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Moisture detector with darlington

  1. Jun 20, 2010 #1
    1. The problem statement, all variables and given/known data

    A pair of bare copper wires a, b is attached to a basement wall to detect the presence of moisture. Design a circuit, using a darlington transistor, to turn on an LED if the current flowing between the wires due to the moisture exceeds 10μA (use a current-sensing resistor to convert the current into a voltage which is then sensed by the darlington transistor). Use a 5V supply. Indicate component values.

    2. Relevant equations


    3. The attempt at a solution

    R = V/I = 5/10 = 0.5MΩ.

    The correct answer for the earthed resistor is 100Ω but I dont understand how to get this.

    Attached Files:

  2. jcsd
  3. Jun 20, 2010 #2
    I suspect the correct answer is not 100Ω but 100KΩ. At what voltage does the Darlington begin to conduct?
  4. Jun 20, 2010 #3
    It conducts when Vin > 1.2V.
  5. Jun 20, 2010 #4
    So if you want to develop 1.2 V with 10 uA, what value resistor do you need?

    (It might be that the Darlington begins to conduct at a Vbe of 1.0 V, at least for this example.)
  6. Jun 20, 2010 #5
    The way I see it is that 10μA splits at the common point where the transistor base line joins with R's line. But I am not sure how to calculate this current which flows through R, so that I can use R = V/I to calculate the correct R.
  7. Jun 21, 2010 #6
    This example assumes all of the 10 uA goes through the resistor. Any current over 10 uA flows into the transistor and assumed to be sufficient to turn on the LED.

    I am a little curious about your usage of R = V/I. R and I represent resistance and current whose units are ohms and amperes respectively. The product of those two quantities is electromotive force which is represented by E. V is the unit of electromotive force. Do your instructors teach Ohm's Law as V = I*R? In order to be consistent why not use V = A*Ω instead?
  8. Jun 21, 2010 #7
    Wouldnt this disagree with the rule that current spilts when 2 lines are joined? Although I do know that if Vin < 1.2 at the base, there is no IB.

    I have been taught it as V = I*R, and what ever component values are in the diagram are used without conversion e.g. if I = 10μA and R = 0.5MΩ then V = 10μA * 0.5MΩ and the "μA" and "MΩ" cancel ( 10^-6 * 10^6) and so you get V = 5V.
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