# Adding the currents of all pins

1. Feb 29, 2012

### Physicslearner500039

Hi all,

i have this clarification i have one voltage output card and it has got 8 output pins and each pin can source around 120mA (5V), but i have to drive one external circuit which requires around 1A. so my question is it possible to short all the 8 output pins so that i can get 8 * 120 = 960mA(~1A) from the pins? is this type of addition possible? In case if it is not possible to achieve this with voltage output card i also have the supplier giving current output card can i achieve the above requirement with current output card? If somebody can show a link of clearing all these concepts it will be great?

regards,
Satya

2. Mar 1, 2012

### Floid

In general, yes it is possible. The problem is that in reality all the 5V outputs won't really be 5V, so what ends up happening is the highest voltage output will tend to source most of the current because there is a voltage drop between it and the other 5V outputs. Also, you would probably want to use a diode on each output so that the lower voltage outputs wouldn't draw current from the highest one (but then this introduces a voltage drop across the diode). You also run into a problem if you turn the power supplies off/on that if one turns on faster than the others it will have a current spike that will cause problems.

It may be a better idea to find a card able to source more current through a single output. Also, if you think you could need to source 1A I would make sure my supply can provide that plus a buffer (like at least 1.2-1.5A).

3. Mar 1, 2012

### StrykerTECH

Connecting all 8 output pins together to drive a single load introduces a host of problems. Putting diodes on each output pin, as Floid suggests, would be a good way to protect the card from this type of use.

You may find it easier to just connect 1 output pin to a 5V relay. Or you could connect one of the output pins to the base of a mosfet.

4. Mar 4, 2012

### Mike_In_Plano

Typically, you design to operate the pins at a derated current, i.e. 75% of max, then add an additional pin for redundancy.

This is not valid for hot socketing (connecting while under power).

If you're concerned about uneven current, you can ballast (add series resistance to) each pin by running a trace that has about 2x the rated pin resistance to each pin prior to joining the individual traces with a heavy trace.

An example:
You need 1 amp. Derated, you need 1/.75 ampacity, or 1.33 amp. If you're working with .125 amp connections with 15 milliohm each, you would need 1.33/.125 + 1 connections, or about 12.

If you were worried about uneven load distribution (which generally, I don't), then you'd want a 2X ballast resistance in series with each pin, or about 30 milliohm. Using 8 mil traces, and 1 oz copper, this comes to about .5" long trace per pin.