0
\$\begingroup\$

I am studying mechanical engineering and is trying to design a product that will contain an electronic component with this Datasheet.

I am trying to figure out is how to calculate the heat dissipation from this component when it is going at full load to get a sort of worst case scenario. I am aware that this could be done by using the power equation, as in this example

P=V*I Where V = Vin-Vout

My problem is that I - not being well versed in electronics - do not understand what values to use from the datasheet. I'm sorry if I haven't provided enough information.

Your assistance would be very appreciated!

\$\endgroup\$
  • \$\begingroup\$ you can't, not enough info in that datasheet \$\endgroup\$ – JonRB Nov 6 '16 at 12:05
1
\$\begingroup\$

The datasheet doesn't really give enough information to calculate power dissipation accurately, and indeed it will probably vary a lot depending on what the part is actually doing. But you want the worst case, and we can work that out.

The key figures are these lines on the datasheet:

  • Operating voltage Vsupply 9 V to 30 V DC
  • Current consumption 70 mA

Now, for a worst case calculation we assume all the energy going in is dissipated as heat in the part. We calculate this as P=IV. I is 70mA, but what do we use as V? Well, the part will work with supply voltages between 9 and 30V. It is up to you to choose what voltage within that range you want to use, but remember that you can't apply more than your chosen voltage to the input pins (see the inputs/outputs details table).

So for example, if you run it at 30V, then the power dissipation could be up to 30V x 70mA = 2.1W. But if you can work with 12V, then it's only 0.84W.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ This makes sense to me. I was suspecting that the datasheet might not be enough, but your solution should be okay to put into the the otherwise rough estimates I am doing. Thank you very much for taking the time! \$\endgroup\$ – ushigeri Nov 6 '16 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.