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I'd like to make a 4 pin pwm fan control using 3v3 uC. Fan is 12v powered and I have a 12v to 3v3 converter to power uC (VCC gate is 3v3 uC power). I want to isolate 3v3 VCC and 12v as much as possible, but without using optocouplers, just with FETs. This fan has a PWM input that shall be used to control it's speed. I need to make some kind of "repeater" for uC 3v3 logical level PWM signal. The output level shall not exceed 5V, so I made the following schematics:

4-wire PWM Fan Control

The "highlighted" part is the "repeater". My questions are:

  1. Is it ok to use such a repeater?
  2. Is it possible to simplify it, for example to make it using just one FET?
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  • \$\begingroup\$ Why do you want to isolate but without an isolator? There's no purpose that repeater is serving. It's not isolating anything. It's just redundant extra parts. Just get rid of both transistors and connect it directly. \$\endgroup\$
    – DKNguyen
    Commented Oct 15, 2019 at 23:30
  • \$\begingroup\$ I want to isolate these parts but without optocouplers. I suppose it is possible with fets. Just want to make it simpler \$\endgroup\$
    – BUKTOP
    Commented Oct 15, 2019 at 23:34
  • \$\begingroup\$ But WHY? For what purpose? Because it's not protecting anything against anything. MOSFET gates are literally atoms thick and isolate nothing. \$\endgroup\$
    – DKNguyen
    Commented Oct 15, 2019 at 23:34
  • \$\begingroup\$ For example, some of my fans are draining power thru pwm control and pwm monitoring pins. Some other sets 5-12 v on these pins. But it is out of the question's scope really. You can just treat it as a theoretical question about electricity. \$\endgroup\$
    – BUKTOP
    Commented Oct 15, 2019 at 23:37
  • \$\begingroup\$ @BblkTOP It isn't. This is what we call an XY problem. Are you trying to say that your control signal does not have enough current capability to drive multiple PWM input of the fans and therefore you need a signal buffer? \$\endgroup\$
    – DKNguyen
    Commented Oct 15, 2019 at 23:37

1 Answer 1

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You have over designed the circuitry of your PWM driver. This should be all you need:

enter image description here

I have shown changed values for the resistors in the voltage divider get a full 5V swing of the signal and to speed up the rise time of the PWM to the fan.

Member Requested Description of Rise Time PWM Distortion

The following waveform shows a PWM waveform with a lazy rise time in the green color. The horizontal red line shows an example 2.0V threshold voltage of where a fan's internal circuit may detect the level change of the input PWM waveform. The blue colored waveform shows the net high level the fan sees for the PWM input. As can be seen if the rise time of the green waveform were way faster there would be much less distortion of the PWM by the fan. This is further confirmed by the fast falling edges of the waveforms.

enter image description here

** Comparison of PWM Rise Times (20kHz PWM) ** These first two simulation pictures show the waveforms for the original resistor values in your circuit 20K * 10K. The values which give a 4V PWM swing.

enter image description here

The following pictures show the modified resistor values that provide for a 5V swing and with improved rise time.

enter image description here

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  • \$\begingroup\$ In that case the pwm signal will be inverted, isn't it? \$\endgroup\$
    – BUKTOP
    Commented Oct 15, 2019 at 23:43
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    \$\begingroup\$ @BbIKTOP - Please unstuck your thinking. Any PWM is normal. In this case when you want 20% program 80%; want 45% program 55%. \$\endgroup\$
    – Michael Karas
    Commented Oct 16, 2019 at 2:15
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    \$\begingroup\$ @BbIKTOP - To get the fan PWM signal swing to be 0V to 5V. The lower values also allow for better rise time of the signal. When the resistor values are too high the rise time becomes longer and it distorts the net duty cycle seen by the speed controller chip in the fan. \$\endgroup\$
    – Michael Karas
    Commented Oct 16, 2019 at 12:52
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    \$\begingroup\$ @BbIKTOP - Rise time description added to the answer. \$\endgroup\$
    – Michael Karas
    Commented Oct 18, 2019 at 11:09
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    \$\begingroup\$ @BbIKTOP - The rising edge is all about capacitance being charged through the upper resistor. The fan input has capacitance, the cable to the fan has capacitance, the circuit on your board has capacitance. I lowered the resistor values based upon my practical experience of years spent designing server computer boards with lots of fan driver circuits on them. You do have to consider the frequency of the PWM waveform. For a given length of rise time the amount of PWM distortion will be way more for a 20kHz PWM that it would be for a 500Hz PWM, In the end (continued) \$\endgroup\$
    – Michael Karas
    Commented Oct 19, 2019 at 0:55

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