I don't understand the formula for the thermal dissipation of the STCS1A. The datasheet in chapter 8.2 states:

Pd = (Vd - Vfb) * Id + (Vcc * Icc)

Where Vd is Vcc - VLED. Here the connection scheme:


simulate this circuit – Schematic created using CircuitLab

When the mosfet turns on, the current flows through the LEDs and the feedback resistor. Internally, the feedback voltage is fed to a comparator that turns off the mosfet when the preset current is reached.

Now, while the mosfet is on, the voltage across the diodes depends on their forward voltage with the actual current. The voltage across the mosfet is due to the Rds_on (ideally it would be a short-circuit) so it also fixed by the current. The only component that would receive the remaining voltage is the feedback resistor.

Of course, at steady-state this would be Vcc - VLED - Vds but this is never reached because the mosfet is turned off when the Vfb is ~ 100 mV in order to achieve the current regulation.

If this is correct, I don't understand why the voltage across drain-source receive the gap between VLED and Vcc. The mosfet is not used in the linear region, it's a PWM control so it's off or completely on.

If I'm wrong and the datasheet is correct (and I bet that's right!) it means that if you use a high voltage supply with few LEDs, the mosfet will dissipate a lot more. But this makes no sense to me. I selected this device because I understood it regulates the current turning off the mosfet whenever the current rises over the preselected value, so the power dissipation (during the on-state) would be just Vds * Id where Vds = Rds_on * Id.

  • 1
    \$\begingroup\$ I don't really understand what the question is. The first term in the power equation are the MOSFET losses (VDS*Id) and the second term is the power drawn from the other pin that is connected to 24V \$\endgroup\$ Mar 20, 2021 at 7:15
  • \$\begingroup\$ The question is: why the Vds in the MOSFET losses is calculated as (Vcc-Vled-VRfb), instead of Rds_on * Id? \$\endgroup\$
    – Mark
    Mar 20, 2021 at 7:52

1 Answer 1


The first contribution to Pd is the power wasted as heat by the MOSFET.

The second contribution to Pd is the power wasted as heat by the led controller.

The led controller and the MOSFET are on the same chip so Pd has two contributions.

The definition of power consumption of a MOSFET is:

Pd(t) = Id(t) * Vds(t)


Pd(t) = Rds(t) * Id(t) * Id(t)

In a lab it's much easier to use the first formula because you can measure currents and voltages.

  • \$\begingroup\$ Exactly, but Rds is a characteristic of the MOSFET. We don't know its value here, anyway we can assume it quite constant when the MOSFET is fully on. Hence, Vds depends on the current only. Instead, if you define Vds as (Vcc-Vled-Vrfb) it depends on the gap between the power supply rail and the voltage across the LEDs. \$\endgroup\$
    – Mark
    Mar 20, 2021 at 12:11
  • \$\begingroup\$ Rds is the ratio between two measurable variables: Vds and Ids. \$\endgroup\$ Mar 20, 2021 at 12:58

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.