I'm building a circuit whose outputs go via a ULN2803 to screw terminals, and I'm looking for ideas to allow me to protect the ULN2803's pins against overload/short-circuit.
The outputs will be LED's, of an undefined number (at the design stage) for each pin.
I could add a resistor between each pin and its screw terminal, however this will affect the brightness of the LEDs.
I could add a little glass fuse to each pin, but then I'd end up with quite a bulky item!
Does anyone know another way of achieving overload protection for the pins that I can investigate? - I'm a little stumped for ideas I'm afraid.
I suspect a fuse on the GND pin would be required for overall protection of the chip.
Rather than put up the full circuit diagram for the project, I've attached a schematic for a single pin to illustrate what I mean.
I think the transistors in that array are rated at 500mA. ( you need to check this )
500mA at 5v corresponds to 10ohms .
Could you use this in series ( use an appropuate wattage resistor ) with your output to limit fault current ? It should make little difference to brightness if you don’t have too many leds per output.
If you use a fuse for all of it , it is better practice to put it in the positive supply line
hammy:
I think the transistors in that array are rated at 500mA. ( you need to check this )
500mA at 5v corresponds to 10ohms .
Could you use this in series ( use an appropuate wattage resistor ) with your output to limit fault current ? It should make little difference to brightness if you don’t have too many leds per output.
If you use a fuse for all of it , it is better practice to put it in the positive supply line
Yup,according to the datasheet the absolute max sink current per channel is 500ma.
I think what you are saying is that if I protect each channel with a 10ohm resistor (or maybe slightly bigger to bring it down a bit) - hardly any resistance at all in the scheme of things - it's unlikely to affect the brightness of the LED's ? I must admit I hadn't though of it in that way - out with my little book methinks to learn about resistor combinations, and a little experimentation to see the effect visually) ...
And since the power to the LED's will come from the same circuit (through another screw terminal), I could position the fuse there..
Note that protection for two or more channels gets more complicated.
If you sink 2*500mA with two channels, then the chip temp could reach about 100C.
A third channel active, and it pops it's top.
Maybe it's better to use an IC socket, and buy a bag of the chips.
Leo..
And do remember that this outdated chip will drop between 1 and 2 Volts depending on your load - that is why it heats up so badly - so it is for example not particularly useful for driving LEDs with a 5 V supply.
If you must use it rather than a TPIC6A595 and are using it for no more than 15 Volts, you should use the updated version.
