12C509 to drive 12V latching relay from 24VDC?
Russell McMahon email (remove spam text)
From: Paul B. Webster VK2BZC <midcoast.com.au> paulb
>Russell McMahon wrote:
>> I would be tempted to use something like the ULN2803 driver IC.
>> 'Tis cheap and has 8 drivers.
> *Cheap!* You say? Could you *please* explain to us where to get
>cheap? List price in the Dick Smith catalogue is $4.95 or $3.96 in
>tens. I think I can get a 12C509 for that price! The ULN2003 which
>probably meant to suggest for this particular project is $3.95 or
I did mean the ULN2803 but other similar chips in the family would do
I THINK the price I got my last lot of 2803's for was between $NZ1
and $NZ2 each in tube quantity - these were from an AVNET VSI
sub-supplier who deals in smaller quantities. My price recall may be
Yes, this is dearer than discrete transistors for 2 channels (as are
. For 8 (which its capable of), depending on current, its a fairly
> The PIC ability to source as much as 20mA means you don't really
Yes, the darlington isn't essential but it comes as part of the IC
(as does the input resistor and catch diode).
>> I would be tempted to use a 12volt relay and a regulator.
> *I* wouldn't when regulation is not required and an 8 cent
Yes. I agree. For some reason I think that I was thinking in terms of
switching the motor itself directly when I wrote this - it was
specified as 200ma so would drop 2.4W if it was a 12 volt one.
I would hope that TO220 regulators would typically cost $A1 from a
reasonable source (but of course you can always go to DSE :-)).
>> Total drive circuit then becomes one cheap drive chip
> Cheap? *Please* Russell, where do you get ULN2803s for $1.00 or
>Assuming the postage is reasonable I'll have a couple of dozen.
As above - check with AVNET re tube quantities - you MAY be
pleasantly surprised (but maybe not).
>> plus 1 resistor perhaps for catch diode energy dissipation
> What's that?
Depends on circuit and timing - if you place the back "anti-spike"
diodes directly across the relay coil, when the relay opens the
current will "circulate " through the coil and diode until the energy
dissipates. With low resistance in the circuit this can slow the
relay release time (time constant is L/R where L is coil inductance).
If you place a resistor in series with the diode the energy will
dissipate faster due to higher R and the relay will turn off more
crisply. This affect can be significant in some applications. The
"downside" is that the voltage spike will rise to "IR" above the
supply voltage (I is relay current, R is total resistance including
coil and resistor). The diode forward voltage drop complicates all
this only slightly.
See also: www.piclist.com/techref/power/batterys.htm?key=12v
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