3 ms·
Correct. A 10% error can be calibrated out if it is constant (in practice it probably varies a bit with temperature). But for measuring thermocouples or strai
by auxym 2y ago
Correct.
A 10% error can be calibrated out if it is constant (in practice it probably varies a bit with temperature).
But for measuring thermocouples or strain gauges, for example, 50 mV (your example) can be 100-500% of the signal, which becomes impractical to calibrate-out (due to maximum output levels, etc).
For these applications, Vos is one of the first things to look at. Another one is the temperature coefficient on the gain. High frequency noise metrics such as CMRR and PSRR are sometimes important if you're looking at high frequency signals, but most of the time mechanical phenomena don't have much interesting content above a few 100s of Hz, and high frequency PS or CM noise can be removed by a simple high pass filter.
- kazinator 2y agoPower supply and common mode noise can be 60 Hz.
- dragontamer 2y ago> A 10% error can be calibrated out if it is constant (in practice it probably varies a bit with temperature). Unfortunately, Vos on cheaper general purpose OpAmps is the kind of thing that varies by... voltage. Ick. > For these applications, Vos is one of the first things to look at. Another one is the temperature coefficient on the gain. High frequency noise metrics such as CMRR and PSRR are sometimes important if you're looking at high frequency signals, but most of the time mechanical phenomena don't have much interesting content above a few 100s of Hz, and high frequency PS or CM noise can be removed by a simple high pass filter. No. CMRR is about DC in the applications I'm talking about. It's weird because CMRR is listed in decibels but it's absolutely a DC spec. If you have a high side current sense circuit with common-mode voltages of 24V +/- 0.05V (ex: 24V power supply that dips to 23.95V at 5Amps), CMRR tells you how accurate you are here. Your typical 60db (btw I need to kill the engineer who decided db measures DC noise/errors....) means that the 24V of common mode voltage (which is the 24V DC power supply in this case) leaks into your measurements. Or in other words: 60db * 24V == 3 decades or 24mV of 'Noise' aka your +/-50mV signal/measurement got completely wiped out by your DC errors. Like 50% error bars on your signal now, good luck with that. That's the real issue with OpAmps. There's surely an OpAmp out there that solves your problems. But it requires knowing the general tradeoffs and picking-and-choosing different parts for different purposes. Secondly, the specs are not intuitive. 60db CMRR sounds like a high frequency issue but becomes DC in this case. You could of course go full isolation (optoisolators) that allows you to shift voltages down to near zero (removing CMRR issues) but that's money and additional parts. You could go low-side voltage sense but this doesn't work for all circuits (most circuits are fine with Vcc error, not Ground errors). So high side current sense is the most flexible and generic well engineered solution. So long as you choose the correct OpAmps. ------- As far as when this could be useful: consider Maximum PowerPoint Tracking for solar. 0-24 V and 0-1 Amps. And a need to accurately measure Amps and Volts from this entire range. (A variable load + voltage converter like a switch-mode power supply + battery can search for the optimal Voltage/Current combination to maximize the Solar Panels power). Yes the microcontroller will do the bulk of the math. But the initial multiplies and subtract is best handled by an OpAmp.
- hn4352 2y ago> It's weird because CMRR is listed in decibels but it's absolutely a DC spec. If you get the Franco book equation 5.27 (my edition is the 3rd) explains why they do that. Long story short: It's a convenient form when CMRR = dVcm/dVos due to the orders of magnitude involved.
- kazinator 2y agoThe moral is that you have to be careful. The ratio and decibels giving CMRR might not be the same as quoted for a given closed loop gain at a given frequency of interest.
- kazinator 2y agoThe difference in brightness between two incandescent light bulbs running on DC could be expressed in decibels. Log10 of the ratio of wattages times ten. Decibels are useful for any intensity scale that occurs in a context where something responds in a logarithmic way, regardless of whether the power phenomenon is steady or pulsating.