5
\$\begingroup\$

I'm planning on shifting the output of an AD8226 instrumentation amplifier by 'about' 0.5 volts using a 2.5 v precision reference and a voltage divider. According to the datasheet this is a bad idea and a buffer must be used:

enter image description here

I understand that if I don't use a buffer I'll probably end up with a slightly different voltage shift and a slight increase in gain. I'm guessing these can be calibrated out by software without a problem. If yes, then why should I use a buffer? Also, there is a note at the last line about a degradation in CMRR. I'm using the device for reading a 0-10 v (very slowly changing) signal in an industrial environment (single ended). Will the CMRR degradation be significant?

P.S. Voltage divider used has two resistors 20k and 4.7k.

\$\endgroup\$
5
\$\begingroup\$

Because "for best performance, the source impedance to the REF terminal should be kept below 2 [ohms]".

If you made a resistive divider to meet that requirement, it would be 4 ohms in each arm, or 8 ohms between power and ground, which would likely consume a lot more power than you'd want to use for this function.

Why does the impedance need to be below 2 ohms?

It's explained in the text you posted:

enter image description here

\$\endgroup\$
  • \$\begingroup\$ I don't mind having that extra gain in the circuit. As I said I would remove it in software by calibration. With the issue of gain resolved (I hope), is there any other reason other than the CMRR degradation that I should consider, or is it safe to assume I can ditch the buffer? \$\endgroup\$ – hadez Apr 11 '16 at 3:38
  • 4
    \$\begingroup\$ If you don't care about CMRR, you can probably ditch the $3 in-amp and use a $0.50 op-amp for a similar function, though it's hard to know without knowing exactly what you're doing. \$\endgroup\$ – The Photon Apr 11 '16 at 4:11
5
\$\begingroup\$

Have a look at the (simplified) internal schematic of the AD8226:

enter image description here

Obviously having any significant (relative to 50K) impedance in series with the REF pin will negatively affect the common mode rejection. The recommendation of 2 ohms represents a degradation in the matching of 0.004%, or about 88dB. The guaranteed CMRR is 86dB at DC (gain = 1). The gain from the positive input is increased, not the negative input.

Does that affect your application? You'll have to run the numbers and see. There's perhaps little point in spending the money for an instrumentation amplifier if you're going to degrade it as any reasonable values of divider resistors will. By the way, there is no guarantee the resistors R3..R6 will be especially stable or accurate in absolute value. The requirement is for matching to achieve CMRR, not absolute accuracy, so you'll also be degrading stability most likely because of mismatch between tempcos.

\$\endgroup\$
  • \$\begingroup\$ Thanks. Two issues: 1- I'm not using the amplifier in differential mode: negative input is tied to ground, unity gain is used (this changes to 1.05 because of external resistors which doesn't make much difference for what I am doing). The inst-amp is used for protection of adc circuit and also converting 10 v analogue input into 2.5 volt for adc. Would this resolve the CMRR issue or will CMRR still be a problem? 2- You stated "you'll also be degrading stability most likely because of mismatch between tempcos". Do you mean the tempcos of the external voltage divider resistors? \$\endgroup\$ – hadez Apr 11 '16 at 6:06
  • \$\begingroup\$ The internal resistors will be mismatched with the external resistors. If it's a 5% difference that may not make a huge difference. If you're using it as a single-ended amplifier, I wonder why you wouldn't just use an op-amp and get better performance for less cost. \$\endgroup\$ – Spehro Pefhany Apr 11 '16 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.