Differential Amplifier Output - Analog Devices Calculator

Hi elsiesemico ,

Good day.

The output common-mode voltage of the circuit is set using the value of the Vocm pin. It is true that the differential mode output voltage, Vout,dm, is equal to VoutP and VoutN. However, the Vout,dm is also controlled by the Vocm value in such a way that the Vdiff is measured with respect to the Vocm. Since the common mode output voltage is forced internally to the voltage applied at the Vocm pin, the Vout,dm or Vdiff should be {2...4...2...0...2} which is the same as that of the ADI DiffAmpCalc. 

Hope this helps.

Regards,
Gilbeys

I agree that Vocm sets the VoutP and VoutN at the selected 2 Volts.

However, isn't it true that the differential output is the result of the substraction between the VoutP and VoutN?

Thus the dc offset set by the Vocm should be eliminated, since it is the same for both pins?

Also, does this mean that when this output applied to ADC with fully differential input then the FSR is limited to 0...4 Volts?

Hi elsiesemico,

Good day. Yes, that is true that the differential output is the difference between the VoutP and VoutN. However, the plot you see on the ADI DiffAmpCalc is centered around the Vocm and does show you how the differential output swings around the Vocm.

In applications where a fully differential op-amp is used as an ADC driver, it is important to consider how the input amplitude and phase of an ADC are balanced. Hence, when driving an ADC, the common-mode voltage is one of the parameters to be reviewed. In such an application, the Vocm is usually set using the reference voltage of the ADC. It is better to know the specification of the ADC first before considering the op-amp. But it is true that the FSR will be around 4 V when you applied it to an ADC. This may help you in the design of an ADC driver using a Fully Differential Output: https://www.analog.com/media/en/technical-documentation/app-notes/an-2555.pdf.

Regards,
Gilbeys

Hi Gilbeys,

Based on the definition provided here [https://www.analog.com/en/design-center/glossary/difference-amplifier.html#:~:text=Difference%20amplifiers%20and%20instrumentation%20amplifiers,variation%20of%20just%20one%20resistor]  as well as in the application note that you provided the differential amplifier outputs the difference between its inputs, thus to my understanding there should not be a dc offset when measuring the differential output. Thus, still I do not understand why in the waveform shown, the differential voltage is swinging around the Vocm and not around 0 Volts.

If this is the case then in all circumstances the differential voltage will exceed the maximum input voltage of the analog inputs of an ADC since in most of the cases the voltage can swing from approximately -Vref to Vref (Vocm+Vout,dm is prohibited).

If you attempt to change the Vocm to achieve a differential output within the -Vref…Vref requirement of an ADC, then the individual outputs (VoutP, VoutN) will violate the absolute maximum ratings for the individual analog inputs of the ADC (i.e. Vin+ to GND, Vin- to GND) which usually are rated from approximately 0 to Vref.

Also in the document that you provided there is not any explicit statement whether the FSR of the ADC is exploited based on the output swing of the differential amplifier and this requirement in not examined.

Hello,

the differential voltage Vdiff=Voutp-Voutn will be in the range -2V to +2V. If you drive a fully differential ADC with it, you can use the complete input range of the ADC.

If you would use an oscilloscope with a differential probe and measure the differential signal, the oscilloscope would show you a voltage symmetric to GND (-2V to +2V), and the GND-line is typically in the middle of the osci-screen. But if you use the GND-Position button of the osci and shift the GND-line upwards by Vcm, the complete signal will be in the upper half of the osci screen.

That's more or less what the software tool does: it plots Vdiff not around the 0V line but around Vcm. That looks nicer (as the centerline for Vdiff, Voutp and Voutn coincides). But it is also in my eyes somewhat misleading, as the real values of Vdiff=Voutp-Vout will range from -2V to +2V.

best regards

Achim

Hi,

Yes, I agree with what Achim had said. The DiffAmp Calculator plots the Vdiff with respect to the Vocm line to compare it with the Voutp, Voutn and Vocm.

Regards,
Gilbeys