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Category: Hardware

In AD8146, if i leave my VOCMC pin as floating, my Common-mode voltage is fixed as +2.5V. And in Pg.No: 15, it is described as resistive divider with an impedance of approximately 12.5 kΩ set this level means a resistive divider of R1=R2=12.5K is pull up with +5V
sets this level. Now I need to set my common mode voltage to -0.6V. So,is it okay to pull up with 8.87K to -5V to generate -0.6V.

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• Based on simulations, I don't think that you will be able to bring the effect of the input offset to the output back to zero by applying a voltage on the Vocm pin. Please see the simulation…

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• Hello ,

To set the common mode voltage to -0.6 V, you can directly control it by applying -0.6 V on the Vocm pin. Just make sure to follow the Input Voltage Range (IVR) of the Vocm pin for proper operation:

For Vs = ± 5 V (Dual Supply)

For Vs = 5 V (Single Supply)

In your case, since you have a Common-Mode Voltage of 2.5 V when the Vocmc pin is floating, I'm guessing that you are using Vs = 5 V (Single Supply). For proper operation, please follow the IVR of the Vocmc pin when VS = 5 V and make sure that the Input voltage is within the Absolute Maximum rating of the pin to avoid permanent damage to the device.

Regards,
Paul

• Sorry for the confusion, I am using Dual Supply mode, +/- 3V3 voltage.

So, my Common mode voltage is 0V, if i leave floating.

My signal input to AD8146 is single-ended with offset of +0.9V. Output is differential.

Now, I want to bring it down to 0V. Is that possible by controlling common mode voltage of AD8146. Because adding AC Caps will  degrade my signal and i don't have enough space to add in Board file.

If my VS is connected to +/- 3,.3V.  What is my maximum input voltage to Common mode pin

• Hello ,

Based on simulations, I don't think that you will be able to bring the effect of the input offset to the output back to zero by applying a voltage on the Vocm pin. Please see the simulation below:

From here, we can see that applying a voltage of 1V on the Vocm pin introduces an offset of about 1 V both on the Vo+ and Vo- pins. Getting the differential output voltage between the two pins, a voltage of about 400 mVpp that has a common mode of about 20mV is obtained which is correct. Assuming Vin = 100mVp (200mVpp). G = 2 (Internal gain)

Removing the voltage on the Vocm pin, when your input has an offset of 0.9 V, this offset will also appear on the part's output pins but with different polarities. (On Vo+ pin, Output offset  0.9 V; Vo- pin, Output offset ≈ -0.9 V)

When a voltage has been applied to the Vocm pin, this voltage will be added to the output offset introduced by the offset voltage in the input. Please see the simulation below:

This is the reason why I don't think it is possible to bring the effect of the input offset to the output back to zero.

Hope this helps.

Regards,
Paul

• Hello ,

Based on simulations, I don't think that you will be able to bring the effect of the input offset to the output back to zero by applying a voltage on the Vocm pin. Please see the simulation below:

From here, we can see that applying a voltage of 1V on the Vocm pin introduces an offset of about 1 V both on the Vo+ and Vo- pins. Getting the differential output voltage between the two pins, a voltage of about 400 mVpp that has a common mode of about 20mV is obtained which is correct. Assuming Vin = 100mVp (200mVpp). G = 2 (Internal gain)

Removing the voltage on the Vocm pin, when your input has an offset of 0.9 V, this offset will also appear on the part's output pins but with different polarities. (On Vo+ pin, Output offset  0.9 V; Vo- pin, Output offset ≈ -0.9 V)

When a voltage has been applied to the Vocm pin, this voltage will be added to the output offset introduced by the offset voltage in the input. Please see the simulation below:

This is the reason why I don't think it is possible to bring the effect of the input offset to the output back to zero.

Hope this helps.

Regards,
Paul

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