desperate to figure 2803 chip out

[kjones9999]

Lot of smart people here-- hope you can help.

I am working on a project for my christmas lights. basically, I havea measurement computing dio48/ctr15 card and opto22 oac5 relays. As these relays are negative true logic (sink to turn on) and the card is standard, i needed someway to invert the logic. I got a 2803 and took each input from the board, ran it through the chip and out to the relay. I did this also for the common (ran it through the chip). I did not connect the bottom 2 pins (9 and 10) on any of the chips. But while i do get some response from the relay leds, it is not enough to switch a load. Do I need to power the lines coming from the dio board? If so where does the + and - go? Does the common from the card get buffered and inverted as well?

The opto22 board does have power supply input. I have tried hooking up a 5v supply with little difference. Specs say a 15ma load switches the relay.

Help-- too poor and stubborn to buy the retail stuff!

 

[betwixt]

Pin 9 is the common ground pin so you must connect it to the negative side of your supply.

Brian.

 

[kjones9999]

thanks--another board mentioned that pin 9 might be the positive side of the supply-- any thoughts?

if i use a separate power supply from the card, where would the positive side of the supply go?

Thanks!

 

[betwixt]

No - absolutely without doubt you should connect pin 9 to ground. It is the common current return from all the outputs so it has to go to the negative (ground) side of the supply.

Pin 10 is the common clamp diode connection, if you are driving relays or similar inductive loads you should connect it to the positive side of the relay supply, this ensures the internal diodes can clamp any spikes produced as the relay is switched off.

The 2003 and 2803 series are simply an array of driver transistors and clamp diodes, they do not have a supply pin as such as there is no circuitry inside them that needs to be permanently powered up.

Brian.