2
\$\begingroup\$

To measure the current "drawn" by a 12V PTC heater element and to be able to switch it on and off from an AVR micro controller, I have come up with the circuit below.

diagram

The PTC heater element has about 2 Ohm cold and 8 Ohm at steady state and is powered by a 12V 10A switching power supply.

I tested the circuit and it works okay, I can measure the expected 600mV (cold) and 150mV (steady state) across the 0.1 Ohm 9W shunt resistor.

Now I suppose I should protect the shunt resistor from overcurrent, i.e. in case the PTC heater element has a short circuit. In code I'm switching off the relay when the measured current exceeds 9000mA and there is a 10A fuse in series with the heater and the shunt.

But what if for some reason the power supply "delivers" only, say, 8A? In that case, neither the software protection nor the fuse will kick in, and the shunt will be severely overloaded. Am I right?

If yes, how would I solve this? Would I have to add some thermal protection for the shunt resistor?

\$\endgroup\$
2
\$\begingroup\$

Why would the shunt be overloaded? P = I^2 * R. At 9A (Worst case) that is 8.1W dissipation. At 8A, it is 6.4W dissipation.

\$\endgroup\$
  • \$\begingroup\$ Oh, I calculated power dissipation completely wrong - still quite new to electronics. Again learned something, thanks! \$\endgroup\$ – Torsten Römer Jun 12 '15 at 17:20
  • \$\begingroup\$ I recently have to think about the same thing. I have to shunt a 400V 3.3kW system and the supplier specify the rating of shunt resistor of max. 5W. After using P=I²R, i realized that at maximum load of 8A, the shunt was dissipating only 3W. \$\endgroup\$ – Natthapol Vanasrivilai Nov 7 '17 at 21:06

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.