Internal Resistance Creates offset at High Common Mode

Hi,

If the common mode voltage is 200V, the 4mV error is normal, it means CMRR of 94dB. See Table 7. Error Budget Analysis Example in the datasheet to learn more about how to calculate the errors. Use such value for Rsense, that gives large enough full scale. For example if the maximum voltage on Rsense is 1V, the 4mV offset means only 0.4% error.

This offset can be reduced (almost removed) by inserting a resistor in the wire connecting Rsense and the positive or negatíve input of the AD8479. In reality both case can occur, since the internal resistor errors are varied from device to device. If you try to simulate this, do not forget to move the in- and in+ labels to the terminals of Rsense, because that is the reference.

Zoltan

Hi Zoltan,

I had already tried the suggestion for adding Rcomp=Rshunt as defined in the datasheet but it did not correct the offset error in the IN+ or IN- signal path.

The datasheet specifies

"To measure low current or current near zero in a high commmon-mode voltage environment, an external resistor equal to the shunt resistor value can be added to the low impedance side of the shunt resistor as shown in Figure 41"

Upon running this simulation by placing Rcomp in the IN+ path I see that the error gets worse as Rcomp increases. Verr(Rcomp=1) = ~4mV
Verr(Rcomp=10) = ~6mV
Verr(Rcomp=100)= ~22.5mV

When I place Rcomp in the IN- path I do see an improvement in voltage error from Vrsense to AD8479 output

Verr(Rcomp=10) = ~2mV
Verr(Rcomp=20) = ~240uV
Verr(Rcomp=30) = -1.6mV

Trying difference resistor combinations Rcomp=21.3 gives me an error closest to 0.

The problem is that the datasheet says that I should be able to use a Rcomp=Rshunt but that is not working in my simulation. I don't know how to derive Rcomp=21.3 so how would I pick a nominal value?  I've attached my latest simulation.  Plot below show simulation with Rcomp=21 to 22 in 0.1ohm increments.

Regards, Tony

8372.AD8479_200VCM.asc

Hi Tony,

AnthonyStockalis said:

datasheet says that I should be able to use a Rcomp=Rshunt

Do not forget, that it is from the section USING A LARGE SHUNT RESISTOR, which is not your case. It only compensates CMRR error introduced by the large shunt resistor, it does not improve the amplifier’s specified CMRR. Look at the functional block diagram. The amplifier’s CMRR comes from the tolerance of the internal resistor network. For example, if the 1M connected to IN+ is a bit smaller, then adding a small external resistor can reduce the output error close to zero, as you’ve seen by the simulation. So, I think the simulation model uses this case. But in reality,  the error of the internal resistors is different from device to device. Therefore you can only compensate it by experimenting.

But I think, it would be better to use larger full scale, such a good CMRR that this amplifier offers should be enough. What is your requirement? What is the current range you want to measure and what accuracy is needed?

Zoltan

Hi Zoltan,

Use case #1 - Measure high side current sense of a >200V power supply with a full scale of ~50mA.

In my application I should be able to do an 0A offset calibration so that could make up for any part to part variation, but I'd still want to have an Rcomp value that gets me close.

Regards,

Tony 

Hi Tony,

I'm sure that there is no theoretical value for Rcomp (other than zero) to compensate the mismatch of the internal resistor network, since there is no intentional CMRR error. This error comes from the manufacturing process, so it can be anything within the specifications. So I recommend to follow the datasheet to choose Rcomp to compensate the effect of Rshunt. Use 200 Ohm for Rshunt to get 10V full scale for the 50mA full scale current.

Zoltan