0
\$\begingroup\$

I have the following circuit built from the 'choppy' pwm controller kit.

enter image description here

However upon testing it, it only ranges the voltage from 10-12V.

How would one analyze the circuit shown to confirm this is the correct output?

\$\endgroup\$
8
  • \$\begingroup\$ How are you measuring the output voltage? What's your load? \$\endgroup\$
    – winny
    Jul 4, 2022 at 13:49
  • \$\begingroup\$ Multi-meter across the output terminals, no load \$\endgroup\$
    – SheerKahn
    Jul 4, 2022 at 13:54
  • \$\begingroup\$ Just tried again with a led strip as load and range was from 4-12V. \$\endgroup\$
    – SheerKahn
    Jul 4, 2022 at 14:01
  • 1
    \$\begingroup\$ (1) I guess, if your Vcc is 12V, then using the pot to get 10-12V looks reasonable. If you want lower voltage output, you you can lower Vcc (but not much lower than IRF540's trig threshold of about 7V (or use IRL540 for a lower trig threshold < 5V). (2) Of course it would be nice if you have a scope. Otherwise, if you can set 50% duty cycle, DC voltmeter would get 1/2 Vcc. (3) I am only wildly guessing. \$\endgroup\$
    – tlfong01
    Jul 4, 2022 at 14:08
  • 2
    \$\begingroup\$ Your standard multimeter will have a hard time reading the average voltage correctly. Check with an oscilloscope, and some resistive load. \$\endgroup\$
    – winny
    Jul 4, 2022 at 14:09

1 Answer 1

2
\$\begingroup\$

Your MOSFET has an internal output capacitance (Coss) in the range of 500 pF. You can think of it like a parallel capacitor between drain an source. If the FET is conducting this capacitor is discharged. If it turns off, this capacitor will be charged via the load resistance.

With a motor as load this charge time is very short and will not affect the voltage reading of the multimeter, but without a load the charge time is much longer, because the charge current through the 10 Mohm internal resistance of the multimeter is in µA range only.

The voltage across the multimeter drops during the charge phase (FET off) but the next conducting period starts long before the charge can complete. As a result the average voltage across the multimeter cannot fall below 10 V.

With a LED as load you probably add some more capacitance and the forward voltage of the LEDs will also not charge this capacitance. You can think of the LED chain as a Z-diode not conducting below Ufwd of the LEDs.

A proper voltage integral reading with a multimeter needs a pure resistive load here with a current of at least 10 mA.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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