Can I use a non-inverting op-amp configuration for driving SAR ADC instead of buffer?

Hi Darrell,

This can be a difficult question to answer as it will depend on the characteristics of the amplifier you use, the acquisition time of the SAR, and the resolution you want. Buffers are typically used because this maximizes the loop gain of the amplifier, which keeps the output impedance as low as possible and results in the best load transient response (Zout = Zo / 1 + AolB). You want the output impedance to be low because this impacts how quickly the amplifier can drive the charge required when the SAR switches the acquisition capacitor in. If you are using our calculator, the tool assumes a buffer is used.

https://www.ti.com/tool/ANALOG-ENGINEER-CALC

Adding circuit gain reduces the loop gain of the amplifier, resulting in a somewhat higher output impedance. Adding a feedback capacitor and limiting the bandwidth of the amplifier as much as possible will improve this though. 

I would suggest simulating the front end of the ADC with your amplifier of choice in the noninverting configuration. As long as you have a good amp model this will be the best way to tune your charge bucket and verify the settling performance. It sounds like you are familiar with this but if not we have a whole precision labs module on SAR ADC drive design and how to simulate the settling into the SAR!

https://training.ti.com/ti-precision-labs-adcs-introduction-sar-adc-front-end-component-selection?context=1139747-1140267-1128375-1139106-1128643

Best,

Zak

Hey Darrell,

Happy to help! One of my old colleagues, Ian Williams, wrote a very good article that steps through the derivation of this equation and demonstrates how higher gains lead to higher Zout. You can find it here: https://www.edn.com/designing-with-a-complete-simulation-test-bench-for-op-amps-part-1-output-impedance/

You'll notice from the equation and curves that closed loop output impedance increases with frequency as the loop gain (the Aol*B term) of the amplifier becomes smaller.

Since switched waveforms are comprised of a fundamental plus many higher frequency harmonics, the output impedance of the amplifier at higher frequencies is going to play a large role in the amps ability to settle back to the desired voltage after that acquisition cap is switched in.

This is one reason why you typically need a driving amplifier with much higher bandwidth than the signal you intend to capture. Higher bandwidth amps are going to have much more loop gain left at higher frequencies and consequently lower output impedance at those frequencies. This is also why the charge bucket is so handy, it allows you to use a much lower bandwidth amplifier than would otherwise be necessary because a lot of the charge can be quickly drawn from the charge bucket cap rather than from the amplifier.

Best,

Zak