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HELP! Vu-meter with LM393

Discussion in 'Homework Help' started by SpeakZ12, Jun 30, 2018.

  1. SpeakZ12

    SpeakZ12 New Member

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    Hi! I'm new here and i need a little help with a project. I need a circuit diagram of a VU-meter, using this components ; x4 - LM393 , x10- 5k and x10 - 10k resistors, x8 LEDs yellow and a 9V battery , 1 condeser microphone with 2 terminal
     
    Last edited: Jul 1, 2018
  2. audioguru

    audioguru Well-Known Member Most Helpful Member

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    You also need a voltage reference IC, current-limiting resistors for the LEDs and a resistor ladder that sets the input voltage level that turns on each LED.
    The LM393 is destroyed if you use an AC audio input (its maximum allowed negative input is only -0.3V) so you must rectify it so it is positive voltage only or add a negative supply so that the supply has dual polarities.

    The LM3915 IC was used for many years as a VU meter but it is obsolete now. Look at its datasheet online. Here is its block diagram:
     

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  3. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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  4. dave miyares

    Dave New Member

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  5. audioguru

    audioguru Well-Known Member Most Helpful Member

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    I designed and built a Sound Level Indicator with an LM3915 IC over 12 years ago. It has a microphone as its input and it has been in my living room most of the time but it has a rechargeable 9V battery so it has been used outside to measure the sound of dogs barking, sirens and noisy cars. It uses a fast acting peak detector to accurately show exact levels and it stretches the duration of short duration sounds to 30ms so that our slow vision can see them brightly. I added a circuit that increases its sensitivity 20dB when sound levels are low so that it has a 50dB range then it shows faint sounds and very loud sounds.
     
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  6. AnalogKid

    AnalogKid Well-Known Member

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    4 393's get you only 8 indicators, ok but pretty coarse. With nothing but 5K (?) and 10K resistors, you cannot build a VU meter because those are not the correct resistor values to create a group of logarithmic-spaced comparator circuits.

    Is this homework, classwork, etc?

    ak
     
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  7. SpeakZ12

    SpeakZ12 New Member

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    Yeah...
     
  8. dave miyares

    Dave New Member

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  9. SpeakZ12

    SpeakZ12 New Member

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    You cand show me a circuit diagram please? :wideyed:
     
    Last edited: Jul 1, 2018
  10. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Since this is your homework, what circuits have you checked so far? Post links to a few and we can advise where to go from there.
    Are your LEDs rated for a specific voltage, i.e. do they have built-in current-limiting resistors?
     
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  11. audioguru

    audioguru Well-Known Member Most Helpful Member

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    My sound Level Indicator circuit has the same circuit as in the datasheet of the LM3915 which is now obsolete. Maybe you can find a fake or defective one on ebay.
    My circuit uses a microphone so I designed and built a mic preamp for it that you do not need. Also, I added a peak detector that you might not need.
    Here is my project:
     

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  12. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Does your 2-terminal condenser microphone have a built-in power source or does it need to draw bias current from the 9V battery via the circuit?
     
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  13. SpeakZ12

    SpeakZ12 New Member

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    Current from the 9V battery via the circuit.
     
  14. SpeakZ12

    SpeakZ12 New Member

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    This is an example. I think is not a good circuit diagram. Show me the way :D
     

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  15. alec_t

    alec_t Well-Known Member Most Helpful Member

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    That's about as simple as a non-microprocessor circuit gets (in the absence of a LM3914 or LM3915 IC) for driving a LED group according to input signal strength. However, (a) it doesn't have a VU's logarithmic response and (b) there is nothing to prevent the input being driven below the negative rail. These requirements were pointed out above. Given the limitations of the components you have, I don't see how you would adapt that circuit to meet the requirements. Consult your instructor to see if the requirements can be relaxed or additional components can be made available.
     
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  16. SpeakZ12

    SpeakZ12 New Member

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    Thank you!
     
  17. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    That circuit is just this: (more or less)
    [​IMG]
    You will need a way to convert the audio signal to dc. Like in post #9 picture 1.
     
  18. audioguru

    audioguru Well-Known Member Most Helpful Member

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    My circuit uses an electret mic because a condenser mic needs a 48V power supply plus a preamp with a very high input resistance. An electret mic is a newer condenser mic permanently stores the 48V in its electret material and has a Jfet inside with a very high input resistance. The Jfet must be powered, my circuit use R1 connected to a regulated and filtered 5V.

    You are powering the LEDs with a 5k series resistor. If the LEDs are 2V old green then the current when the battery is new is only (9V - 2V)/5k= 1.4mA which is dim. If the LEDs are 3.2V new bright green and the battery has dropped to 6V then the current is only 0.56mA which is very dim. Use Ohms's Law to calculate a suitable resistor.

    Your circuit has a wire that is shorting the battery.
    Your circuit is missing a regulated reference voltage to feed the resistor ladder. Its reference voltage drops as the battery voltage runs down. Your circuit is also missing an important supply bypass capacitor that all electronic circuits must have.
    The top LED will light when the battery is new and the input signal is higher than +7.2V. Where will that come from? It is too high for an earphones signal or from line level and its -7.2V will destroy the LM393 ICs without rectifying the signal.
    The datasheet for the LM3915 LED bar graph IC says that the LEDs will be a dim blur if a peak detector is not used.
     
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