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Questions on an auto fan circuit (741)

  • 05-12-2011 1:02am
    #1
    Registered Users, Registered Users 2 Posts: 936 ✭✭✭


    Posted this on the engineering forum but thought a few here might be able to help

    I have a presentation to do on this circuit

    auto-fan.thumbnail.gif

    I know pretty much everything about it, the thermistor has a negative coefficient of resistance, so that when the temperature increases current flows into the 741, thus switching the transistor and activating the relay bringing on the fan.

    My question is how does the feedback loop work with the 741 op amp being supplied from a single supply? I was told that in order to power the 741 op amp from a single supply a capacitor was needed to remove the DC bias that's required to make the supply all positive?


Comments

  • Registered Users, Registered Users 2 Posts: 1,134 ✭✭✭FarmerGreen


    Ignore R3 for a a minute
    U1 is being used as a comparator.
    R2 and R5 set pin 3 (non-inv) to half rail voltage ie 6v.
    R1/P1 with Th1 form another potential divider, so that when pin 2 (inv ip) is below 6V the output of U1 goes to nearly +12V.
    Likewise when the non-inv ip pin 2 goes above +6V then the output
    of U1 goes to nearly 0V.

    The purpose of R3 is to provide hysteresis or dead band.
    As the op of U1 goes high, R3 gives positive feedback to pin 3,
    meaning that the temperature will have to drop a few degrees before
    the relay drops out.
    This is to speed the operation of the relay on a slowly changing
    temperature, and it also gives you noise immunity to prevent Re1
    chattering.
    This circuit is devoid of caps (bad), suggest 50uF between +12V at
    the relay coil and E of Q1.
    100nF between U1 pin 7 and 4, 10nF between U1 Pin 2 and 4.


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