# Signal attenuation to increase ADC range

Hello again, I need some help with an ADC (MCP3903).
I am trying to increase its analog input range by x2 by halfing the
signal voltage. Originally, I tried this (http://www.ti.com/lit/an/sbaa244/sbaa244.pdf)
by using the external resistor to create a voltage divider but it did not work for some reason.

Does anyone know of a method of attenuating the input signal without having to connect it to ground?
(I assume it would increase noise in the signal).

Thank you in advance for your replies.

Did not work in what way? The voltage was not attenuated? It was attenuated, but by the incorrect amount? You got zero signal?

Originally, I tried this (http://www.ti.com/lit/an/sbaa244/sbaa244.pdf)

There are a couple of “funny” things about that… It’s differential and the ADC seems to have a “known” high impedance which is being used as part of the voltage divider. (You are using a different ADC.)

Your ADC is also differential, but is your source differential? What’s the impedance of your source?

A “regular-old” [u]voltage divider[/u] will probably work. And if your source isn’t high-impedance the resistors in your voltage divider should typically sum-up to around 10K. (Higher impedance is more prone to noise pick-up and lower impedance will draw more current from the source.)

With a voltage divider the signal and noise should be attenuated together so the signal-to-noise ratio shouldn’t be affected, unless the ADC (or whatever follows the voltage divider) is generating noise.

If your signal is differential, you can make your own differential voltage divider with 3 resistors. That’s 3 resistors in series with the balanced output coming off the middle resistor. The top & bottom resistors should be the same and the attenuation is the middle resistance to the total resistance.

Using 1k resistors had a small risk of blowing the MCP3903 protection diodes in that circuit - check
its still working with normal voltages. Protection diodes are very tiny and are often only rated for
1mA or thereabouts - they are normally only designed for static protection, nothing more.

Then try a 10k+10k resistive divider (single-ended). Having 10k in series with the inputs is much
more robust to overvoltage, but if you suspect regular overvoltage or under-voltage events, its
good practice to bolster on-chip protection diodes with external schottky diodes.

The MCP3903 has 16/24 bit resolution.

Do you really need better?

Any resistive divider to be accurate to this level would need resistors of enormous accuracy - very expensive.

And only to achieve 1 more bit.

What's the problem?

Allan

without having to connect it to ground?
(I assume it would increase noise in the signal).

It will not increase the noise.

You could a,ways use an op-amp with a gain of 0.5

Klagemauer:
Does anyone know of a method of attenuating the input signal without having to connect it to ground?
(I assume it would increase noise in the signal).

If you are doing differential, use a differentiatial divider, otherwise yes you'll inject common-mode
noise into your measurement.

If single-ended then ground is by definition completely noise-free as its 0V.

Too large resistance values will add Johnson noise, whether this matters depends on your noise bandwidth,
so you have to figure that out - basically if your signal of interest has a particular bandwidth it always
pays to low-pass filter and remove noise at higher frequencies. And of course you'll probably need an anti
aliasing filter anyway.

Too large resistance values will act as a low-pass filter anyway, so you really need to know what
your source impedance, desired bandwidth and accuracy are.

polymorph:
Did not work in what way? The voltage was not attenuated? It was attenuated, but by the incorrect amount? You got zero signal?

Originally it just gave me complete nonsense so I switched to a different power supply and now it works.

DVDdoug:
There are a couple of "funny" things about that... It's differential and the ADC seems to have a "known" high impedance which is being used as part of the voltage divider. (You are using a different ADC.)

Your ADC is also differential, but is your source differential? What's the impedance of your source?

In this case, my source is does not have a differential input. What I would like to try however is make use of the common mode to offset the sampling range.( If id didn't misunderstand how this works)

Unfortunately I don't have the input impedance right now but Ill try to make it available as soon as possible.

DVDdoug:
If your signal is differential, you can make your own differential voltage divider with 3 resistors. That's 3 resistors in series with the balanced output coming off the middle resistor. The top & bottom resistors should be the same and the attenuation is the middle resistance to the total resistance.

Would this still make sense when using the ADC with a single signal input (other one is connected to ground)

Grumpy_Mike:
It will not increase the noise.

You could a,ways use an op-amp with a gain of 0.5

Would using an op-amp in this case have advantages?

MarkT:
Too large resistance values will add Johnson noise, whether this matters depends on your noise bandwidth,
so you have to figure that out - basically if your signal of interest has a particular bandwidth it always
pays to low-pass filter and remove noise at higher frequencies. And of course you'll probably need an anti
aliasing filter anyway.

Too large resistance values will act as a low-pass filter anyway, so you really need to know what
your source impedance, desired bandwidth and accuracy are.

Right now I tried tripling my range so I used 750kohm resistors in combination with the 350kohm
internal resistance of the ADC. That did work, and I am using capacitors to filter the noise.
Do you think the resistance is unusually high?

What do you guys think would be the best solution?
-The voltage divider to ground.
-The voltage divider in relation to the adc's internal resistance?
-An op-amp?
-Would I still be able to offset the input range by by connecting -ve to a different voltage?

Thank you all for your responses.

The internal 350k resistance of the ADC will be very ill-defined and unstable I suspect. The datasheet says
it depends on the clock speed so this is a switch-capacitor effect.

You probably need to drive this ADC direct from a quality opamp like an AD8656 (5V rail to rail low noise high speed).

Otherwise you risk getting poor performance with a naive divider, it seems to assume low input impedance.

For a gain of 0.5 use a voltage divider into the opamp as a unity-gain buffer.

Klagemauer:
Originally it just gave me complete nonsense so I switched to a different power supply and now it works.

Then why did you even mention that? Why not start with where you are now? You gave the clear impression that you did not have a working circuit.

Instead, it sounds like it is working, that you only need to attenuate the input.

This is why I waited for you to respond.

MarkT:
The internal 350k resistance of the ADC will be very ill-defined and unstable I suspect. The datasheet says
it depends on the clock speed so this is a switch-capacitor effect.

You probably need to drive this ADC direct from a quality opamp like an AD8656 (5V rail to rail low noise high speed).

Otherwise you risk getting poor performance with a naive divider, it seems to assume low input impedance.

For a gain of 0.5 use a voltage divider into the opamp as a unity-gain buffer.

That sounds promising, thank you. However I am not too familiar with op-amps. Would this be the correct configuration for the device? With both resistors at 1kohm?

However I am not too familiar with op-amps. Would this be the correct configuration for the device? With both resistors at 1kohm?

Yes that is right.