LTC2492 input range

Hi Nickbaxev ,

We will look into this, I'll contact the product owner and get back to you.

Regards,

JC

Hi Nickbaxev,

If you bias COM at 1.5V, You will have an input range of 0-3V on each of the CH0-CH3 inputs by setting SGL=1 in the MUX address.

Many thanks for the answer! I assume you refer to the situation (c) of Figure 38 of the datasheet.

In that case, would it be safer to bias COM at say 1.36V, so that the differential signal will always be under Vref/2? I made my input readings less than Vref peak to peak to ensure that no reading will be close to being out of range.

And, how important is the stability of the biasing? I assume that since the reading is differential a simple resistor divider from Vref (say 15k/18k resistors and a 100nF cap to ground) to COM would do the trick. Or do I need something more stable and clean?

EDIT: I feel that not selecting precision resistors for the biasing would ruin the accuracy, since if input signal is stable but COM moves the differential input will be false.

With COM at 1.25V, the input range will be 0-2.75V. You really don't gain anything by doing that. You cannot go below ground.

The COM pin is just like any of the analog inputs. There will be a current glitch during each sample period so you need to have a low impedance input or you need to have the impedance matched with the analog input being measured.

I understand that the full differential range is Vref (so 3V), but if I make COM = 1.5V, then if CH0 = 0, wouldn't that result in Vdifferential being -1.5V? So if COM = 1.5001V and CH0=0, wouldn't that throw the differential measurement out of range? From your answer I realise that this is not the case, I just don't understand why.

I feel that making COM = 1.5V needs excellent DC precision AND stability, but making COM = 1.25V would require just stability, because the differential reading would never approach the -0.5*Vref margin. Why is that wrong?

I am not challenging your input, I just don't understand how the 3V range is what matters (maybe I am missing something basic about ADC operation). I am eager to be mistaken and learn though! 

(Many thanks for your clarification about COM node impedance too)

You are correct about if COM=1.5001V and CH0=0V. From that standpoint setting COM to 1.25V will give you some margin.

So if I am not wrong, the key is that inputs are truly differential (as stated in the datasheet), so CH0 for example can only go Vref/2 up or below COM, so if I keep COM at 0, my positive only inputs would need to be Vref/2 max to be read. Since they go to 2.5V they would go out of range once they surpassed 1.5V (Vref/2). So the proper way as ghoover suggested is to bias COM at a convenient voltage instead. 

Figure 38 actually explains this pretty well, but for some reason I did not catch it up. Many thanks!