# Recommended 2.5V common mode?

I'm using the ADS1675 ADC in a project of mine. It has a differential input which is +/- 3V. I'm confused by the datasheet that says the recommended common mode voltage is 2.5V. Why would you want a common mode voltage in a differential signal? And this is essentially just a DC offset right, so the actual input range relative to the ADC ground will be between -.5V and +5.5V?

So essentially I will need to add a 2.5V supply or reference and use that as ground for the input stage as in the picture below?

Also they use -4V and 9V as the amplifier rails. I get that this is like +/- 6.5V raised by 2.5V.. So would I just use the 2.5V reference as ground for a +/- 6.5V supply?

• Presumably because the supply rails have strict requirements for AVDD of 5V and AGND of 0V. So they want the input signal to be centred in the centre of the analogue supply rails. – Tom Carpenter Nov 24 '15 at 3:22
• NOTE that in that circuit you show, that is NOT the ADC, it is an ADC input driver/amplifier - note the part number. If you read the documentation it says that due to the high speed an input driver is required. – Tom Carpenter Nov 24 '15 at 3:24

The ADS1675 is a high speed ADC which runs of a $5\mathrm{V}$ Single-Ended power supply. If you note the following tables from the datasheet:

The differential input must be referenced to mid point of the supply rails which in this case requires a common mode voltage of $2.5\mathrm{V}$.

You then have an input full scale range of $\pm3\mathrm{V}$. I can only presume there is some attenuation internally which means that this full scale range is brought to within the specs of the device (need to clear that up with some more reading).

The circuit you show is for a THS4503 which is a high speed fully differential amplifier. The purpose of this is to either buffer or amplify the incoming signal to the full scale range of the ADC, and critically to drive the sample/hold capacitance of the ADC input.

Because of the high speed, you need a pretty strong driver to ensure that the capacitance of the sample capacitor in the ADC is driven quickly to the level of the incoming signal. This requires a fairly high current to drive the capacitor at high frequencies. So if you have a weak signal source it is important to amplify it before driving the ADC.

• So I'll need +5V, +3V, +2.5V, -4V, and +9V rails?? Seems like a lot.. Also should the preamp and vga I will need to use be referenced to 2.5V as well you think? – THEMuffinMan7 Nov 24 '15 at 3:41
• @THEMuffinMan7 You don't necessarily need to use that op-amp. There are others around. I'd look and see if you can find one which would do the job with a +5V/0V supply. You can also find ones which will add a common mode offset to the output allowing the input to be referenced to 0V (though these will require some form of negative supply rail). – Tom Carpenter Nov 24 '15 at 3:43
• Ok. The range of the ADC is +/- 3V though so wouldn't I be loosing out on dynamic range if Im not using something with a dual supply that can do +/- 3V? – THEMuffinMan7 Nov 24 '15 at 3:56
• Sorry for all the follow up questions but I was also wondering why the datasheet for that op amp they recommend says that it supports 14-bit operation? The ADC will be running at 23 bits.. ti.com/lit/ds/symlink/ths4503.pdf Thanks for all your help :) – THEMuffinMan7 Nov 24 '15 at 3:59
• @THEMuffinMan7 14bits at 40MHz. It's higher than that at lower frequencies (4MSPS means you are only interested in up to 2MHz). There is a graph here. Interestingly it is not in the datasheet, but it is on the product page. – Tom Carpenter Nov 24 '15 at 4:22