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Thread: 741 op-amp circuit

  1. #1 741 op-amp circuit 
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    hi everyone i have been trying to learn about 741 op-amp ICs and came across a very simple circuit but i can't get it to work. I am using 9V DC batteries instead of 12V and have connected positive and negative of each battery with the other and the other terminals to IC's input. That is how this is suppose to work right?? plz see the circuit given below and thank you for your help


    http://www.electroschematics.com/wp-...-indicator.gif


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  3. #2  
    Cooking Something Good MacGyver1968's Avatar
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    When you say, "It doesn't work"...please be a little more specific. You could power it with 2 9 volt batteries in series, so you can get both +9Vdc and -9Vdc for your circuit.


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  4. #3  
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    i mean that it is suppose to turn on one led at a time when either positive or negative polarity is provided to R1 but it turns on both leds at the same time when i give power to it and has no change when probe is connected to R1. so basically it is just switching in the leds when provided with DC power
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  5. #4  
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    With using only a single 9 volt battery, your ground and -V are the same. It may be screwing things up, since the circuit is designed for +V and -V. Take two 9 volt batteries and connect the negative terminal of battery 1 to the positive terminal of battery 2, and use this for your ground. the positive terminal of battery 1 is your +V and the negative terminal of battery 2 is your -V.
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    yes that is what i have done but it doesn't seem to be working
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  7. #6  
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    Do you have a voltmeter or multi-meter?
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  8. #7  
    Cooking Something Good MacGyver1968's Avatar
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    I haven't worked with a 741 since electronics school in '94...so I'm a bit rusty, but there are others here that really know their shit. Also, if you could explain how the circuit is being used.
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  9. #8  
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    yes and i have checked the voltages and after a bit of twerking now it's lighting up both leds when provided with positive voltage at pin 3 but does not light up either of the leds at negative voltage
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  10. #9  
    Cooking Something Good MacGyver1968's Avatar
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    Hmmm....what does fiddling with the pot1 do? Fiddling with pots is always a good approach.
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  11. #10  
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    this circuit is supposed to detect polarities of the input signal. when provided with positive voltage at pin 3 D1 will light up while negative voltage at pin 3 will cause D2 to light up
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    i am unaware of what fiddling with pots is
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  13. #12  
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    P1 is a potentiometer...or adjustable resistor. "fiddling" or playing around with it's setting can might help tune to the circuit into working.
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  14. #13  
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    yes i have tried it but no change
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  15. #14  
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    i am starting to think that this circuit is faulty
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  16. #15  
    Cooking Something Good MacGyver1968's Avatar
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    It could be...just hang out a bit...There are other very knowledgeable EE's on the board that might see something I'm missing. Like I said, it's been almost 20 years for me.
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  17. #16  
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    Quote Originally Posted by fine View Post
    i mean that it is suppose to turn on one led at a time when either positive or negative polarity is provided to R1 but it turns on both leds at the same time when i give power to it and has no change when probe is connected to R1. so basically it is just switching in the leds when provided with DC power
    If it is lighting up both LED's. then one of your LED's is in backwards.
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  18. #17  
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    see^
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  19. #18  
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    Quote Originally Posted by billvon View Post
    Quote Originally Posted by fine View Post
    i mean that it is suppose to turn on one led at a time when either positive or negative polarity is provided to R1 but it turns on both leds at the same time when i give power to it and has no change when probe is connected to R1. so basically it is just switching in the leds when provided with DC power
    If it is lighting up both LED's. then one of your LED's is in backwards.
    i have checked both leds several times and have tried changing directions also but to no avail
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  20. #19  
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    Quote Originally Posted by fine View Post
    i have checked both leds several times and have tried changing directions also but to no avail
    Then try this.

    Disconnect R4 from the op amp. Attach that (now disconnected) side of R4 to +V. One LED should go on. Now attach it to -V. The other LED should go on. If not, one of the LED's is backwards.
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  21. #20  
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    Quote Originally Posted by fine View Post
    Quote Originally Posted by billvon View Post
    Quote Originally Posted by fine View Post
    i mean that it is suppose to turn on one led at a time when either positive or negative polarity is provided to R1 but it turns on both leds at the same time when i give power to it and has no change when probe is connected to R1. so basically it is just switching in the leds when provided with DC power
    If it is lighting up both LED's. then one of your LED's is in backwards.
    i have checked both leds several times and have tried changing directions also but to no avail
    That's because your circuit is not producing a DC output -- it is producing AC. The LEDs are telling you this.

    Why is the output AC? You have chosen resistor values that produce an enormous voltage gain. That means that any noise on the input, or on Venus, will be amplified to LED-lighting voltages.

    Reduce your gain and the problem will go away (reduce the 100k resistor to something much smaller, like 10k or even less). Or even better, reverse the + and - inputs to the op-amp. This will convert the circuit from an amplifier to a thing called a Schmitt trigger.

    Btw, remove the potentiometer. It is just a source of noise injection for your circuit. If you need precision, you are using the wrong op-amp. Tweaking with a pot will not really help that much.

    See, e.g., Horowitz and Hill, The Art of Electronics.

    ETA: Make sure that your wires are all short. Also, you did not show any capacitors to filter the supply, so I assume you have none. Add them. Recommended: Around 10 microfarads; one each to ground from each battery, located as close to the op-amp as practical. Make sure that you get the polarities right, or the capacitor will stop capacitating.
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  22. #21  
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    the leds are working like they are supposed to
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  23. #22  
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    Quote Originally Posted by fine View Post
    the leds are working like they are supposed to
    Above you say that both LEDs are always on. Are you now saying that they aren't?
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  24. #23  
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    bump
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  25. #24  
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    tk421 knows his stuff. I can't add anything.

    Try all that he suggests, then see if it works.

    OB
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  26. #25  
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    thanks for all your help guys. Just one last final question. i have removed the potentiometer and have placed both the leds correctly, now when i give DC power the negative activated led is on even when there is no input at pin 3. When i provide pin 3 with positive voltage the negative activated led switches off and the positive activated one turns on as it should. When pin 3 is grounded both the leds turn off as should be the case, but when no input is given the negative led is still on. I have tried reducing the circuit gain from 150 to 50 to 2 but the result is the same. Any explanation would be greatly appreciated
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  27. #26  
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    Quote Originally Posted by fine View Post
    thanks for all your help guys. Just one last final question. i have removed the potentiometer and have placed both the leds correctly, now when i give DC power the negative activated led is on even when there is no input at pin 3. When i provide pin 3 with positive voltage the negative activated led switches off and the positive activated one turns on as it should. When pin 3 is grounded both the leds turn off as should be the case, but when no input is given the negative led is still on. I have tried reducing the circuit gain from 150 to 50 to 2 but the result is the same. Any explanation would be greatly appreciated
    The circuit is working properly now -- congratulations! Its behavior should be properly indeterminate if you allow pin 3 to "float" (i.e., have an uncontrolled potential). When you have no input at all, then there is nothing dependable determining the voltage on that pin.

    To fix this last little problem, add a resistor from pin 3 to ground. That will ensure that, in the absence of any explicit connection, the voltage on pin 3 will be well controlled. A value of 10k to 100k should do the job.
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  28. #27  
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    Fixin' shit that ain't broke.
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