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'[EE]: Using a PIC to control medium current loads.'
2001\05\13@010932 by Nick Veys

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Wondering how to get a PIC to control a 1-2A load.  The PIC I'm working with
is a F84a so there's no way it can source/sink that much current.  I've
thought if transistors/relays/SCRs but I really have little experience with
that stuff, and I couldn't find much beyond that in a web search.  I sure
people have ran into this before so I'm looking for suggestions, any
really...

Thanks!

spam_OUTnickTakeThisOuTspamveys.com | http://www.veys.com/nick

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2001\05\13@013226 by Spehro Pefhany

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At 11:52 PM 5/12/01 -0500, you wrote:
>Wondering how to get a PIC to control a 1-2A load.  The PIC I'm working with
>is a F84a so there's no way it can source/sink that much current.  I've
>thought if transistors/relays/SCRs but I really have little experience with
>that stuff, and I couldn't find much beyond that in a web search.  I sure
>people have ran into this before so I'm looking for suggestions, any
>really...

Perhaps you could part with a bit more information on the load voltage and
the
type of load?

Is it a 150W light bulb you are running off of 120VAC, a 2A 12V solenoid,
a fractional horsepower 24V DC motor? etc. etc.

Best regards,
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2001\05\13@013241 by Alexandre Domingos F. Souza

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>Wondering how to get a PIC to control a 1-2A load.  The PIC I'm working with
>is a F84a so there's no way it can source/sink that much current.  I've
>thought if transistors/relays/SCRs but I really have little experience with
>that stuff, and I couldn't find much beyond that in a web search.  I sure
>people have ran into this before so I'm looking for suggestions, any
>really...

       http://www.epanorama.net - you will get good info if you look also for stepper motor control.

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2001\05\13@013947 by Nick Veys

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At 11:52 PM 5/12/01 -0500, you wrote:
>Wondering how to get a PIC to control a 1-2A load.  The PIC I'm working
with
>is a F84a so there's no way it can source/sink that much current.  I've
>thought if transistors/relays/SCRs but I really have little experience with
>that stuff, and I couldn't find much beyond that in a web search.  I sure
>people have ran into this before so I'm looking for suggestions, any
>really...

Perhaps you could part with a bit more information on the load voltage and
the
type of load?

Is it a 150W light bulb you are running off of 120VAC, a 2A 12V solenoid,
a fractional horsepower 24V DC motor? etc. etc.

-----------------

Sure thing, it's simply a 12VDC fan, the ones in question can draw between
.5A and .7A and I would like to be able to hook 2 up to each line, hence the
1-2A...  Hope this helps!

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2001\05\13@014007 by Bob Blick

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>Wondering how to get a PIC to control a 1-2A load.  The PIC I'm working with
>is a F84a so there's no way it can source/sink that much current.  I've
>thought if transistors/relays/SCRs but I really have little experience with
>that stuff, and I couldn't find much beyond that in a web search.  I sure
>people have ran into this before so I'm looking for suggestions, any
>really...

Hi Nick,

Depending on your load there are a few simple ways. If your load is DC that
is good. If your load uses the same ground as the PIC, that makes it
easier. If you can leave positive connected to the load and switch the
negative, that is even easier. If you answered yes to all three then you
can use an NPN transistor or N-channel MOSFET, and one or two more resistors.

                   +
--------            |
       |         LOAD
       |          /
       |    220  |
 PIC   |---/\/\--|
       |         |>
       |           \
--------          GROUND


Cheers,

Bob

P.S. You can also use a darlington array if you need several outputs, they
have resistors built in.

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2001\05\13@015452 by Dan Michaels

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At 11:52 PM 5/12/01 -0500, you wrote:
>Wondering how to get a PIC to control a 1-2A load.  The PIC I'm working with
>is a F84a so there's no way it can source/sink that much current.  I've
>thought if transistors/relays/SCRs but I really have little experience with
>that stuff, and I couldn't find much beyond that in a web search.  I sure
>people have ran into this before so I'm looking for suggestions, any
>really...
>

Go to the bottom of the piclist archives page, and enter any number of
related topics into the search engine:

http://www.infosite.com/~jkeyzer/piclist/

"motor control" brings up 1613 entries.
"1A load" brings up 4379 entries.
"heavy loads" ...
"large currents" ...
on and on

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2001\05\13@015508 by Alexandre Domingos F. Souza

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>easier. If you can leave positive connected to the load and switch the
>negative, that is even easier. If you answered yes to all three then you

       Hmmm...Why switching the negative is easier???

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2001\05\13@015514 by Alexandre Domingos F. Souza

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>Depending on your load there are a few simple ways. If your load is DC that
>is good. If your load uses the same ground as the PIC, that makes it
>easier. If you can leave positive connected to the load and switch the
>negative, that is even easier. If you answered yes to all three then you
>can use an NPN transistor or N-channel MOSFET, and one or two more resistors.
>
>                    +
>--------            |
>        |         LOAD
>        |          /
>        |    220  |
>  PIC   |---/\/\--|
>        |         |>
>        |           \
>--------          GROUND

       To be a little more pratical: Out of the pic, goes to a 1K resistor to the base of the BC556 transistor. Emitter goes to the earth. Colector goes to the negative of the fan. Positive of the fan goes to the +vcc. A diode goes with the anode connected to the transistor side, and the catode to the +vcc side. Here you have your pic-controlled-fan :o)

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2001\05\13@015524 by Spehro Pefhany

picon face
At 12:38 AM 5/13/01 -0500, you wrote:

>
>Sure thing, it's simply a 12VDC fan, the ones in question can draw between
>.5A and .7A and I would like to be able to hook 2 up to each line, hence the
>1-2A...  Hope this helps!

For a typical 12V brushless fan, you can use a power MOSFET, but it has to
be a "logic" type to allow it to be driven directly from the PIC (preferably
through a resistor, and you should put a diode across the fan (reverse
biased),
a 1N4004 should suffice.

For example, an IRL3102 (about $1.35 in 10's from digikey)

                   x----|>|----x
             D     |           |
            |------x---[fan]---x- +12
PIC --Rs --| |
        G   |---  0v
             S

Rds(on) of this part is 13m Ohms, so the power dissipation at 2A is
only 52mW (you won't need a heat sink!)

Best regards,

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speffspamKILLspaminterlog.com             Info for manufacturers: http://www.trexon.com
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2001\05\13@023021 by Chris Carr

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> At 11:52 PM 5/12/01 -0500, you wrote:
> >Wondering how to get a PIC to control a 1-2A load.  The PIC I'm working
> with
> >is a F84a so there's no way it can source/sink that much current.  I've
> >thought if transistors/relays/SCRs but I really have little experience
with
{Quote hidden}

the
> 1-2A...  Hope this helps!
>
I presume that the current draw you are quoting above is with the motor
running. Generally, the critical current parameter that you have to cater
for
is the start-up current this can easily be 5 or more times the running
current
when the motor is started under load.

Regards

Chris Carr

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2001\05\13@023043 by Spehro Pefhany

picon face
At 02:52 AM 5/13/01 -0300, you wrote:
>>Depending on your load there are a few simple ways. If your load is DC that
>>is good. If your load uses the same ground as the PIC, that makes it
>>easier. If you can leave positive connected to the load and switch the
>>negative, that is even easier. If you answered yes to all three then you
>>can use an NPN transistor or N-channel MOSFET, and one or two more
resistors.
{Quote hidden}

resistor to the base of the BC556 transistor. Emitter goes to the earth.
Colector goes to the negative of the fan. Positive of the fan goes to the
+vcc. A diode goes with the anode connected to the transistor side, and the
catode to the +vcc side. Here you have your pic-controlled-fan :o)

Just a couple points on this..

1)      BC556 is a PNP TO-92 transistor. The above circuit requires an NPN.

2)      The BC556 is only good for 100mA Ic, not 1-2A

3)      With 1K we'll be getting no more than 4.3mA base current, that means
       the transistor wouldn't (even typically) be saturated (even if was
      good for 1-2A. I like to see Ic/Ib ~= 20 for BJTs in hard saturation,
       so you need 100mA of base current, requires another transistor
       (though the BC556 is too wimpy even for this). Maybe a 2SC8550
       TO-92 PNP BJT driving a TIP31 through a 43R 1W resistor. (dissipates
       about 1/2W with fan on).  Then your 1K base resistor would be ok.

4)      I agree with the need for a diode across the fan.

Best regards,

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2001\05\13@023849 by Alexandre Domingos F. Souza

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>Just a couple points on this..
>1)      BC556 is a PNP TO-92 transistor. The above circuit requires an NPN.

       I gotta sleep, BC546 :o)

>2)      The BC556 is only good for 100mA Ic, not 1-2A

       Yep, right.

>3)      With 1K we'll be getting no more than 4.3mA base current, that means
>        the transistor wouldn't (even typically) be saturated (even if was
>       good for 1-2A. I like to see Ic/Ib ~= 20 for BJTs in hard saturation,
>        so you need 100mA of base current, requires another transistor
>        (though the BC556 is too wimpy even for this). Maybe a 2SC8550
>        TO-92 PNP BJT driving a TIP31 through a 43R 1W resistor. (dissipates
>        about 1/2W with fan on).  Then your 1K base resistor would be ok.

       Wasn't a couple? :o) Yes, the transistor was (badly) calculated from my mind. I think I gotta have some sleep :o(

>4)      I agree with the need for a diode across the fan.

       Ufa! At least one :o)

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2001\05\13@090632 by Olin Lathrop

face picon face
> Sure thing, it's simply a 12VDC fan, the ones in question can draw between
> .5A and .7A and I would like to be able to hook 2 up to each line, hence
the
> 1-2A...  Hope this helps!

The basic topology is an NPN power transistor in common emitter
configuration.  The emitter to ground, the base to the PIC pin thru a
resistor, and the collector to the low side of the load.  The high side of
the load goes to the 12V supply.  Don't forget the flyback diode accross the
load in reverse else you will fry the transistor when you shut it off.

That's the basic concept.  Draw it out and understand that first.  ...
OK, now here's an additional wrinkle.  You can't rely on the power
transistor having enough gain to take 20mA from the PIC into its base and
allow 2A to flow thru its collector.  That would require a current gain of
100 and push the PIC to its limit.  To fix this, we use almost the same
topology but insert an extra transistor between the PIC and the base of the
power transistor.  I'll call the power transistor Q1 and the new transistor
Q2.  The PIC output drives the base of Q2 directly.  Tie the collector of Q2
to the PIC power supply.  The emitter of Q2 feeds the base of Q1 thru a
resistor.  The emitter and collector of Q1 remain as before.

A current gain of 20 sounds like a safe design value for this type of power
transistor (Q1).  If you want 2A out, that means the Q1 base current needs
to be 100mA.  When the PIC pin is high, we can assume it is at 5V.  Figure a
.7V drop accross the base-emitter of Q2 and .8V accross the base-emitter of
Q1.  That leaves about 3.5V accross the resistor.  For 100mA at 3.5V you
need 35 ohms.

Q2 can be almost any small signal transistor.  If it has a current gain of
100, then the PIC will only need to source 1mA when the fan is on.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, EraseMEolinspam_OUTspamTakeThisOuTembedinc.com, http://www.embedinc.com

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2001\05\13@113152 by Herbert Graf

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> >easier. If you can leave positive connected to the load and switch the
> >negative, that is even easier. If you answered yes to all three then you
>
>         Hmmm...Why switching the negative is easier???

       Mostly because in the MOS world nmos is smaller and cheaper for the same
current flow (because of the higher mobility of n doped silicon IIRC),
similarly in the BJT world. TTYL

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2001\05\13@114642 by Spehro Pefhany

picon face
At 11:29 AM 5/13/01 -0400, you wrote:

>        Mostly because in the MOS world nmos is smaller and cheaper for
the same
>current flow (because of the higher mobility of n doped silicon IIRC),
>similarly in the BJT world. TTYL

Also you don't need any level shifting.. assuming the +12 and +5 are tied
together on the minus ends. Otherwise, switching the high side might be
easier.

Best regards,
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
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speffspamspam_OUTinterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
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2001\05\14@190510 by Peter L. Peres

picon face
For newbies I'd recommend a logic level VFET in TO220 case driven directly
by a PIC, for low voltage and no separation. This will switch up to about
30A without problems.

For mains for newbies I always recommend a solid state relay (optically
insulated). This is also driven by a PIC pin and will drive up to 25A at
mains with insulation. About $25.

Then there are relays.

Objections ?

Peter

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2001\05\15@032431 by Raymond Choat

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Is the VFET used for AC or DC? I will need to soon run 24volt DC 3amp
relays.
Thanks Wrong Way Ray (Raymond Choat)

{Original Message removed}

2001\05\16@151037 by Peter L. Peres

picon face
> Is the VFET used for AC or DC? I will need to soon run 24volt DC 3amp
> relays.
> Thanks Wrong Way Ray (Raymond Choat)

The VFET is used in DC but it can be used in low voltage AC with a normal
rectifier bridge and a little trick. You usually use either a vfet or a
relay, not both. Some VFETs have lower Rdson than relays that cost the
same ... ;-).

Peter

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