Darlington transistor switch

[David M]

Hi,

I have a relay that I want to drive from a PIC micro. The PIC has a
5V supply - the relay has a 12VDC coil.

The circuit is basically this (fixed pitch font):

+5v +12v
| |
\ |
/ 10K pullup resistor $
\ $ (12V relay coil)
| /
PIC IO pin ------- /\/\------------- | (Darlington Driver)
5K6 \
|
Gnd

Will this circuit work with the 5v supply on the base of the
Darlington switching a load with a 10 to 12V supply ?

Regards

David

 

[David M]

On Fri, 24 Sep 2004 10:42:30 GMT, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

On Fri, 24 Sep 2004 21:38:45 +1200, the renowned David M
news^#NO@SPAM^moorhouse#.net.nz> wrote:

On 24 Sep 2004 05:23:30 GMT, cfoley1064@aol.com (CFoley1064) wrote:


I have a relay that I want to drive from a PIC micro. The PIC has a
5V supply - the relay has a 12VDC coil.


Hi, Dave. It will work, but it's customary to use a pulldown for NPN and
pullup for PNP. Try this (view in fixed font or M$ Notepad):

VCCVCC
+ +
| |
- C|
^ C|
| C|
| |
'--o
Output Pin |
PIC ___ |/ NPN
o-|___|-o-| Darlington
5.6K | |
.-. |
10K| | |
| | |
'-' |
| |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

By the way, don't forget the diode to protect the transistor.


Thanks for your help Chris.

The Darlington is a ULN28903A (npn array) which has a built in
protection diodes.

It's interesting that I need a pulldown rather than a pullup as I
built a similar circuit a few years back. That circuit had a LM339
voltage comparator driving the Darlington and used a pullup.

Why don't you use a ULN2803A (8 drivers) or ULN2003A (7 drivers) which
has the resistors inside? I'm not familiar with the part number you
mention above.

Sorry - my typo - should be 2803A.

The LM339 has an open-collector output. Most PIC outputs are
push-pull, some have one pin that is open-drain.

Yes, I now understand. Electronics is a hobby for me.

Thanks & Regards

David