# Calculating Power Requirements Darlington Array

Hi all, I am confused as to how to determine the power dissipated in a Darlington array. Take for example the ULN2068 array. It is rated as quad 1.5a. I know that the maximum current depends on the number of drivers operating and their duty cycle. What I don't know is how they work out the power dissipated in the package. I know that P=VI , I know I, but don't know what value to use for V. I imagine it must be the voltage across the driving transistor, but I don't know where to find that.

The "V" is the collector-to-emitter saturation voltage, listed as VCE(sat) in the datasheet. It varies from 1.13V to 1.6V (worst-case) as the load current varies from 500mA to 1.25A.

So the worst worst case of 1.6V while sinking 1.25A means a power dissipation of 1.6*1.25=2W. That's PER DRIVER, so if all 4 of them are doing this, the chip would be dissipating 8W (but not for long!)

Since it's a bipolar Darlington transistor driver, you could get a pretty good approximation by just assuming a 1.4V drop across the transistor.

You should definitely google for "ULN2068 datasheet", though, and look at the voltage drop curves and at the warnings for total package dissipation. It's normal for each driver to be rated in terms of "maximum current without burning out prematurely", but to have a lower overall rating that says only some of them can be running at/near maximum at any given time. I.e., if you try to drive 4 1.5A loads at once for more than a very brief period (a few seconds. Maybe even less), you'll probably burn out the chip.

You can buy heatsinks designed to attach to DIPs. If you're going to be driving heavy loads, you should seriously consider using them.

Ran

Thanks for the quick reply guys. Now I know how to work out the power. I did look at the data sheets but was unclear, until now!