AD8138
Production
The AD8138 is a major advancement over op amps for
differential signal processing. The AD8138 can be used as a
single-ended-to-differential amplifier or...
Datasheet
AD8138 on Analog.com
I am working with a legacy design using an AD8138 to drive an ADS5281. I'm trying to understand the following circuit:
The feedback resistors are symmetric at 1k2, but why are R65 and R63 split as they are, and what is the purpose of the R74/R68/R66 network? Why are the input resistors not simply symmetric?
C1ROH is an analog input from a back-plane connector. The input range is supposedly +/- 3V. The ADS5281 has 2 Vpp input range.
Hi tgbe ,
Good day. To further help you with your inquiry, I would like to clarify some details of the question. May we clarify where did you get the circuit presented above? Also, is the input range +/- 3V is equivalent to C1R0H? Lastly, what was your intended output level?
Regards,
Gilbeys
Hi Gilbeys,
thank you for your reply. The circuit is from a legacy design we're working on, where the original designer is not available anymore. I'd like to understand the implications of the resistor circuits for a possible redesign.
The The C1ROH input range is supposedly +/- 3V. The output should match the ADC's range, which is Vcm (1.5 V) +/- 0.5 V for each input, or 2 Vpp differential swing.
Thanks,
Thomas
Hi tgbe ,
Good day. The resistors at the input should be symmetrical to avoid mismatch errors. You might want to use ADI's DiffAmpCalc to redesign your circuit. Here's the link for download: https://www.analog.com/en/resources/interactive-design-tools/adi-diffampcalc.html. I hope this helps.
Regards,
Gilbeys
Hi Gilbeys,
thanks, that is what I thought. We'll have to redesign this circuit.
Thank you,
Thomas
Hi,
I believe that the question has been answered already so I am closing this thread. In case there will be additional questions, feel free to reply to this thread. Thank you!
Regards,
Paul
Hi,
I believe that the question has been answered already so I am closing this thread. In case there will be additional questions, feel free to reply to this thread. Thank you!
Regards,
Paul