Window Detector Circuit

Discussion in 'Homework Help' started by Zaraphrax, Aug 14, 2009.

  1. Zaraphrax

    Thread Starter Active Member

    Mar 21, 2009
    47
    3
    G'day all,

    I'm working on a project, and as I'm a bit new to electronics, I'd just like to have someone else look over my answer before I commit.

    I have the following circuit:

    [​IMG]

    I have calculated the voltage ranges that each of the LEDs will be on and off.

    LED1:

    Reference voltage (negative terminal of U1) = 7.5 V

    So this means that it will be on between 7.5V and 15V?

    LED2:

    Reference voltage (negative terminal of U2) = 3.75V

    Will be on between 3.75V and 15V.

    Is that correct? Thanks.
     
  2. bertus

    Administrator

    Apr 5, 2008
    15,647
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    Hello,

    Your calculations are not correct.
    Look at the resistor values R1, R2 and R3.
    They are bound between ground and + 15 Volts.
    Calculate the voltage accross each of the resistors.

    Greetings,
    Bertus
     
  3. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,158
    If your wanting a window detector, with the LEDs illuminating when the voltage is outside the boundaries you decided upon, then your diagram is also incorrect.
     
  4. bertus

    Administrator

    Apr 5, 2008
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  5. Zaraphrax

    Thread Starter Active Member

    Mar 21, 2009
    47
    3
    Sorry for the bump, looking for some more guidance.

    Ok, I've revised my solution.

    Would this be correct?

    15V across the rail. Disregarding Vin for the moment.

    1) Voltage across R1 - 15 * (10000/30000) = 5V

    For U1, this means the lower voltage reference is 5V. The upper voltage reference will be Vin. This means the comparator will pull the load to HIGH between 5 and Vin, provided that Vin > 5. If Vin < 5, it will be on outside of this range.

    2) Voltage accross R2 = 5 * (10000/30000) = 1.6667V

    For U2, the upper voltage reference is Vin and the lower voltage reference is 1.6667V. The LED will be on between 1.6667V provided Vin > 1.6667V. If Vin < 1.66667V, it will be on outside of this range.

    Once again, not sure if this is correct (I'm thinking I might have it backwards...).
     
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Did you have a look at the page I gave?

    There you would find this picture:

    [​IMG]

    For the calculation of the reference voltages:
    What is the current flowing through the resistors R1,R2 and R3?
    If you multiply the found current by the value of the resistors, you wil find the voltage accross the resistors.

    Greetings,
    Bertus
     
    Last edited: Aug 21, 2009
  7. Zaraphrax

    Thread Starter Active Member

    Mar 21, 2009
    47
    3
    I did indeed. I *should* have written on here that is the voltage DROPPED, not the voltage across each of the resistors.

    So R1 will be 10V and R2 will be 5V. (apparently I need more sleep). We aren't given a current, but now that I've got the voltages etc I can calculate it using Ohm's law, I suppose (but I don't know if I really need it). But, is my theory right?

    LED1 will be on between 10V and the input voltage (when greater than 10V), and when the input voltage is less than 10V, it will be on outside of that range.

    Ditto LED2 except at 5V?

    That's the main concept I'm having trouble understanding.
     
  8. bertus

    Administrator

    Apr 5, 2008
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