18F2455 controlling relays

As metioned earlier by kchriste don't forget the 100nf capacitor across the pic power pins 19 / 20 - should be fitted as close to the pics pins as practicable - would be good to fit one across pins 9 /10 of the 2803.

Apart from good design pactice to fit one to each ic used, they will help with the poor usb power rails.

hi,
I would add a 100nF on the PIC, pins 19/20.

My only concern is the 5V relays, you say will operate at 4V.

So starting at 5V minus the 0.7 [possibly 0.8V] loss across the diode = 4.3V

and the approx 1V loss across the ULN in Vce[sat] thats 3.3V.!!!!

I wouldnt design a circuit that marginal...

Whats the voltage on the input side of the 5V reg.???

I'll put in some caps across the pins 19/20 on the PIC and 9/10 on the ULN.
Thanks for the suggestion.

ericgibbs said:

So starting at 5V minus the 0.7 [possibly 0.8V] loss across the diode = 4.3V

Click to expand...

Maybe I'll bin that diode and put in a changeover/SPDT switch or perhaps some jumper pins or something....

You could also consider using 12V relays as they usually work with as little as 9V across the coil.

Where is the power supply part of your design? Are you powering the relays from USB?

blinky465 said:

I'll put in some caps across the pins 19/20 on the PIC and 9/10 on the ULN.
Thanks for the suggestion.

Maybe I'll bin that diode and put in a changeover/SPDT switch or perhaps some jumper pins or something....

Click to expand...

hi,
As you have only 4 relays to drive, a transistor drive/relay would be a better option than a ULN, because of the tight voltage margin.
A 2N2222 transistor will give a Vce of about 0.2V and without the series +5V diode, it would give approx 4.8V across the the 5V relay coils.