2
\$\begingroup\$

Hi all I have a question regarding using a MOSFET for a constant current source. I have found this question with some answers but it wasn't clear to me yet.

I want to create a constant current LED driver with a MOSFET shunt and opamp configuration. I have designed the schematic below with a 1 ohm sense resistor and a BUK6D43-60E MOSFET. The forward voltage of the LED's will be between 34.5 and 45 volt and the current will be 1 ampere. So the voltage over the MOSFET will be around 12.5 volt maximum. The LED's will be pulsed with a frequency of <100 Hz and a pulse duration with a maximum of 300 microseconds, this will be a dutycycle of 3%. From the datasheet below I got a Zja of around 4-6 K/W. I know Tjunction = Tambient + Rj-a * P, now Tambient has a maximum of 60C and Rj-a = Zja = 4-6 K/W. Now what do I have to use for P? Is this (Vcc-Vleds-Vshunt) * If or is this (Vcc-Vleds-Vshunt) * If * dutycycle? Because this would be the difference between 135C and 62.25C.

Datasheet schematic

\$\endgroup\$
16
  • \$\begingroup\$ P will be the power dissipated in the FET: the drain current TIMES the voltage across the source-to-drain. \$\endgroup\$ – analogsystemsrf Mar 14 '19 at 8:41
  • \$\begingroup\$ Your MOSFET "only" dissipates power during the 300us pulse. As the DuCy is always less than 3% the average power will be only 3% of the dissipated power during the pulse. The power dissipation in the MOSFET is Vds * Id = 12.5 V * 1 A = 12.5 W. Make sure the MOSFET can handle that (it has a higher maximum power dissipation). Since the pulse is short, the heat can average out so 3% * 12.5 W = 0.375 W. \$\endgroup\$ – Bimpelrekkie Mar 14 '19 at 8:52
  • \$\begingroup\$ I think you're a decade mistaken regarding the Zja. 300 usec = 3 10^(-4), so, Zja is about 1.3 K/W \$\endgroup\$ – Huisman Mar 14 '19 at 9:23
  • \$\begingroup\$ @Bimpelrekkie Thank you for your input. I didn't know if the dutycycle was already in the Zj-a. \$\endgroup\$ – BobLee Mar 14 '19 at 9:41
  • \$\begingroup\$ @Huisman I also realized this but since there is no graph in my region i decided to take the closest one possible. \$\endgroup\$ – BobLee Mar 14 '19 at 9:42

Your Answer

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

Browse other questions tagged or ask your own question.