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Interpreting the RPM feedback signal from a 3-wire fan

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lorhve

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Hi,

I have a trouble interpreting the RPM signal from a 3-wire fan here. It seems that the RPM feedback signal puts a high voltage on the pin of the micro-controller to which is connected (i.e. 6V). Due to that the micro-controller behaves erratically so I have it disconnected for the time being. In the meantime, I'm really trying to figure it out how I can possibly have it work.

So to clear the explanation of the situation. When the fan runs, the yellow line (RPM signal) gives out a clean 5V square wave signal. However, when the fan is turned off, it has an idle DC voltage of about 6V. I'm not really sure why it is happening so. Someone suggested that I put a voltage divider in between but that mitigates the pulse signal (square wave) voltage level and it won't trigger the edge interrupt on the controller.

I'm using a regular DC brushless 12 fan and PIC18F4525. I'm attaching an image of the circuit part as well. Moreover, I kind of got the idea of designing the FET and everything off another circuit but I was wondering why the FET is connected to the ground pin of the fan instead of 12V and 12V is directly fed into the fan. I guess I don't understand how a normal DC 12V fan works and it would be greatly appreciated if someone could thoroughly explain to me.

Thanks loads in advance,
Jeff
 

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On the PIC end, put a 4.7K pull-up to +5V. Put a 1N4148 diode, cathode to the Fan Speed signal, anode to the PIC pin. That way, the fan signal can go as high as it wants without effecting the PIC. When Fan low, the signal will pull the PIN pin low enough to be a "LOW". When fan High, the pull-up resistor will pull the PIC pin to 5V, and no higher.
 
Thanks a lot for the advice.

Just to clarify. Am I connecting the diode before/after the pull-up? Where should the diode be located relative to the pull-up resistor?
 
What does D7 do?

No back-EMF protection is required on a DC brushless fan because it's built-in to the inverter electronics which drive the motor.
 
I did what MikeMl suggested and it did not fix the problem at all. In fact, it got worse. 6V is still there when the fan is idle and the when the fan is running now there is only 5V DC signal with no square waves.
 
Use a resistor in series with the fan lead, and then use a 4V7 zener after the resistor down to ground. This way your original signal will be untouched until it goes over 4.7v.

Edit: You will still get a voltage at idle, however it will be 4.7v.
 

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I did what MikeMl suggested and it did not fix the problem at all. In fact, it got worse. 6V is still there when the fan is idle and the when the fan is running now there is only 5V DC signal with no square waves.


Did you hook it up like this?
 

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No, I didn't hook it up that way.
Were you showing up an example of a incorrect connection or was I supposed to connect like that?
The diode is reversed in my setting.

What happens now is that, the pin on the MCU is simply 4.9V when the fan is idle, which is good. And, when the fan is running, the PWM gets sent to the PIC but its PWM signal has a DC offset of about 0.3V or so. So, its mean is read (by an oscilloscope) as 2.8V instead of 2.5V (which is what it should be).

I also cut the RPM signal from the fan in half so I could measure the current flowing in the line (without any components attached, straight from the fan to the PIC). The current measured is about 12mA, if that gives anyone any idea.
 
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