Signal conditioning at amplifier input stage - Page 1
Forum section: electronics / projects

author: MihaiSs
author_rank: Contributor
timestamp: August 26, 2023, 10:02:12 am
content: Hello everyone,I was going through one of the Texas Instruments papers regarding high voltage signal conditioning for low-voltage ADCs:https://www.ti.com/lit/an/sboa097b/sboa097b.pdf?ts=1692964757665&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FOPA2277At page 4, there is a circuit which got my attention (Circuit 2: Single-Supply/Single-Port Approach).From my understanding of the circuit, there is a 1.5V reference voltage created and buffered through the OPA335 and fed into the N1 node and this basically creates a voltage divider between the input voltage and a fixed 1.5V. Why is this method used like this?Is there an advantage of the circuit instead using a simple voltage divider at the input directly to GND? Also, is the buffer really needed?I'm trying to wrap my head around it since I'm sure I'm missing some important reason.Thanks a lot!

author: moffy
author_rank: Super Contributor
timestamp: August 26, 2023, 11:32:11 am
content: Quote from: MihaiSs on August 26, 2023, 10:02:12 amThe input is divided with respect to Vcc/2 or 1.5V in this case so that if you have 0V on the input you get 1.5V on the output of the amplifier. This allows the opamp to swing from 0V to 3V for a -10V to +10V input, so it is both voltage dividing and level shifting at the same time. It has converted a bipolar signal into a unipolar one. As the app note states "...Signal conditioning for low voltage ADCs", which are mostly a single unipolar supply.P.S. If you wanted to remove the 1.5V buffer you could keep the 1k/1k divider and connect a 2.51k resistor to the divder in place of the R4/3.01k.

author: MihaiSs
author_rank: Contributor
timestamp: August 27, 2023, 05:02:11 pm
content: After I read your message I put everything on a paper and now it all makes sense. It's quite a clever way to both divide the signal and shift it towards the positive swing of the amplifier when you don't have a bipolar supply available. Thanks a lot for the explanation!One question regarding the removal of the buffer: why the decision to decrease from 3k to 2.51k?

author: moffy
author_rank: Super Contributor
timestamp: August 28, 2023, 11:08:31 am
content: Quote from: MihaiSs on August 27, 2023, 05:02:11 pmBecause the Thevenin equivalent of a 3V supply divided by two 1k resistors is a 1.5V voltage source and a 500R resistor in series i.e. the two 1k resistors in parallel. So you need to drop the 3.01k resistor value by that 500R to 2.51k because they add together i.e 2.51k + 500 = 3.01k.

author: MihaiSs
author_rank: Contributor
timestamp: August 29, 2023, 09:54:18 am
content: Quote from: moffy on August 28, 2023, 11:08:31 amUnderstood, it basically boils down to keeping the same ratio between the 1.5V supply and the +-10V input at node N1.Thank you very much!

author: Terry Bites
author_rank: Super Contributor
timestamp: August 29, 2023, 12:34:11 pm
content: Its a bog standard differential ampliifer circuit.If thats not obvious then you're getting way ahead of yourself.https://www.electronics-tutorials.ws/opamp/opamp_5.htmlRef is buffered. Lets not get carried away, stand down Norton and Thevenin!

author: MihaiSs
author_rank: Contributor
timestamp: August 29, 2023, 01:09:19 pm
content: Quote from: Terry Bites on August 29, 2023, 12:34:11 pmMy first post was not reffering to the differential amplifier itself. I was reffering to the method of conditioning the input signal and then I asked a second question regarding the buffer. I don't think we are getting ahead of ourselves with this discussion. I find it quite fruitful.