Author Topic: 2nd. order Butterworth as attenuator  (Read 2078 times)

0 Members and 1 Guest are viewing this topic.

Offline IvoSTopic starter

  • Frequent Contributor
  • **
  • Posts: 310
  • Country: us
2nd. order Butterworth as attenuator
« on: September 21, 2014, 07:51:28 pm »
I need to generate clean 1kHz 1mV sinusiodal signal. I am using Arta soft on my notebook using headphone output. I can adjust generator in Arta soft to somewhere -21dB and lower the volume on the notebook and that gets me to 1mV out. The problem is the noise floor and therefore to take accurate measurement of 1mVAC. I have couple of multimeters they have 1uV resolution but the noise floor is a killer.
So I was thinking to get 100mV out from my notebook, which produces reasonably clean signal and then use LM49710 in inverting configuration with G=-100 to get me to 1mV.
I was trying to use FilterLab soft to make a 2nd order(or higher order) Butterworth low pass filter to reject unwanted noise but the problem is that options are only for gain 1+. Does it mean there is no chance to have low pass Butterworth as attenuator?
 

Offline T3sl4co1l

  • Super Contributor
  • ***
  • Posts: 21674
  • Country: us
  • Expert, Analog Electronics, PCB Layout, EMC
    • Seven Transistor Labs
Re: 2nd. order Butterworth as attenuator
« Reply #1 on: September 21, 2014, 09:20:39 pm »
1mV / 100mV = 0.01 gain, not -100.  Use a resistor divider.

If you still need a cleaner signal, you can filter it, but mind you will still be left with the noise in the passband.

Also, try the laptop on battery power, AC adapter disconnected.  The adapter often adds a lot of noise for some reason.

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline macboy

  • Super Contributor
  • ***
  • Posts: 2254
  • Country: ca
Re: 2nd. order Butterworth as attenuator
« Reply #2 on: September 24, 2014, 02:16:07 pm »
I would construct a precision 100:1 divider. You can verify the divider by either measuring it with DC or by measuring the resistor ratios directly (99:1). Then send a 100 mV signal into it, and trust that you have 1 mV coming out. Equipment that can accurately measure 1 mV is out of reach of most people. You can measure the 100 mV much more easily. Buffer the output of the divider to prevent voltage sag due to the load.
 

Offline David Hess

  • Super Contributor
  • ***
  • Posts: 16612
  • Country: us
  • DavidH
Re: 2nd. order Butterworth as attenuator
« Reply #3 on: September 24, 2014, 08:05:22 pm »
I would construct a precision 100:1 divider. You can verify the divider by either measuring it with DC or by measuring the resistor ratios directly (99:1). Then send a 100 mV signal into it, and trust that you have 1 mV coming out. Equipment that can accurately measure 1 mV is out of reach of most people. You can measure the 100 mV much more easily. Buffer the output of the divider to prevent voltage sag due to the load.

A simple divider is also easy to capacitively compensate if necessary for a flat frequency response.

I have done this exact thing to generate precisely known low voltage waveforms but I used a precision switched RF attenuator which neatly avoids any compensation issues.  I calibrated the attenuation at high signal levels were I could accurately measure the input and output and then relied on the now calibrated attenuation to know the miniscule output signal level based on the measured input signal level.
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf