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Author Topic: 74HC595B1R->ULN2803A  (Read 765 times)
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Hi,
I want to connect a 74HC595 shift our register to the ULN2803. Do I need to use resistor between output shitft register and Darlington array?

 I am a little bit confused, because I found some sketches without resistors between them, but as the 74HC595 is limited in current and the Darlington transistor traditionally needs to have a resistor at the base NOT to draw too much current,,,  
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Do I need to use resistor between output shitft register and Darlington array?

No resistors needed.

The ULN2803 is not just a darlington array it has some input logic, see the data sheet.
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This datasheet at TI is pretty clear on page 2
http://focus.ti.com/lit/ds/symlink/uln2803a.pdf

There is a 2.7K resistor in series with each base.
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Resistor is built in. Stick your +5 volt signal on it and go. If you're using more than +5 on the outputs you may want to consider current limiting on the outputs for some devices such as LEDs.

From the datasheet.
Applications include relay drivers, hammer drivers, lamp drivers, display drivers (LED and gas discharge), line
drivers, and logic buffers. The ULN2803A has a 2.7-k series base resistor for each Darlington pair for operation
directly with TTL or 5-V CMOS devices.
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If you haven't already bought the parts

http://www.allegromicro.com/en/Products/Part_Numbers/6278/

Both functions in one part. 8-)
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Another all-in-one is from TI:

http://focus.ti.com/docs/prod/folders/print/tpic6595.html

I haven't used it though.  I like the MCP23017 chips from Microchip for I/O expansion, but that may not be enough current for you.
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