6 ms·

Vos doesn't matter until it does. I find it surprising to see Vos mentioned as likely unimportant right next to CMRR, as CMRR and Vos are innately related in so

Vos doesn't matter until it does. I find it surprising to see Vos mentioned as likely unimportant right next to CMRR, as CMRR and Vos are innately related in some circuits. In particular, any low voltage current sense circuit is going to require very precise Vos. Let's say you have a 0.01 Ohm current-sense resistor on a 5-Amp or so circuit. Your current-sense is now in the range of +/- 0.05Volts (!!!!). So a Vos of 0.005 would represent a 10% error. Likely too much for most applications. In effect, CMRR has become hyper-sensitive to Vos in this particular use case, to the point that Vos is suddenly the most important statistic. Fortunately, there are specially designed low-offset chopper or auto-zeroing OpAmps like MCP6V26 or whatever out there. MCP6V26 has Vos of 2uV, or in relatable terms... 0.000002 Volts (!!!!). Meaning it is more than sufficient at reliably making this current-sense application. Indeed, you can drop down to 0.0001Ohm resistance and still have high accuracy (and power savings compared to the earlier assumption). ------ Alas, nothing is ever free in life. Chopper Amps have noise issues and other designs have very very low bandwidth (which is truly an important statistic for most circuits). Choosing a chopper amp specifically is making a Vos tradeoff with Bandwidth. So only choose if you know what you are doing (aka, dealing with very low voltages and needing the precise zero).

nine_k 2y ago
For current measurement, I'd rather use a Hall sensor. But to measure the signal from that sensor a sensitive opamp is needed, and your comment applies :)
- blackguardx 2y ago
  Using Hall-effect sensors to measure current isn’t that common when currents are in the uA to mA range. I’ve also never seen them used even for Amp range measurements when the application is purely on-board power supply meaurement.
  - nine_k 2y ago
    The example uses 5A. For mA range, of course not.
    - kevin_thibedeau 2y ago
      High current lab supplies routinely use current shunts for their self-measurement.
- analog31 2y ago
  Or a high side current sense amplifier... https://www.st.com/en/amplifiers-and-comparators/tsc1031.html https://www.st.com/en/amplifiers-and-comparators/tsc1031.htm... Hall sensors drift.
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 ago
  Power supply and common mode noise can be 60 Hz.
  - kragen 2y ago
    And its harmonics, if you're standing near a fluorescent lamp.
    - kazinator 2y ago
      Yes, but I'm trying to argue that the frequency can go at least as low as 60.
      - kragen 2y ago
        Sure, or 50 most places. Notching that out is often worthwhile.
- 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.
Tade0 2y ago
Back in college I was told to not think too hard about Vos, as it's typically temperature dependent so you need to assume it's non-zero or will be in some circumstances and compensate with feedback anyway.
- kragen 2y ago
  How do you compensate for Vos with feedback?
  - hn4352 2y ago
    You probably don't, but your opamp might. https://www.analog.com/en/resources/technical-articles/to-chop-or-auto-zero-that-is-the-question.html https://www.analog.com/en/resources/technical-articles/to-ch... Technically you probably could do it externally in most cases but it would require a bunch of extra stuff, and be a pain, so usually it's best to use the stuff built into the amplifier itself.
    - kragen 2y ago
      Oh, sure. You could totally build a chopper op-amp out of two discrete op-amps. I'd never thought about actually doing that...
      - dragontamer 2y ago
        I've seen designs that use a Chopper Op-Amp to actually auto-zero a power-OpAmp, effectively transferring the low Vos characteristics to a different OpAmp with completely different characteristics.
        kragen 2y ago
        If the op-amp has a stable Vos I feel like maybe you could zero it by hand with a trimpot? You just need a button to short the inputs together while you're trimming it.
kragen 2y ago
I was puzzled about that too, so I appreciate you confirming what I thought. I'd think the main benefit of using lower-value current-sense resistors in this application would be that the resistor would heat up less, so its resistance would be more stable?
- dragontamer 2y ago
  Accuracy is likely secondary. I expect that most applications are good with 1.5 digits (aka 95% accurate or so). You don't want to blow the entirety of this 5% allowable error on one micro-spec of one component, but I don't expect that most people especially need lots of accuracy here. The issue is that any circuit with 1 to 5 amps of current is a serious amount of power, meaning power efficiency is likely one of the top priorities. A 5-Amp circuit with a 0.01 Ohm sense resistor wastes 250mW on the resistor alone, likely more than the entirety of your microcontroller!! You can actually run an entire Linux capable microprocessor + Low-power DRAM off of that kind of power!! Dropping down to 0.0001 Ohms uses 1/100th the power or 2.5mW. which is likely a more reasonable cost.
  - kragen 2y ago
    That depends on what "most applications" means. I remember a paint program I saw in high school that some kids had written to draw sprites and backgrounds for their video games. The documentation explained that it could only edit 320×200×256 images, but that that should be adequate for "most projects". Depends on the context! In the contexts I'm thinking of, I would think that, if your load is drawing 5 amps of current at 3.3 volts, which is 16.5 watts, an extra 0.25 watts in the 10mΩ current shunt is not likely to be a big problem. And if it's 5 amps at 48 volts or 240 volts, it's even less of a problem, relatively speaking. I guess you're thinking of different contexts, contexts where the power-measurement system is paid for from a different budget than the load, but I can't figure out what they are.
    - dragontamer 2y ago
      You're sniffing out the fact that I didn't have an exact application in mind when I wrote my earlier posts, lol. But yes, you are correct on this front. The more I post on this subject, the more I'm "backwardsly-targetting" a solar-powered MPPT circuit. Maximum Power Point Tracking circuits improve your solar-panel's efficiency by changing the current (through the use of a buck-boost converter, changing the voltage-and-current downstream). Or maybe you have excess current sunk ionto a battery of some kind. Either way, you have some kind of configurable-load and can therefore maximize the solar panel's Voltage/Current curve characteristics to seemingly magic energy out of nothingness. If it costs you 250mW to just *sense* the current and run the calculations, it becomes much harder to justify the small gains of any MPPT circuitry. ------ But yes, I'm changing the target application to suit my argument style. Apologies on that but I think you can forgive me on this!! The point of MPPT is to magic more energy out of nearly nothingness so efficiency is of great concern here!
hn4352 2y ago
I was kinda shocked by the Vos comment as well. On CMRR, in some mathematical treatments it's modeled as a change in offset voltage with respect to common mode, which indirectly effects output voltage of course so at the end of the day it's the same result. (See: https://www.google.com/books/edition/Design_With_Operational_Amplifiers_And_A/nG1zCgAAQBAJ?hl=en&gbpv=0 https://www.google.com/books/edition/Design_With_Operational... highly recommended ) It's also odd that the 741 was dismissed, as it should be, but the TLV9301 was not recommended. This part is specifically called out on TI's 741 page as what to use instead in 2025. Not only does it perform better in basically every possible spec, it's also a drop in replacement for most, if not all, applications. https://www.ti.com/product/LM741 https://www.ti.com/product/LM741 https://www.ti.com/product/TLV9301 https://www.ti.com/product/TLV9301 TLV9301 ($0.5) is also cheaper than a MCP6272 ($0.88)
kazinator 2y ago
Amplification of the "common mode" means that if we tie the + and - inputs of the op-amp together, so that it's not doing its job (amplifying the difference), it's possible that the op-amp amplifies the voltage present on the combined + - node; the voltage between that and the ground reference. Ideally, the amplifier should not do that, but they do. The gain is not nearly as much as the differential gain, so I think the CMRR (common mode rejection ratio) is something like how many decibels down from the payload gain is this unwanted common mode gain. Common mode gain can cause or contribute to an output offset. But if the amplifier circuit is biased such that + and - are kept very close to zero, that should not be the case. We normally care about the common mode rejection because common mode noise makes the + and - inputs swing together, causing the amplifier to amplify noise, like RFI/EMI. In your current sense example, although we have a 10% error, it is a systematic error (a displacement), not 10% uncertainty in the value. It is an inaccuracy, not an imprecision. You're assuming that the amplifier just buffers the output of the current sense resistor, so that the +/- 0.05 V across the resistor is translated to the output of the amp via unity gain. We could configure the amplifier to amplify the signal, so that it becomes +/- 5V (or whatever). Then the Vos of 0.0005 will be a 0.1% inaccuracy.

nine_k 2y ago

For current measurement, I'd rather use a Hall sensor. But to measure the signal from that sensor a sensitive opamp is needed, and your comment applies :)

blackguardx 2y ago
Using Hall-effect sensors to measure current isn’t that common when currents are in the uA to mA range. I’ve also never seen them used even for Amp range measurements when the application is purely on-board power supply meaurement.
- nine_k 2y ago
  The example uses 5A. For mA range, of course not.
  - kevin_thibedeau 2y ago
    High current lab supplies routinely use current shunts for their self-measurement.
analog31 2y ago
Or a high side current sense amplifier... https://www.st.com/en/amplifiers-and-comparators/tsc1031.html https://www.st.com/en/amplifiers-and-comparators/tsc1031.htm... Hall sensors drift.

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 ago
Power supply and common mode noise can be 60 Hz.
- kragen 2y ago
  And its harmonics, if you're standing near a fluorescent lamp.
  - kazinator 2y ago
    Yes, but I'm trying to argue that the frequency can go at least as low as 60.
    - kragen 2y ago
      Sure, or 50 most places. Notching that out is often worthwhile.
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.

Tade0 2y ago

Back in college I was told to not think too hard about Vos, as it's typically temperature dependent so you need to assume it's non-zero or will be in some circumstances and compensate with feedback anyway.

kragen 2y ago
How do you compensate for Vos with feedback?
- hn4352 2y ago
  You probably don't, but your opamp might. https://www.analog.com/en/resources/technical-articles/to-chop-or-auto-zero-that-is-the-question.html https://www.analog.com/en/resources/technical-articles/to-ch... Technically you probably could do it externally in most cases but it would require a bunch of extra stuff, and be a pain, so usually it's best to use the stuff built into the amplifier itself.
  - kragen 2y ago
    Oh, sure. You could totally build a chopper op-amp out of two discrete op-amps. I'd never thought about actually doing that...
    - dragontamer 2y ago
      I've seen designs that use a Chopper Op-Amp to actually auto-zero a power-OpAmp, effectively transferring the low Vos characteristics to a different OpAmp with completely different characteristics.
      - kragen 2y ago
        If the op-amp has a stable Vos I feel like maybe you could zero it by hand with a trimpot? You just need a button to short the inputs together while you're trimming it.

kragen 2y ago

I was puzzled about that too, so I appreciate you confirming what I thought. I'd think the main benefit of using lower-value current-sense resistors in this application would be that the resistor would heat up less, so its resistance would be more stable?

dragontamer 2y ago
Accuracy is likely secondary. I expect that most applications are good with 1.5 digits (aka 95% accurate or so). You don't want to blow the entirety of this 5% allowable error on one micro-spec of one component, but I don't expect that most people especially need lots of accuracy here. The issue is that any circuit with 1 to 5 amps of current is a serious amount of power, meaning power efficiency is likely one of the top priorities. A 5-Amp circuit with a 0.01 Ohm sense resistor wastes 250mW on the resistor alone, likely more than the entirety of your microcontroller!! You can actually run an entire Linux capable microprocessor + Low-power DRAM off of that kind of power!! Dropping down to 0.0001 Ohms uses 1/100th the power or 2.5mW. which is likely a more reasonable cost.
- kragen 2y ago
  That depends on what "most applications" means. I remember a paint program I saw in high school that some kids had written to draw sprites and backgrounds for their video games. The documentation explained that it could only edit 320×200×256 images, but that that should be adequate for "most projects". Depends on the context! In the contexts I'm thinking of, I would think that, if your load is drawing 5 amps of current at 3.3 volts, which is 16.5 watts, an extra 0.25 watts in the 10mΩ current shunt is not likely to be a big problem. And if it's 5 amps at 48 volts or 240 volts, it's even less of a problem, relatively speaking. I guess you're thinking of different contexts, contexts where the power-measurement system is paid for from a different budget than the load, but I can't figure out what they are.
  - dragontamer 2y ago
    You're sniffing out the fact that I didn't have an exact application in mind when I wrote my earlier posts, lol. But yes, you are correct on this front. The more I post on this subject, the more I'm "backwardsly-targetting" a solar-powered MPPT circuit. Maximum Power Point Tracking circuits improve your solar-panel's efficiency by changing the current (through the use of a buck-boost converter, changing the voltage-and-current downstream). Or maybe you have excess current sunk ionto a battery of some kind. Either way, you have some kind of configurable-load and can therefore maximize the solar panel's Voltage/Current curve characteristics to seemingly magic energy out of nothingness. If it costs you 250mW to just *sense* the current and run the calculations, it becomes much harder to justify the small gains of any MPPT circuitry. ------ But yes, I'm changing the target application to suit my argument style. Apologies on that but I think you can forgive me on this!! The point of MPPT is to magic more energy out of nearly nothingness so efficiency is of great concern here!

hn4352 2y ago

I was kinda shocked by the Vos comment as well. On CMRR, in some mathematical treatments it's modeled as a change in offset voltage with respect to common mode, which indirectly effects output voltage of course so at the end of the day it's the same result. (See: https://www.google.com/books/edition/Design_With_Operational_Amplifiers_And_A/nG1zCgAAQBAJ?hl=en&gbpv=0 https://www.google.com/books/edition/Design_With_Operational... highly recommended ) It's also odd that the 741 was dismissed, as it should be, but the TLV9301 was not recommended. This part is specifically called out on TI's 741 page as what to use instead in 2025. Not only does it perform better in basically every possible spec, it's also a drop in replacement for most, if not all, applications. https://www.ti.com/product/LM741 https://www.ti.com/product/LM741 https://www.ti.com/product/TLV9301 https://www.ti.com/product/TLV9301 TLV9301 ($0.5) is also cheaper than a MCP6272 ($0.88)

kazinator 2y ago

Amplification of the "common mode" means that if we tie the + and - inputs of the op-amp together, so that it's not doing its job (amplifying the difference), it's possible that the op-amp amplifies the voltage present on the combined + - node; the voltage between that and the ground reference. Ideally, the amplifier should not do that, but they do. The gain is not nearly as much as the differential gain, so I think the CMRR (common mode rejection ratio) is something like how many decibels down from the payload gain is this unwanted common mode gain. Common mode gain can cause or contribute to an output offset. But if the amplifier circuit is biased such that + and - are kept very close to zero, that should not be the case. We normally care about the common mode rejection because common mode noise makes the + and - inputs swing together, causing the amplifier to amplify noise, like RFI/EMI. In your current sense example, although we have a 10% error, it is a systematic error (a displacement), not 10% uncertainty in the value. It is an inaccuracy, not an imprecision. You're assuming that the amplifier just buffers the output of the current sense resistor, so that the +/- 0.05 V across the resistor is translated to the output of the amp via unity gain. We could configure the amplifier to amplify the signal, so that it becomes +/- 5V (or whatever). Then the Vos of 0.0005 will be a 0.1% inaccuracy.