render it unnecessary; the
darlington should never get hot and if it switched on spuriously during
start-up it shouldn't matter anyway.

 And as Richard said, many Darlingtons include two B-E resistors (one
for each junction) and the ULN2003 arrays of course include the series
resistor and "kickback" diodes as well.  There might even be
applications where it is easier to use two or three gates in such a
package in parallel.
--
 Cheers,
       Paul B.

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