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I'm looking to sample a 0 to +15 V analog input signal, and it is not sinusoidal, rather it is impulse driven. I don't need to sample at a high rate (< 1kHz), but I need to sample over the full range of the signal. The way I see there are two options:

  1. Buy a more expensive ADC with a +/- 10 V range and try to bias the input to fit into that swing. This would require two voltages supplies though I think. I may be wrong though...
  2. Attenuate the input signal to make the signal swing fit in the range of normal low-cost ADCs

While 2) seems more difficult in the design, it certainly seems to have a better cost benefit based on what I've seen from Analog and Linear's offerings.

By attenuating the signal, do I risk losing anything though? I was thinking that if the ADC has the same sample bit width as that of the larger swing ADCs, the samples could be scaled digitally in software so that it appears the initial signal voltages are sampled.

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    \$\begingroup\$ 'negative gain' sort of implies polarity inversion.. unless you specify it in decibels. Sounds like you want a fractional or sub-unity gain, like for example 0.25 or 0.667, to reduce a 0 to 15 swing to something smaller. \$\endgroup\$ – JustJeff Jun 11 '11 at 14:28
  • \$\begingroup\$ Thanks for pointing that out. You're right, I should be talking about gains like 0.5 and 0.1 rather than "negative". \$\endgroup\$ – Dr. Watson Jun 11 '11 at 23:55
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Elementary, Watson. You sortof had the idea with #2, except that you don't want a negative gain but rather a gain between 0 and 1. In other words, you want to attenuate the 0-15 V input signal to match the input range of your A/D.

This is easily accomplished with two resistors in a "resistor divider" configuration. If your A/D has a native range of 0-5V, then you want to divide the input voltage by 3. This can be accomplished, for example, with 2K Ohms in series followed by 1K Ohms to ground.

Anything you do to a signal will always change it slightly. In this case, some of the high frequencies will be lost. However, at impedances of 10s of K Ohms, this won't be a problem with a sample rate of 1KHz or less. That implies a upper frequency limit of 500Hz maximum, rather less in practise. Even 100s of K Ohms used in the resistor divider should be able to pass such low frequencies without losing the part you care about.

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  • \$\begingroup\$ some pedantic weenie is bound to point out that a voltage gain of 1/3 is -9.5dB, so in that sense is a negative gain .. but you have my +1 \$\endgroup\$ – JustJeff Jun 11 '11 at 14:33
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    \$\begingroup\$ @JustJeff - if he writes -9.5dB he still has a lot to learn about being pedantic. -9.542425dB would be more like it :-) \$\endgroup\$ – stevenvh Jun 11 '11 at 14:55
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    \$\begingroup\$ @stevenvh - it may be possible to be pedandtic and pragmatic simultaneously =P \$\endgroup\$ – JustJeff Jun 11 '11 at 15:04
  • \$\begingroup\$ Scintillating, Holmes! \$\endgroup\$ – Dr. Watson Jul 17 '11 at 16:43
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Like Olin said, a voltage divider will do nicely. Keep in mind however that the ADC input impedance is in parallel to the lower resistor and that it may influence your divider ratio; ADCs often have rather lower input impedance. You can buffer the divider's output with an high input impedance buffer like an opamp.

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The LT1677 can operate from a single supply and has rail-to-rail input and output. Since you don't need to sample at high sample rates I presume that the bandwidth of the LT1677 is wide enough to cope with the impulses.

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    \$\begingroup\$ you can even put resistors in the feedback path to reduce the voltage swing at the ADC input. you could even use a potentiometer to do this and make the circuit tweakable, and the variations in impedance won't be reflected back to the source. \$\endgroup\$ – JustJeff Jun 11 '11 at 14:36
  • \$\begingroup\$ @JustJeff - You could, but why use extra components if you already have the resistor divider at the input? \$\endgroup\$ – stevenvh Jun 27 '11 at 4:59

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