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Resistors and?

Discussion in 'General Electronics Chat' started by Electronman, Apr 3, 2009.

  1. Electronman

    Electronman New Member

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    Thanks, Ron,

    A question which comes to my mind is that if the R4 and R3 have this relationship (R3/(R3+R4)=0.75
    Then every pair of resistors will work instead of those both originals while their relationship is 0.75, right?

    By the way I guess with 100% modulation I will lose my carrier while I need it, right?

    So what's your idea if I am able to use your current modification so that it handles a dynamic mic? I do not know if that gain (150) is sufficient for a dynami8c mic?
     
  2. Roff

    Roff Well-Known Member

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    0.75 is not optimum for TL072, because it does not have rail-to-rail outputs. The ratio for TL072 is about 0.69.

    No the carrier is still present with 100% modulation.

    I don't know how much output a dynamic mic has,
     
  3. Electronman

    Electronman New Member

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    But is my assumption right?

    Ops. So how we remove carrier freq? I know that balanced modulators have a pot to remove it. I was thinking that those modulators do so with making the modulation index to 100%?
     
  4. dave

    Dave New Member

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  5. Roff

    Roff Well-Known Member

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    Yes, the ratio is all that is important, but you must take the resistor values into account when choosing the value of the AC coupling cap that brings in the audio.



    All this time you have been making an amplitude (unbalanced) modulator. If you are wanting to experiment with air nonlinearity and the demodulation effect caused by it, I think, from what I have read, that you want an unbalanced modulator.
    Maybe this is not what you are trying to do.:confused:
     
    Last edited: May 4, 2009
  6. Electronman

    Electronman New Member

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    Are you meaning that I must consider the R4 and R3 when I chose the cap or you were meant about R10?
    I never underestood for what aim those AC coupling (the resistor in series with that cap) are for.

    Thats exactly what I want to do. I do not need a balanced modulator but before reading your other post I was think that 100% modulation means that the carrier will be removed at the spectrum. thats why I asked you if I have the carrier yet with 100% modulation.
    It seems I was wrong about that. and as you told my modulator has the carrier at its output even with 100% modulation index (that's what I want).
     
  7. Roff

    Roff Well-Known Member

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    I am not going to try to teach you the fundamentals of RC circuits. You can learn a lot about electronics by reading something like the E-book at AllAboutCircuits.
     
  8. audioguru

    audioguru Well-Known Member Most Helpful Member

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    A dynamic mic has about the same output level as an electret mic, about 5mV to 15mv RMS when speaking in a normal conversational level at a distance of about 20cm.

    An electret mic and the 10k resistor that powers it from 9V is about 3.3k ohms so its preamp should have an input impedance of at least 16k ohms. An low noise opamp with FET inputs will have low noise.

    A dynamic mic is an impedance of about 150 ohms to 600 ohms and should have a preamp with an input impedance of 1.2k to 3k. A transistor or inverting low noise opamp can be used.
     
  9. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi agu,
    You been off air for a few days.:)

    Thought you might find a use for this LTSpice.zip
     

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  10. audioguru

    audioguru Well-Known Member Most Helpful Member

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  11. Electronman

    Electronman New Member

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    I just did not understand your meaning at your other post.
    I did not know which resistor did you reffering too. I Do know that C2 is for bypassing the AC signals to ground but can not understand where is that AC signal on the inverting pin of the Op-AMP?? I asked that question but could not undersrand the replies! maybe the AC signal comes from the output of the op-amp through the feedback pot?
     
  12. Electronman

    Electronman New Member

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    Will have low oise...?!
    Did you use voltage divider and used the voltage across the mic to obtain that number i.e 3.3k?

    Why I am not able to use Ron's last mode and connect an input resistor to the opamps which he has used on it (U1) instead of using another opamp or a tansistor?
     
  13. Electronman

    Electronman New Member

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  14. Roff

    Roff Well-Known Member

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    On an op amp with negative feedback, the voltage on the inverting input will be equal to the voltage on the noninverting input, as long as the signal is within the closed loop bandwidth of the circuit.
     
  15. Electronman

    Electronman New Member

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    So?...?
    It seems I was not able to tell my question because I do not know why you replied by the above..?
     
  16. Roff

    Roff Well-Known Member

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    I quoted your question, and replied to it.

    Here is the specific part of your question that I answered:
     
  17. Electronman

    Electronman New Member

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    Regarding to your last modification,
    The U1 op-amp has a non inverting pin which seeks the Ac signal generated by V5 (the sine wave source). Because the inputs of an op-amp use a differential amplifier so the AC input signal in each of inputs of the op-amp will affect the other input, right? If so, I think I can understand the main reason of using that capacitor. The higher the value for that cap the more AC goes to the ground and more amplified signal (by the non inverting input pin) we will have at the output (because the inverting pin has lower AC signal which affects the output too) right?
    That’s why when I remove that cap I lose much power at the output of the opamp.
     
  18. Roff

    Roff Well-Known Member

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    The voltage at the inverting pin is (almost) exactly the voltage at the noninverting pin. That principle is what makes an op amp such a powerful and versatile tool.
    When the impedance of C2 is equal to the resistance of R10 in the circuit below, the signal at the output of U1 will be -3dB (0.707) relative to the midband gain. Below that frequency, the gain will be cut in half for each halving of the frequency. The formula for the corner frequency is
    F(-3dB)=1/(2*pi*R*C).

    In this case, R10=10kΩ, C2=1µF, so
    F(-3dB)=1/(2*3.14*1e4*1e-6)
    F(-3dB)=15.9Hz.
    As the frequency goes lower, the gain goes lower. As the frequency goes higher, it will flatten out at
    Gain(midband)=(R11/R10)+1

    I explained previously the factors (gain-bandwidth product divided by midband gain) that determine the upper -3dB frequency, where the gain again rolls off, but as the frequency goes higher.
     
  19. Electronman

    Electronman New Member

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    Ron,
    I put a 18V transformer (it was a 1.2A transformer) to your last circuit, I used a 7815 regulator chip with a 2200uF cap on its input and a 1000uF cap on its output. I have some of 'hum' and a bit of noise at the output of the modulator now. The sound appears to be low and high by itself frequently. What's the solution?
    Before training a transformer I ran the circuit with a commercial regulated power supply at 14V without much problem.
     
  20. Roff

    Roff Well-Known Member

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    Can you post aa complete schematic of the power supply, including the transformer, rectifier, regulator, and caps?
     
  21. Electronman

    Electronman New Member

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    I tried a 104 Cap on both the input and the output of the regulator but did not observe any difference.
     

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