Turning on/off LED with 555 and thermistor circuit

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

Apr 19, 2010
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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
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Well for example if we assume this circuit:

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

Last edited: Jun 27, 2010
4. Ron H AAC Fanatic!

Apr 14, 2005
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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?