band-pass lowering the signal

Discussion in 'General Electronics Chat' started by picozero, Oct 21, 2011.

  1. picozero

    Thread Starter New Member

    Oct 21, 2011
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    Hi all!
    I am designing a standard second order bandpass for the range [0.5-5] Hz :
    [​IMG]

    High-pass section:
    C1 ,C2 = 1 microF
    R1 = 450 kOhm
    R2 = 225 kOhm

    Low-pass section:
    R1,R2 = 48 kOhm ≈ 47 kOhm
    C1 = 0.5 microF
    C2 = 1 microF

    I am using an LM386, and non electrolitics capacitors.
    The input signals is suppose to be a square wave of max amplitude of 2V.
    The power supply +Supply is meant to be 9V and +Supply/2 4.5 (a 9V battery and a voltage divider made by two equal resistors).
    The gain is unitary. The signal I am using to test is very small (from the frequency generator of the oscillator).

    I have done it and tested with an oscillator and a frequency generator (integrated in my amplifier).
    Varying the frequency it looks like the passband is working (at 1 kHz the signal is almost completely attenuated and there is no more square wave).

    Now we go to my questions:
    The amplitude is reduced :eek:.
    1) How can this be possible if my gain is unitary ?
    I don't have a graduated oscilloscope so I cannot check how much is the dicrease, but it looks like 1 Volt/Div.
    Is there any explanation for this ? The behavior appears even if I remove the High-part section.

    2) Is the filter really working ? I mean, can be that the attenuation is an effect related to the tesion decrease ?

    3) Is there anything besides the AC-DC switch in the oscilloscope I should change when I play with AC signals ?

    Thank you in advance.
     
    Last edited: Oct 21, 2011
  2. crutschow

    Expert

    Mar 14, 2008
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    3,229
    1) At what frequency is the signal reduced? What do you mean "it looks like 1 Volt/Div"?

    2) What is "tesion decrease"?

    3) The oscilloscope input should be set to DC. You only use the AC setting if you need to block a DC bias on the signal.
     
  3. Audioguru

    New Member

    Dec 20, 2007
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    1) an LM386 is not an opamp. It is a little power amplifier with built-in biasing and built in negative feedback for a gain of 20 or 200. Its datasheet says that it oscillates at a high frequency if its gain is less than 10. Use real opamps for your circuit to work properly.

    2) You have the values of C1 and C2 in your lowpass filter mixed up. For a Butterworth response, C2 must be double the value of C1.

    3) The source that feeds your circuit must have a fairly low impedance like the output of an opamp.
     
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  4. picozero

    Thread Starter New Member

    Oct 21, 2011
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    Thank you for your quick answers.

    Yes, the values of C1 and C2 were swapped. I corrected the original post.

    Regarding the LM386, I am using it at 1Hz which I think it is a very small frequence.. anyway, the reason is that for what I am planning to use the filter I cannot use a dual power supply. Can you advice me any single supply op.amp ? Or I can create virtual ground as in the picture even in a 741 ? I am not familiar with operationa amplifier single supply

    The other answers:
    I tried 1Hz, 10Hz , 100Hz, 1KHz, 10KHz, 100KHz...
    at 10Hz and 100Hz the signal is reduced and from 1KHz on the signal is almost disappeared.

    The input should be set at DC even if I am looking at periodic Square wave ?
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    An LM386 power amp will not work in your circuit.

    Any opamp (it does not need a single-supply opamp) will work in your second circuit (a single supply circuit) except some 741 opamps. The 741 opamp is 43 years old and is designed to use only a 30V supply. Some 741 opamps work from a 10V supply but many do not. Use a newer and better opamp. Some opamps work fine from a 9V supply that drops to 6V when the battery voltage runs down.

    The circuit will pass 0.5hz to 5Hz and will attenuate lower anmd higher frequencies at -12dB per octave. So 10Hz will have an amplitude that is only 1/4 the amplitude at 2Hz. 100Hz will have an amplitude that is only 1/320. 1kHz will have an amplitude that is very small.

    It doesn't matter if the square-wave input has a DC voltage because the input capacitor on the highpass filter blocks DC.
     
  6. picozero

    Thread Starter New Member

    Oct 21, 2011
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    Ok I believe you, I try with a 741 otherwise I go and by some TL071, I will let you know

    but I'd like to understand:
    Why does the lm386 won't work ? I cannot understand it from the datasheet.
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    It is a power amp that drives a speaker, it is not an opamp.
    1) It has built-in biasing so it works with a single supply and with its inputs at 0V. It will not work in your circuit that makes its input at +4.5V.

    2) It has built-in negative feedback and will not work in your circuit with a gain of 1.

    Opamps do not have built-in biasing and do not have built-in negative feedback and most work fine with a gain of 1.
     
  8. picozero

    Thread Starter New Member

    Oct 21, 2011
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    Ok, Thank you.

    Summary of the experiments.
    I tried it in single and dual supply configuration using a UA741, a NE553 and a LM358.
    First of all for some strange reasons C1 in the low-pass must be smaller than 220 nF. Otherwise it won't work. Therefore I adapted the values accordingly (C2 the double and R1 and R2 the expression with the frequence). People are welcome to explain to me why I am suppose to use caps smaller than 0.5 uF.

    All the three op.amps worked properly filtering if used in a dual power supplier configuration.

    Only the LM358 worked as a single supply. I have done exactly as in the figure and there is no way to make it work as a single supply for the UA741 or the NE553.
    However, It seems that the LM358 reduces a bit the negative part of the waves I am using to test it.

    Moreover I don't see any changes if I add the additional cap Cout. Is it just for decoupling the circuit ? Why is it not present in the dualsupply configuration ?

    I am reading some guide about the single supply but I haven't figured out yet the answers to my questions. Anyway using the LM358 the circuit is behaving more as planned. The original problem was the IC in use. Thanks
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    There is no "NE553". Maybe you used a dual NE5532 or a NE5534 that will not work properly with a gain of 1?

    0.5uF is not a standard value. 0.47uF is a standard value. The circuit should work perfectly with two 0.47uF capacitors in parallel for C2 and one capacitor for C1.

    How did you make the "+ Supply/2" (+4.5V) for the single-supply circuit?
    All three opamps (except maybe not the old 741 opamp) should work in the single-supply circuit.

    The LM358 is 'low power" so its output low has low current. What is the load resistance of your circuit?

    It blocks DC. The opamp output in the single-supply circuit is at +4.5V. The opamp output in the dual-supply circuit is 0V so the output capacitor might not be needed.

    I think your load resistance is too low (it should be 2k ohms or more) and your +4.5V is wrong.
    Post a schematic showing how you made the +4.5V and say what is the load resistance.
     
  10. picozero

    Thread Starter New Member

    Oct 21, 2011
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    I used a NE5532.

    No, unfortunately I tried that and it doesn't work. But it can be something related to the value and input I am using. I agree that these values should work.

    Load resistance ? I am using right now only this circuit (or even just the low-pass to test), the oscilloscope and the wave generator. I guess the only load resistance is the internal resistor of the generator probe (almost nothing I assume far below 2k ohms ). Should I put a resistor between the wave generator probe and the input of the circuit.

    I have done as manual, using a voltage divider. Is it wrong ?
    I posted the schematics, now only with the low-pass. Since I am oversimplifying the circuit to understand this single supply business.

    [​IMG]
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    What is "1muF" which you show for C1 and C2?
    Why are the values of R1 and R2 backwards now? They were correct before.

    Why did you change the lowpass and highpass filters around so the input is now lowpass? Then the signal generator might short the "virtual ground" (+4.5V) to ground. Or now you must add an input coupling capacitor.

    Your schematic shows the voltage divider using very high value resistors which allows the input signal to feed into the 225k resistor. A "virtual ground" is supposed to be a low impedance. Try 1k or 100 ohm resistors for the voltage divider.
     
  12. picozero

    Thread Starter New Member

    Oct 21, 2011
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    Thank you for the reply.
    Sorry, maybe I misunderstood seriously how to measure this thing when it is single supply.:confused:

    I prefer to stick with the low-pass, since it is simpler (that's why I am not caring if the values in the High-pass are correct).

    According to this picture from a texas instruments guide:
    [​IMG]The input is connected to the virtual ground (as you noticed I should have included a cap).
    Now what do you mean exactly with
    "Then the signal generator might short the "virtual ground" (+4.5V) to ground." ?

    where are suppose to be connected the terminals of the probes ?(is this what you meant ?) The negatives still to the "real" ground and the positive one as input of the circuit and the other at the output. Right ?
     
  13. Audioguru

    New Member

    Dec 20, 2007
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    Your new circuit is a good lowpass filter. The "virtual ground" is at +4.5V (when the supply is 9V) and it biases the input of the opamp so its output can swing symmetrically up and down from +4.5V.

    Without the input coupling capacitor then the output of the signal generator might short the +4.5V of the virtual ground to 0V if it has an output that is at 0VDC.

    Copnnect the red probe to the output of the output coupling capacitor and the black probe to 0V (circuit's ground).
     
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