1. This forum section is a read-only archive which contains old newsgroup posts. If you wish to post a query, please do so in one of our main forum sections (here). This way you will get a faster, better response from the members on Motherboard Point.

What Fan controler would you trust to control your CPU cooler fan ?

Discussion in 'AMD Overclocking' started by larrymoencurly, Jun 1, 2004.

  1. We Live for the One we Die for the One ()

    www.cpemma.co.uk has plans for both linear and switching fan speed
    controllers, but the linear ones are a lot simpler and don't cause
    pulsation noise or electrical interference (one person said that his
    Vantec controller prevented his floppy drive from working).

    I used a couple of one below (legible with nonproportional font, like
    Courier), but you can make an even simpler one from an LM317
    adjustable 3-pin regulator chip (max. voltage will be only about 10V;
    you can raise this to about 11.4V with a low-dropout equivalent chip).
    This circuit is almost like from an Enlight (Sirtec?) PSU, and one
    nice thing about it sets a lower limit for the fan voltage (change
    ratio of R1 to R2 to change this limit); it will also make sure that
    the fan runs even if the thermistor wire breaks. Capacitor C2 across
    the thermistor was added to apply full voltage to the fan at start-up
    because some fans will run at 4-5V but won't reliably start with less
    than about 6V. If you want manual adjustment, add a variable resistor
    across the thermistor, something rated 5-10 times the room temperature
    resistance of the thermistor. This controller and other linear ones
    are so small and cheap that you can build one for each fan, and that
    way you don't have to worry about blowing up transistors (but bipolar
    transistors like these usually short when they blow, so the fan should
    simply run at full speed).

    ^ +12V
    | | | |
    + | | | |
    | | | |
    C1 ----- | \ |
    10uF ----- | R1 / |
    16V | | 390 \ |
    | | / |
    | | \ |
    ----- | | /
    --- | | |/ Q1
    - \ |-----| 2SC1384
    / R3 | | 1A, 1W
    \ 91K | |
    / \ |\
    \ R2 / >
    | 560 \ |
    | / |
    | \ |
    | | |
    | Q2 | |
    | 2N3904 / o fan+
    | |/
    | |
    |-----------| o fan-
    | | |
    ------| |\ |
    | | > |
    + | | | |
    C2 | \ | |
    50uF ----- / thermistor | |
    16V ----- \ (10K @ 25C) | |
    | / | |
    | \ | |
    | | | |
    | | | |
    larrymoencurly, Jun 1, 2004
    1. Advertisements

  2. Just to note that if the thermistor wire breaks R3 will turn Q2 full on
    placing the fan at about 4.3 volts (neglecting fan current draw) so it goes
    to minimum speed (if it runs that low).

    Maximum voltage on the fan would depend on fan current and the gain of Q1
    as there must be enough voltage across R1 to provide Q1's base drive and
    the base drive needed is the collector current divided by the transistor
    gain. Fan voltage would be that plus the base-emitter drop of about .6V.

    I.E. FanV=12-([390*If/Gq1]+.6)

    Where If is fan current
    Gq1 is the gain of Q1.

    The gain for a 2SC1384 is 85 minimum but the curves indicate 200 'typical'
    at 200ma and 25C (gain goes up a bit with temperature so that actually helps).

    So, if we look at a 2.4 watt 80mm fan (the Sunon I'm using) that's 200ma
    (at 12V) and, using the 'typical' gain, that means 1 ma is needed through
    R1, which is .39V across it (V=IR). Add the .6V Vbe drop and we get 12-.99
    for 11.01 volts max. However, the transistor only guarantees 89, which
    would end up as .88+.6=1.48 for a max fan voltage of 10.52

    Note that the 2SC1384 specs a base-emitter saturation voltage of typically
    ..85 but that's at 50ma and since we wouldn't be saturating it I would think
    the 'rule of thumb' .6 for a 'typical' NPN Vbe would be closer.

    So you'd likely get around 11V max (with 'typical' numbers) but you
    couldn't guarantee more than 10.52V. Or maybe 10.27 if we used the typical
    ..85 Vbe that's in the 2SC1384 specs and only 9.92 if we used the guaranteed
    Vbe max of 1.2V. (they don't specify Vbe vs current).

    I just thought that would give an interesting insight into design problems
    as the 'might lose' portion, due to specification tolerances, is larger
    than the circuit design loss.

    Of course, the fan isn't going to pull max current at the reduced voltage
    so you'd get a little better than that but without a fan current/voltage
    curve you can't calculate it out.

    A less powerful fan will get a higher 'full on' voltage because it doesn't
    take as much R1 current to drive it and, conversely, a higher power fan
    will be driven less when 'full on' because of the increased R1 current
    needed, and corresponding voltage drop.
    David Maynard, Jun 1, 2004
    1. Advertisements

  3. I appreciate your analysis, and I'm going to change a resistor value
    to keep the fan above 5V at all times.
    larrymoencurly, Jun 2, 2004
  4. Glad it helped.
    David Maynard, Jun 2, 2004
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.