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I saw this application example in the LM324 datasheet.

enter image description here

In a case like this:

enter image description here

where R1/R2 is equal to R4/R3 and R4/R3 is large (330 like above), the first op-amp seems to act like a unity gain buffer; its output isn't much different from its non-inverting input.

Also, the inputs (and thus, the output) to the op-amp are already offset by Vcc/2 and buffered. Is there still any need for op-amp U1A above? Can both inputs just be fed into U2A directly without compromising the "High Input Z" quality (or anything else) of the original circuit? I'd also like to understand what is meant by "CMRR depends on this resistor ratio match" in the first image.

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2 Answers 2

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Circuit 1 uses two non-inverting inputs and their high input impedance as a differential amplifier suitable for low voltage sensing without significant loading current but poor CMRR due to R matching needed. Each input impedance is implied by the Op Amp input bias current.

Circuit 2 has several design violations ( including U1B-out to 240V Neutral and trying to measure AC current with poor CMRR and no isolation) It also is not the same as CCT 1.

Note how above CCT 1 is pleasing to the eye and figure out why that is so. :)

Further info below

enter image description hereenter image description here FWIW https://meettechniek.info/diy-instruments/arduino-wattmeter.html

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  • \$\begingroup\$ Thanks. I checked out the link but, except for the extra opamp stage in my circuit (which I want to get rid of), the diode protection and cap filtering in their circuit, I don't see how my circuit is much different from theirs. They connected an (unbuffered) reference to the COM/NEUTRAL as well as one side of the shunt and fed both to the opamp to amplify the difference by 10, feeding the result to the ADC. Or did I miss something? \$\endgroup\$
    – SoreDakeNoKoto
    Jan 22, 2018 at 0:06
  • \$\begingroup\$ Proper isolation would require a transformer which I dont want because of circuit size and cost. I understand it's also common for meters to not have galvanic isolation...a sort of compromise. \$\endgroup\$
    – SoreDakeNoKoto
    Jan 22, 2018 at 0:07
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    \$\begingroup\$ "transformer which I dont want because of circuit size and cost." How much will your hospital bills cost after you electrocute yourself? How much will hospital and legal bills cost after someone else gets electrocuted? And I'm not kidding here. \$\endgroup\$
    – WhatRoughBeast
    Jan 22, 2018 at 1:43
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    \$\begingroup\$ @TisteAndii - "It's a compromise" is not a response to my question. And the app notes are intended for people who know what they are doing. You clearly don't and you should not be taking chances connecting directly to the power line. You don't understand the hazards so you cannot be trusted to avoid them. \$\endgroup\$
    – WhatRoughBeast
    Jan 22, 2018 at 2:12
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    \$\begingroup\$ @TisteAndii - "You don't know what you're doing" PLUS It can kill you is, or ought to be, sufficient reason to start asking someone local for help. It is, or ought to be, sufficient reason to do a lot more research on how people get injured doing this stuff. I am not trying to demean or insult you. I am trying to keep you out of trouble. \$\endgroup\$
    – WhatRoughBeast
    Jan 22, 2018 at 3:47
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Firstly, the CMRR comment is understood by looking at your first circuit. Since the output of the left-hand op-amp with respect to ground is 2 x Vin, the first stage will saturate at 1/2 of your positive rail. So if your common mode input exceeds 1/2 the rail, the circuit doesn't work properly. You can see that the ratio of resistors limits the common mode.

Your reasoning for the arrangement of your second circuit is good from a theoretical standpoint. The problem is that when measuring sense resistors, using a current-carrying ground will introduce errors. The inductance and resistance of your ground path means that there will be voltage differences between different points in your ground path when current is applied or when it changes. Therefore, the designer of circuit #2 uses U1:A with a separate sense line running from a ground reference point (ideally close to the ground side of the sense resistor) to compensate for local changes on the sense resistor's ground with respect to the op amp ground, which might otherwise push U2:A into saturation.

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  • \$\begingroup\$ Hmm. So if I understand you, it's best to work with CCT 2 as is (maybe with a cap on the VCC/2 reference to help filter the voltage differences you mentioned?), at least till I get an instrumentation amp like others have recommended? \$\endgroup\$
    – SoreDakeNoKoto
    Jan 23, 2018 at 13:04

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