Circuit loading input offset - Page 1
Forum section: electronics / beginners

author: sy
author_rank: Regular Contributor
timestamp: February 29, 2024, 09:25:43 am
content: Hello,I am confused about how this offset circuit works, and how to choose appropriate resistor values.I imagine that the 10K is dropping most of the supply voltage compared to the 1K at the wiper, to minimize loading effects, but I don't get how the circuit interacts with the rest of the input to minimize loading at the input?Could someone please help explain?Many thanks!

author: antoniobeta
author_rank: Contributor
timestamp: March 01, 2024, 10:44:10 pm
content: Can you give us more information about the circuit, where is it used or where are you planning to use it?

author: sy
author_rank: Regular Contributor
timestamp: March 01, 2024, 11:34:55 pm
content: Hello antonio,To give more context this circuit is applied at the input terminal of the op-amp to correct for the input offset voltage. So say for an inverting amplifier configuration, the bottom leg of this circuit would connect to the inverting terminal of the op-amp.

author: shapirus
author_rank: Super Contributor
timestamp: March 02, 2024, 12:37:57 am
content: Something doesn't add up here, it looks like you may be trying to wire it in a wrong way. More context on the schematics (including the op amp and the input signal terminal) would shed more light on this.Additionally, you will find useful info on (supposedly) this topic inhttps://www.ti.com/lit/an/snla140d/snla140d.pdf, starting from page 12.

author: Vovk_Z
author_rank: Super Contributor
timestamp: March 02, 2024, 09:18:29 am
content: 10 k resistor gives you a >= 10 k additional impedance but not <=250 R from just 1 k only.Quote from: antoniobeta on March 01, 2024, 10:44:10 pmLook at a datasheet of some precision opamp with a trimming circuit. Pretty common circuit for a fine tuning of a zero offset.

author: Terry Bites
author_rank: Super Contributor
timestamp: March 04, 2024, 09:11:34 am
content: The current in R2 is small compared with the current in the pot. This has a small effect on the output voltage.This not ideal and introduces a bit of non linearity in the pot voltage vs wiper position.This is often tolerable. If you are offsetting an opamp say and V1 and V2 are stable, the circuit works ok.If the current required by the thing needing the offset voltage is small R2 can be increased.See equivalent circuit showing the node voltages and currents.You can accurately calculate the voltages using the superposition theorem for a T-network.Seewww.electronics-tutorials.ws/dccircuits/superposition-theorem.html

author: Zero999
author_rank: Super Contributor
timestamp: March 04, 2024, 11:03:12 am
content: That circuit makes no sense on its own.Here's the full circuit.P1 forms a variable potential divider. The voltage at the wiper, which is connected to R3 varies between -15V and +15V, depending on the potentiometer setting. R3 and R4 form another potential divider, which divides it by 1501, giving a an adjustment range of just under +/-10mV. The impedance seen at the potentiometer's wiper is near zero at either extreme and 5k when at the centre position, which increases the divider ratio to 2001, so the potentiometer less sensitive when it's set near zero, which isn't a bad thing.