Uln2803 - as relay driver

Hi, I am looking forward to design a PCB where I can hook up 8 relays, and drive them by using ULN2803 & my PIC.

I tried on the internet and looked up the datasheet as well which left me with slight uncertainty, which is that do I need to place a resistance between the PIC & the input to ULN2803?

Datasheet says the base input already has a 2.7K built in. I just want to reconfirm if I need one externally also.

My circuit will be simple : (Planned circuit) - Port Pin -> ULN2803 input -> Relay ground (i.e. relay gets 12V common - every output of ULN2803 will be connected to the other point which will ground the 12V)

Yep, you got it right. No need for input current limiting resistor. If you google ULN2803 and Stepper, or ULN2803 and Relay you will find several schematic pictures that confirm this.

have fun!!

No need for the resisitor between the Pic pin and ULN pin. Direct connection.

And for your question, for 12 relays, use two ULN2003 instead of ULN2803.
You will have 14 outputs.
And, do not trust the diodes in ULN chips, use external diodes accross the relay coils.

You can also parallel outputs on the ULN chip to achieve more current handling. Each output will sink 500mA, so two in parallel will sink 1 amp. One post I read suggested stacking and soldering two DIP ULN chips, one on top of the other to achieve this increased current while still maintaining the same number of outputs.

There is also TPIC595 "POWER LOGIC 8 BIT SHIFT REGISTER".

With just 3-pins on PIC, you can drive as many outputs as you want and the load per output is 500mA.