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Fan Timer Circuit...

Discussion in 'Electronic Basics' started by pipco, Feb 13, 2006.

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  1. pipco

    pipco Guest

    I've been doing a lot of reading today on 555 timer circuits, and I'm
    not sure if this type of circuit is the best for my application. I'm
    hoping someone can steer me in the right direction. I originally was
    going to try and use a PIC, but I don't really want to invest in a
    programmer and although I know I could probably do it that way, I'm
    afraid the learning-curve would be beyond what I could handle right

    I'm trying to use a timer to turn a small computer fan on for a short
    time period and off for a longer time period, then repeat. I'd like it
    to be adjustable if possible (both the on and off periods). It would be
    something like: on for 1 minute, off for 5.

    My goal is to power the whole thing with a 9-volt battery.

    I'm not really an electronics guy, although I can build a circuit from
    a diagram. I think part of the problem I'm running into is trying to
    figure out the values for the resistors and caps.

    If someone could either point me to a diagram on the web or can talk me
    through some component selections, I would appreciate it.

    This is the closest thing I could find:

    I understand that the time values will be appoximate, I don't need a
    ton of accuracy.

    Also, if someone thinks this would be better handled by a PIC and wants
    to write and program a chip for me, I would be interested!

    Thanks in advance...

  2. John Fields

    John Fields Guest

    Try this: (view in Courier)

    | | | |8 | | |8
    [10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
    | | | 2|_ Vcc |3 | | 2|_ Vcc |3
    +-[0.1µF]-+-----|--O|T OUT|-[0.1µF]-+----|--O|T OUT|--+
    | | | 6| | | | 6| | |
    | O | +---|TH 7555| | +---|TH 7555| |
    | O | | 7|_ _| | | 7|_ _| |
    | | +--O|D R|O-+V | +--O|D R|O-|-+V
    | | +| | GND | | +| | GND | |
    | | [100µF] +---+---+ | [510µF]+---+---+ |
    | | | |1 | | |1 |
    GND>-+---------|-----+-------+-------------|----+-------+ [0.1µF]
    | | |
    | |
    | +--------------+
    | |
    +--A | +V +V
    |NAND Y--+ | | |
    +-|--B | | +-------+ O
    | | | | |K | |
    | +--A | | [DIODE] [COIL]- - - -|
    | |NAND Y--+ | | | O-> |
    +-|--B | | +-------+ |
    | | | | | |
    | +--A | | C |
    | NAND Y--+---|--[1000]--B NPN [FAN]
    +----B | | E |
    | | | | |
    | A--+ | GND GND
    +----Y NAND |
  3. John Fields

    John Fields Guest

  4. pipco

    pipco Guest


    Thanks for posting that... unfortunately, there are parts of the
    diagram that I don't really understand. That obviously has nothing to
    do with the diagram, but of my abilities at reading ascii schematics
    and my knowledge of electronics. I think this is way beyond my
    abilities. I'm looking for something simple and idiot-proof... you
    know, Rat Shack shopping list and all. I'm guessing that kind of
    hand-holding is not really possible in this kind of environment.

    Thanks though!

  5. bw

    bw Guest

    John's circuit will do exactly what you want.
    It's just a dual 555 with modified output to control a relay/switch for your
    Not sure why he used the buffer, a transistor can drive the relay.
    It should be possible to get a single 555 to do the same thing.
    You can buy the same thing, already made, called a "lab" type program-able
    There are all kinds, Cole-Parmer company might have what you are looking
    Also try your local electrical supply house.
  6. Bob Monsen

    Bob Monsen Guest

    I can do this for you. Send email. rc<surname>

    Replace <surname> with monsen.

    Bob Monsen

    I feel that you are justified in looking into the future with true
    assurance, because you have a mode of living in which we find the joy
    of life and the joy of work harmoniously combined. Added to this is
    the spirit of ambition which pervades your very being, and seems to
    make the day's work like a happy child at play.
    Albert Einstein (1879 - 1955), (referring to America)
  7. pipco

    pipco Guest

    Thanks. I never thought of looking for a kit timer... that might be the
    best option.
  8. John Fields

    John Fields Guest

    Actually, It's two 7555s, A dual 7555 (7556) would work just as
    well, but it would pin out differently:

    | | | |14 | | |
    [10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
    | | | 6|_ Vcc |5 | | 8|_ Vcc |9
    +-[0.1µF]-+-----|--O|T OUT|-[0.1µF]-+----|--O|T OUT|--+
    | | | 2| 7556| | | 12| | |
    | O | +---|TH U1A| | +---|TH U1B| |
    | O | | 1|_ _|4 | | 13|_ _|10|
    | | +--O|D R|O-+V | +--O|D R|O-|-+V
    | | +| | GND | | +| | GND | |
    | | [100µF] +---+---+ | [510µF]+---+---+ |
    | | | U1A |7 | | U1B | |
    GND>-+---------|-----+-------+-------------|----+-------+ [0.1µF]
    | | |
    | |
    | +--------------+
    | |
    +--A | +V +V
    |NAND Y--+ | | |
    +-|--B | | +-------+ O
    | | | | |K | |
    | +--A | | [DIODE] [COIL]- - - -|
    | |NAND Y--+ | | | O-> |
    +-|--B | | +-------+ |
    | | | | | |
    | +--A | | C |
    | NAND Y--+---|--[1000]--B NPN [FAN]
    +----B | | E |
    | | | | |
    | A--+ | GND GND
    +----Y NAND |
    A transistor _does_ drive the relay. It's identified as "NPN" on the
    schematic, and could probably be something like a 2N4401.

    The buffer (three NANDs in parallel are just to increase the drive
    capability to the base of the transistor and, more importantly, to
    use up the spares. ;) Depending on the OP's supply voltage, (which
    can be anything up to 18V for the 7555) a 4011 wouldn't be a bad
    choice for a quad dual-input NAND.
  9. pipco

    pipco Guest

    Email sent! Thanks.
  10. John Fields

    John Fields Guest

    Sure it is. That's what this newsgroup is for, and we do it all the

    BTW, forget the circuit I posted. The circuit you referenced should
    work, and here's how to figure out the timing components you'll

    From the referenced URL:

    "The charge time (output high) is calculated by:

    T(on) = 0.7 Ra Ct

    The discharge time (output low) is calculated by:

    T(off) = 0.7 Rb Ct"

    The common element is the capacitor, Ct, and we have to guess at
    what the values of the components will be to start,(!) so if we
    start with the OFF time and guess at 500k ohms for the pot (1 megohm
    set at the half-way point of the pot so you can go more than or less
    than 5 minutes) we can take:

    T(off) = 0.7 Rb Ct

    and rearrange it to solve for C:

    T(off) 300s
    Ct = -------- = ------------ = 857µF
    0.7 Rb 0.7 * 500kR

    Now, one of the problems associated with RC timers is that for a
    long timing interval (large C and/or R) the leakage current of the
    cap and the current needed to trigger the chip might be large enough
    to keep the cap from charging to the voltage needed to trigger the
    timer or might be great enough to take a much longer than expected
    time for the cap to charge up to the trigger point.

    The 7555 needs 500pA for the TRIGGER input and 500pA for the
    THRESHOLD input, so since they're parallelled for this application,
    that'll be a total of 1nA needed for the chip's bias, which is
    insignificant for what we're trying to do.

    We won't have any problem discharging the cap, but the leakage
    current problem will come up when we're trying to charge it up to
    the threshold voltage, which will be 2/3 Vcc.

    Assigning a Vcc of 9V to the circuit yields:

    | | Vth|
    | | 6V|
    [Ct] [RL]
    | |

    Where RL is what looks like the cap's internal parallel resistance,
    which will allow charge (leakage current) to flow "around" the cap
    instead of being accumulated by the cap.

    6V is the timer's threshold trip point, so with a 9V supply and no
    leakage current, the current flowing through R1 at the instant E2
    got to 6V would be :

    Vcc = Vth 3V
    I = ----------- = ------- = 6µA
    Rt 500kR

    Such will still be the case if there _is_ current in RL, but it will
    take longer to charge Ct because the charge which it would
    accumulate without RL there will be diminished.

    However, there will be a point where it will be impossible for Vth
    to _ever_ get to 6V, and that will be when the value of RL is low
    enough to keep Vth below 6V.

    Looking at RT and RL as a voltage divider:


    The smallest value of RL which can be used to get a Vth of 6V can be
    determined by:

    Vth Rt 6V * 500kR
    RL = ----------- = ------------ = 1E6R = 1 megohm
    Vcc - Vth 9V - 6V

    With Vth needing to get to 6V and RL equal to 1 megohm, that means
    that when the cap starts charging up it'll start accumulating the
    full 6µA of current because it'll look like a dead short. However,
    as it accumulates charge, the voltage across it will rise and,
    consequently, so will the voltage across RL. That means that as the
    voltage across the parallel pair rises, RL will be diverting more
    and more of the current away from Ct, as time goes by, with the
    result that it'll take the cap forever to charge up to 6V!

    One way around this problem is to find a low-leakage cap for Ct, but
    for large capacitances that can get expensive.

    Consider: Looking at the 857µF cap we got for your 5 minute OFF time
    and trying to get the leakage down to even 10% of 6µA will leave you
    looking for something like a cap with a leakage current of 0.6µA at
    6V, if you're working with a 9V supply. Good luck...

    The other way out is to put something reasonable in there for Ct and
    then to use whatever you have to for Ra and Rb to get the ON and OFF
    times to be what you want.

    For your application, what I'd do would be to use a 1000µF +/- 10%
    aluminum electrolytic for Ct and a couple of 1 meghohm pots set at
    about the center of their travel for Ra and Rb, then fire the thing
    up and see what happens.

    If you wire the thing up right it'll work, but chances are the
    timing won't be where you want it to be, so diddle with the pots
    until you get what you want.

    One caveat: You may want to put something like 1000 ohms in series
    with Ra so that if you've got it cranked down to zero ohms
    accidentally, when DISCHARGE (pin 7) goes low you won't be trying
    to short the power supply to ground through the internal MOSFET
    connected to pin 7.
  11. John Fields

    John Fields Guest

    Error: with the cap looking like a dead short when it starts
    charging, the initial current into the cap will be:

    E 9V
    I = --- = ------- = 18µA
    R 500kR
  12. pipco

    pipco Guest


    Thanks for the time and effort on this one. I think you've cleared up
    most of my questions... I'll give it a shot and see what happens.

    Thanks again!
  13. Eric R Snow

    Eric R Snow Guest

    Greetings Pip,
    I am a rank amateur when it comes to electronics. But I (or should I
    say Velleman) may have an answer. They make a timing kit, P/N MK111,
    or maybe PMK111, that has a relay and two adjustments. One is pulse,
    the other pause. It runs off 12 volts. Your computer power supply can
    power this. I have one of built one of these and it was easy. Follow
    this link: . The thing is
    about 5 bucks. I have never bought anything from this outfit though, I
    got my timer from a local store. Velleman kits are sold by many
    Eric R Snow
  14. pipco

    pipco Guest


    Thanks for the link. I've done a lot of searching for kits since bw's
    post the other day, and I've seen this one and many other timing kits
    based on the 555 chip. This is the one I've settled on: The one you linked to
    only has ON times up to 5 seconds (great for firing a camera). The
    canakit one has both the ON and OFF times up to 30 minutes.

    Of course, there is a possibility that I'll just build the circuit
    myself based on my original post and the info from John... and there is
    a possibility that I'll use a PIC based solution (with help, of


  15. pipco

    pipco Guest

    Okay, replying to myself here...

    I've been trying to figure out some calculations to determine how long
    this setup might last. According to, a 9V alkaline
    averages 3.6 watt-hours. If I'm using a 12V .09Amp fan, but running it
    at 9V, that comes out to .81watts. Does that mean that my fan set up
    would last just over 4 hours running non-stop? Let's say that I run it
    only 1/3 of the time, I'm looking at 12 hours of usage on one battery?
    Or, are my calculations totally incorrect?

    If I switched to a 5Vdc .15Amp fan running at 4.5V (3 AA batteries),
    that would get me about 39 hours if I only ran it 1/3 of the time.

    Someone tell me my calcs are off... otherwise, it doesn't make sense to
    run a small fan off of a battery for any length of time!
  16. Chris

    Chris Guest

    Hi. Mr. Snow is a gifted amateur -- he's got a lot of common sense.

    Your calculations are also good. But 3.6 watt-hours is optomistic, and
    is probably based on a smaller load than 90mA. Your 9V battery
    probably won't even last that long with a 90mA load.

    If this is a PC, you have both 5V and 12V available on the power
    connectors. If you can live with 5 seconds on, 25 seconds off, you can
    use the Velleman MK111 kit (perfect choice for a newbie -- 1 soldering
    iron difficulty level), the 12V supply, and drive a 12V muffin fan with
    the relay output, and you'll be good to go.

    Good luck
  17. pipco

    pipco Guest

    Thanks for the response. Actually this is fully battery powered. I'm
    not using a computer or a computer power-supply. I'm not using a AC-DC
    adapter either. Needs to be a self-contained battery powered unit. So,
    this actually might not work anyway!
  18. Hangglider

    Hangglider Guest

    Hey pipco, John Fields helped me out a week or two ago and is one of
    the nicest, most helpful guy on this group.

    John, thanks for taking the time to help us new guys out and for your
    patience with our inexperience (or stupidity in my case! :)
  19. Bob Monsen

    Bob Monsen Guest

    My emails to you have been returned as undeliverable, so I'll answer here.

    If you use D cells, rather than a 9V battery, you can run your 12V
    computer fan at a slower speed. I have a Cooler Master 12V 80mA brushless
    fan that I've been playing with. I can get it to start at about 4V, and
    run continuously down to about 2.3V. At 4.5V, the fan uses about 35mA. So,
    if you just hooked up my fan to 3 D cells (MN1300 duracell cells, to be
    exact, I'm looking at the datasheet) they would run for 380 hours.
    However, as the voltage goes down, so does the current, so that gets
    stretched out to possibly 450 hours. That is 19 days. It'll run down to
    about 2.3V, which is what the graphs go down to.

    Now, you could use my PIC circuit, to run it at 1/10 duty cycle (meaning 1
    minute on, 10 off). Sadly, since the starting voltage is higher, the
    circuit won't be able to start the fan once the batteries go below 4V,
    which will happen fairly quickly. Darn. One solution would be an
    additional battery (like your 9V) that got switched in just to start it,
    then switched out. That would allow us to get out to about 180 days,
    max... say 6 months.

    You could also use a smaller fan. Your application requires only very mild
    stirring of the air inside the box, so you could use a tiny heatsink fan,
    for example. Or, you could forgo the fan entirely, and use a resistor to
    heat up a bit of air at the bottom of the box, creating a convection
    current. You won't need much energy to keep the air circulating inside. A
    single D cell with a 220 ohm resistor across it will provide 10mW (down to
    5mW at the end of its life) of heat. It'll last a year, won't heat the
    inside too much, and will probably be enough to cause circulation to
    prevent the humidity problem you've been seeing.

    Also, you mentioned using a solar powered fan, since the top is clear
    plastic. I've played with a little solar cell (4x4 in, 3.8V output) that I
    can get to run the fan directly by shining one of those handheld 1,000,000
    candlepower spots at. A bigger or more efficient solar panel would be even
    better. Sadly, the light will heat up the interior of your box.

    I'd try the resistor before anything else, since you can measure the
    humidity at the top and bottom. If that doesn't work, try a smaller
    resistor, maybe 100 ohms (22mW down to 12mW at end of life). If you want
    the pic program, fix your return email address...

    Bob Monsen

    Nothing in the world makes people so afraid as the influence of
    independent-minded people.
    Albert Einstein (1879 - 1955), 1879-1955
  20. Rich Grise

    Rich Grise Guest

    It's not stupidity, unless it's born of stubbornness. :)

    The only dumb question is the one you don't ask. :)

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