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I've recently attempted a repair on a cooling fan circuit in a hard drive enclosure that has me somewhat baffled. The fan is an AD08012HX207000 with a connection for two wires. The circuit that drives the fan supplies 12V to the positive pin, and the negative pin is driven by the circuit below. The circuit appears to take the 3.3V PWM signal from the MCU, reduces it to ~1.6V to drive the base of a low voltage, high current BJT transistor. The transistor provides a path to ground when the base is driven high, and the fan can spin.

I'm not really sure what the 47uF capacitor is for. Smoothing out the PWM transition when the fan changes speed? When I probe the capacitor I see a square wave at 12V, and my sense is that the capacitor is charging to 12V, at which point the fan is off because the voltage on both sides is the same, and then discharging, allowing the fan to spin again, but I have no idea why it's connected via a 4.7k resistor to MCU power.

Here's a shot of my oscilloscope probing the pin of the capacitor that is connector to the fan negative:

enter image description here

If I remove the capacitor from the circuit, and connect the fan negative directly to the emitter of the transistor, I see this:

enter image description here

I guess my questions in order of importance are:

  1. Why do it this way? Why not just use a mosfet to either drive the 12V positive or the path to ground? Does this save money somehow?

  2. Why the voltage divider on the PWM signal? I checked the datasheet for this transistor and 3.3V seems fine.

  3. What function does the capacitor serve?

[edit] just to clarify a bit, what I'm trying to "repair" is the fact that the fan spins very slowly when connected, and only on the highest setting. I've actually tested this fan with a simple PWM circuit coming from a 555 timer, and it doesn't really seem to like being drive by PWM very much. It basically maintains a constant speed until ~50% duty cycle and then turns off, which makes me think that it has some sort of internal regulation that is interfering. The manufacturer sent me this fan as a replacement, and my growing suspicion is that they sent me the wrong one and the circuit is actually working fine. I've lost the original, so I can't use it for comparison. All of that aside, I want to understand how this circuit works for my own edification.

schematic

simulate this circuit – Schematic created using CircuitLab

Turns out the schematic above was incorrect. I've updated it to indicate that the capacitor is connected across the collector/emitter in case someone comes across this in the future.

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    \$\begingroup\$ Your last update just shorted the transistor+capacitor out. Fan would be always on. \$\endgroup\$ – Wesley Lee Apr 12 at 17:38
  • \$\begingroup\$ Agh, yes thank you. I believe I fixed it now. \$\endgroup\$ – flimsy Apr 12 at 21:46
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  1. Low-side switching is easier to deal with using ordinary logic levels. High-side equivalents need a level shifter, be they FETs or BJTs.

  2. R2 is a pull-down to set the default state to ‘off’ when the MCU starts up but hasn’t yet configured the pin.

  3. No idea. Seems like it won’t work. If the cap were connected across Q1 collector / emitter that would work. I’ve used this technique many times, it reduces switching noise kicked back into the power supply.

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  • \$\begingroup\$ I've added a screenshot of my oscilloscope probing the capacitor, if that helps clarify what's going on there. \$\endgroup\$ – flimsy Apr 12 at 17:03
  • \$\begingroup\$ Actually, nm, I just poked around more and the capacitor is connected across the collector / emitter, so that answers that! I was confused by that resistor back to 3.3V, which I still don't really understand. \$\endgroup\$ – flimsy Apr 12 at 17:13
  • \$\begingroup\$ You could also try eliminating the cap and just connect the fan low side. That's the more typical connection. \$\endgroup\$ – hacktastical Apr 12 at 17:19
  • \$\begingroup\$ The resistor to 3.3V is required because you need a way to pull the output node high when the transistor is OFF. In the scope plot, you can see that the falling edge is quite steep, but the rising edge is an RC curve. This type of PWM will work well with older style brushed motor fans, although the frequency of the pulse train can be a factor. This is not recommended for brushless fans which have internal controllers that need to boot up. \$\endgroup\$ – Troutdog Apr 12 at 20:12
  • \$\begingroup\$ That resistor could expose the MCU power supply to the fan voltage. Not a good idea. As far as whether this will work with a BLDC fan, it depends on the driver chip. Using the cap to smooth the voltage as shown makes it more likely to work than doing a pure chop (you get a sawtooth), but it sacrifices some control range at the lower PWM settings. \$\endgroup\$ – hacktastical Apr 12 at 20:34

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