Pulling too hard with a pull-up resistor?

[dinofx]

I have a circuit which operates on 12V, and takes a PWM input signal. According to Intel's specs for PWM fan control, the control should allow itself to be "pulled up" to 5.25 volts. I am taking the input and feeding it into an RC filter and op-amp adder/subtractor. In the end, the input signal is connected to the +12V power on my circuit by way of two resistors (around 100K each IIRC). I'll attach a schematic.

The question is, is this going to cause problems for the PWM driver circuit? I am basically pulling it up to 12 volts, aren't I? In tests, the PWM voltage stays at 5 volts @100% duty cycle.

 

[Papabravo]

If the spec sheet says pull it up to 5.25 volts why would you think for a moment that it was reasonable to do otherwise?

A simple level shifter will solve your problem. Do a Google search one.

 

[dinofx]

Well, I didn't have the specs at all when I created the circuit. I found out afterwards. Why, if I'm "pulling up" to 12V, does my motherboard's signal not exceed 4.2 volts, at 100% duty? The voltage going into the opamp is somewhere around 8.0V.

 

[audioguru]

The 10nF capacitor in the opamp's negative feedback loop causes it to be an integrator. Therefore it makes triangle waves that will cause the Mosfet to ramp up and down slowly and it will become very hot.
A PWM circuit needs to switch the Mosfet on and off very quickly so it doesn't become hot.

 

[Papabravo]

I don't know what the driver on your motherboard looks like, but I don't find the observation that the voltage does not exceed 4.2 volts very significant. That's just the combination of a five volt output and a twelve volt supply with the current limiting of the resistors. Is it working correctly? I doubt it.

I see your problem with the absence of the datasheet. On further reflection a simple common base level shifter may control the opamp input. It will give an active pulldown but the resistors would provide only a passive pullup. The opamp will have a great deal of trouble turning the N-channel MOSFET on and off through a 47K resistor. That's about 250 uA which is not enough current to move the charge on and off the gate. You probably need at least several tens of milliamps.

You may need a circuit, called a gate driver, that will drive the MOSFET gate from 0 to 12v or higher and back again. Although it will add some expense to the circuit there are gate drivers that have a charge pump to produce a voltage that is higher than the +12V motror supply so that in the ON State, the MOSFET is turned on hard.

The IR2101 and IR2304 have both been discussed on other threads in this forum. One of these or a similar part may be useful to you.

I don't know if they could be placed between the opamp and the FET or not.

BTW what is the integrator for?

 

[dinofx]

The 10nF capacitor in the opamp's negative feedback loop causes it to be an integrator. Therefore it makes triangle waves that will cause the Mosfet to ramp up and down slowly and it will become very hot.
A PWM circuit needs to switch the Mosfet on and off very quickly so it doesn't become hot.

I'm actually using .047uF for more "dampening", and the whole purpose of the circuit is to get hot. It is a variable resistor to convert PWM to smooth, high current (1.5 amps@12V), analog output.

 

[dinofx]

I don't know what the driver on your motherboard looks like, but I don't find the observation that the voltage does not exceed 4.2 volts very significant. That's just the combination of a five volt output and a twelve volt supply with the current limiting of the resistors. Is it working correctly? I doubt it.

The circuit is working exactly like I want. I am just curious about the pull-up issue.

Isn't 4.2 significant? I was thinking perhaps the PWM driver is turning a single transistor on and off, power by the +5V power source. Wouldn't the voltage drop across one cheap transistor be about 0.8V? The real question is what type of damage can I do with the amount of current that could go through 180K worth of resistors? We're talking about .06mAh. What kind of PWM circuitry has the ability to alternate between ground and 4.2V?

Actually during prototyping, with absolutely no pullup resistor, I created a resistor-capacitor filter between just the PWM output and ground. I measured 4.2V at 100% duty even without pulling up on the signal. So it's almost like it "buffered" or whatever is the appropriate term, perhaps an op-amp voltage-follower like circuit.

 

[dinofx]

This is not an integrator. It's a low-pass filter which feeds into your standard op-amp adder/subtractor resistor network, with a feedback capacitor recommended by Ron H to prevent oscillation that might be caused by the MOSFETs capacitance, or something like that.

Here is Ron's original suggested circuit:

https://www.electro-tech-online.com/attachments/pmos_amp-png.6985/

All I did was combine the two capacitors into a single capacitor.