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Hello electronics experts,

I do not yet have enough reputation to post more then two images. For the images please see: http://imgur.com/a/dizis enter image description hereenter image description hereenter image description hereenter image description here I have a single ended signal that i want to convert to be differential so that I can use the full scale of the differential ADC (ADS1278).

The single ended signal that i want to use is 0V - 2.5V. The ADC input range goes from +Vref to -Vref, i have the Vrefp tied to 2.5V and the Vrefn tied to ground. To convert the signal a THS4521 differential amplifier is used.

Now from what I know, when the single ended signal looks like image 1 the differential signal should look like image 2.

But when I use the single ended to differential schematic from the amplifiers datasheet image 3 (Vocm is set to 1.25V), The output of the amplifier gives me the signal of image 4.

I can see that the pk-pk voltage of this resulting differential signal is the same as the single ended signal (2.5V). But is is different from what I thought that was going to happen (image 2). Can anyone explain to me why this happens or what I am not understanding about differential signals?

Thank you, Sisco

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  • \$\begingroup\$ What is the DC offset voltage on your input signal? \$\endgroup\$ – Andy aka Jan 22 '16 at 11:15
  • \$\begingroup\$ The English word "I" is always spelled upper case. \$\endgroup\$ – Olin Lathrop Jan 22 '16 at 12:14
  • \$\begingroup\$ I have edited the question to have all upper case "I" \$\endgroup\$ – Sisco Jan 22 '16 at 12:22
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Well, this is exactly what should happen ! The reason is as follows:

Your input signal is a sinewave between 0 - 2.5 V

I could also say that your input signal is a 1.25 V (2.5 V /2) DC voltage + a sinewave with an amplitude of 1.25 V.

The thing that confuses you is the 1.25 V DC ! It is converted to differential as well, resulting in the DC voltages you see at the outputs.

If you want to prevent this you need to make that DC voltage at the input 0 V. This might be impractical.

Fortunately what you can also do is shift the DC reference voltage of the single to differential converter. I see that is has a Vocm input, if this is a common-mode level input then you could apply 1.25 V DC to it (the DC voltage I mentioned above) then the single to differential converter will consider 1.25 V as the commonmode level and the outputs will get the same DC voltage.

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  • \$\begingroup\$ Thank you for the clear asnwer! The Vocm input sets the output common-mode voltage which is from what I understand the DC voltage at which the differential voltage (Vdiff+ - Vdiff-) is zero. When connecting the negative input to 1.25V, would this give me the same effect? \$\endgroup\$ – Sisco Jan 22 '16 at 12:28
  • \$\begingroup\$ You mean instead of connecting ground at the left side of Rg2, connect that to 1.25 V ? Yes that would indeed give the same effect. You can explain this by assuming Vin is 1.25 V DC and the left side of Rg2 is also 1.25 V, then the opamp's in and outputs will also be at 1.25 V. \$\endgroup\$ – Bimpelrekkie Jan 22 '16 at 12:36

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