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Electronics - Op-amps: when will the LED light up

  1. Jan 1, 2012 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    http://img193.imageshack.us/img193/6627/circuitg.jpg [Broken]

    So I got this circuity thingy and I'm told that each LEDs only lights up at 1.4 V. I am asked what should be the vaue of Vin1 where the green LED lights up.

    Now, I see I have 2 comparators here. Each one is 15 volts Vcc (either -15 or +15). So, why does the value of Vin1 matter? I just see what's bigger, V+ or V-, and I anyway get 15 volts, so no matter what the LED will light up.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 1, 2012 #2

    vk6kro

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    Notice that the inputs of the comparators are different for the two devices.

    Also, the 3 series resistors in the circuit have 30 volts across them so how much voltage is at the inputs of each of the comparators?

    What would happen if you tied the two inputs together and increased the input voltage from -15 volts to +15 volts?
     
  4. Jan 1, 2012 #3

    Femme_physics

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    What makes you say that?

    It doesn't matter whether it's plus or minus, voltage is just the difference in potential, if I know my stuff correctly. So, what matters is the number.
     
  5. Jan 1, 2012 #4
    Hello for 2012
    These op amps have no feedback so the output must be +15 or -15 (I am discounting that V at the +/- inputs may be EXACTLY equal!!!!!!!)
    The output depends on which is greater V at the -input or V at the +input
     
  6. Jan 1, 2012 #5

    Femme_physics

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    I know, but they don't ask for Vout, they ask a very weird question - "the range of voltage value of Vin1 where the green LED (the upper one) lights up". And we need to explain our solution. The way I see it, it lights out no matter what because it gets either 15 volt or -15 volts. Regatdless, it's more than enough.
     
  7. Jan 1, 2012 #6
    Can you see what the voltage is at the - input of the top op amp?(VK6 has given a clue)
    Then you need to look at what the effect of different voltages on the + input will have on the output.
    You are correct to identify these op amps as 'comparators'
     
    Last edited: Jan 1, 2012
  8. Jan 1, 2012 #7

    Femme_physics

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    Yes, voltage divider!

    http://img17.imageshack.us/img17/5968/image201201010016.jpg [Broken]

    Well, if Vin1 is more than 7.5V than the voltage is -15V

    If Vin1 is less than 7.5V than the voltage is +15V
     
    Last edited by a moderator: May 5, 2017
  9. Jan 1, 2012 #8

    vk6kro

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    The LED will only light if the output of the comparator is positive by more than 1.4 volts.

    Each resistor drops 10 volts, so, relative to the - 15 volt supply, what is the voltage due to the series resistors at the two comparators?

    This is the homework section, so we can't just do it for you, however this is a well known circuit and you are missing the point of it.

    That voltage divider calculation is not correct. You have 3 resistors with 30 volts across them.
     
  10. Jan 1, 2012 #9
    Your potential divider calculation is not correct. 3 equal resistors in series across 30V
     
  11. Jan 1, 2012 #10

    Curious3141

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    As others have said, the voltage divider calculation is incorrect. 3 equal resistors, a total of 30V (+15V -(-15V)). And the top of the bridge is at +15V while the bottom is at -15V. After figuring out the drop per resistor (hint, they're all equal since the resistances are equal), you need to do the signed addition of voltages here to find the actual voltage at each point. Keep everything with reference to the ground voltage (0), which is not actually shown here. So you can have positive and negative voltages.

    In simpler voltage bridge problems, the top is at +V, and the bottom is at ground (0), with just 2 resistors. So what you did would be appropriate there. This is a more complicated case, but luckily all 3 of the resistances are equal making it easy.
     
    Last edited by a moderator: May 5, 2017
  12. Jan 1, 2012 #11
    First it should be noted that no connection between ground and the power supply is shown so technically the power supply is floating and the LEDs would never light. However since that is too simple of an answer we will assume the power supply is grounded at halfway between the two supply voltages and the power supply voltages are relative to ground, not each other. This means the difference between the two supplies must be (+15 - 0) - (-15 - 0) = 30 volts.

    Now with three equal resistances between +15V and -15V what must be the voltage across each resistor be? What will be the voltage at the junction of R1 & R2 relative to ground? (In other words, subtract the voltage across R1 from +15V.) Likewise at the junction of R2 & R3?
     
  13. Jan 2, 2012 #12

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  14. Jan 2, 2012 #13
    You have got it
     
  15. Jan 2, 2012 #14

    vk6kro

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    Now this is no longer homework, you might like to know that this is a "window" circuit.

    If you tie the two inputs together, the bottom LED will light if the input voltage is less than -5 volts and the top LED will light if the input voltage is greater than +5 volts.

    This just depends on the ratio of the series resistors, so you could set this up to give you warning messages or turn on alarms.

    If you were running a camping light off your car battery, you could sound a warning if the voltage was getting too low to start your car.

    Notice the similarity between your circuit and this one, which shows the insides of a LM3914 chip.
    200971341723454.gif

    This chip drives up to 10 LEDs depending on the input voltage, but it works the same way as your circuit.
     
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