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David_

Hello.

I have trouble grasping this, MCP3553 is a single 22bit differential input ADC. It can be supplied with a VDD voltage of 2.7V-5.5V with VSS being ground.
The input range is Vref- to Vref+... If Vref = 3V, -3V to +3V but it says that the minimum input is VSS - 0.3V?

As a real circuit problem I am about to use a LTC2442, it can be used as ether 4 single ended channels or two differential.
Will I get more bang for my buck if I implement a single to differential stage and use the two differential input channels as opposed to using two of the single ended straight away?
(not counting the cost of the extra stage)

Regards

Hi,

An example: Vcc = 3.3V, VRef=3.0V:
--> input voltage range is +/- 3V.
It is the difference between the two inputs. So one could be 0.1 V then the other must be 3.1V for the ADC to see 3.0V
And the one is 3.1V and the other must be 0.1V
With a ref of 3V both input voltages needs to move in opposite direction to acieve full i put range.

The you could one input channel fix to 1.6V, then the other channel has an (gnd referred) input range of 0.1 ...3.1V

*******
Two differential vs four single ended.

Sometimes signals are not related to Gnd.
Prfessional audio studio equippment for example work with differntial signalling.
Imagine in one cable there are two wires. One carries the audio signal +A the other wire carries the inverted audio signal -A.
If now the cable is influenced with noise N (either capacitive or inductive coupled), then both cables are influenced the same way.

Wire1: signal = +A + N
Wire2: signal = -A + N

If you now feed both channel to differential input of an ADC, the the ADC sees:

AdCSignal = Wire1 - Wire2 = (+A +N) - (-A + N) = A + A +N -N = 2A
Only Audio signal remains, the influenced noise is cancelled out (ideally)

Klaus

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