Turning on/off LED with 555 and thermistor circuit

Discussion in 'Homework Help' started by TsAmE, Jun 27, 2010.

  1. TsAmE

    Thread Starter Member

    Apr 19, 2010
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    0
    Using the thermistor characterized in the data sheet, and a 555 integrated circuit as a threshold detector, design a circuit to light an LED when the tempreature exceeds 80°C. Draw a circuit diagram with component values marked on it. Estimate the temperature at which the LED would turn off (correct answer attached).

    Now what I want to know is:

    *How do you know that the supply rail = 9V?

    *How do you calculate R = 8K?

    I know that you are suppose to use Vout = V.Rth / Rth + R but only knew that Rth ≈ 3.8k when the temperature = 80°C.
     
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  2. Georacer

    Moderator

    Nov 25, 2009
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    Do you want the LED to flash above 80oC or to stay lit?
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,963
    1,098
    Well for example if we assume this circuit:
    [​IMG]

    http://www.fairchildsemi.com/ds/LM/LM555.pdf

    And we find in datasheet that supply voltage for 555 must be in the range 4.5V to 16V.
    So to to light an LED when the temperature exceeds 80°C
    Voltage on pin 2 (trigger node) must be lower then 1/3 of the supply voltage.
    Rth for 80°C is equal 3.8KΩ so to get 1/3 voltage on pin 2 R1 mus be large then R2 > V1/Vth = R1/Rth = 2/3/1/3 = 2*Rth (if voltage on Rth is equal 1/3 the voltage on R1 mus be 2/3 of a supply voltages).
    Or with voltage divider equation
    Vout = Rth / ( R1 + Rth ) * Vcc ---> R1 = (Rth* Vcc - Rth* Vout)/Vout= ( Vcc/Vout - 1) *Rth = (1/0.333 - 1 )*Rth = 2*Rth
    So your digram is not quite right.

    And real working circuit should look something like this
    [​IMG]
     
    Last edited: Jun 27, 2010
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
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    Is the answer you posted your attempt at the answer, or is it the actual answer as given by the person or book that posed the problem?
     
  5. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    To say lit.

    Its the actual answer given.
     
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