My hardware design team have given me circuit to measure current consumed by load.I have to measure current consumed by Load using MCU ADC. Can someone explain what exactly this OP amp is doing ?
I am from software side & not a hardware expert.
\$\begingroup\$
\$\endgroup\$
4
-
\$\begingroup\$ It measures the current through R33, amplifies the signal and sends it downstream to/as IU. \$\endgroup\$– winnySep 16, 2016 at 11:45
-
2\$\begingroup\$ Why are you asking us? Since your hardware design team created this, they should be able to trivially explain how it works. \$\endgroup\$– Olin LathropSep 16, 2016 at 12:03
-
\$\begingroup\$ I am from software side Then why would you need to know how this works ? As a HW designer I'd tell a SW designer like you that all you need to know is that at output UI there will be a voltage proportional to the current. Just connect it to the ADC and read the value. You would not ask the HW team to write the code to read out the ADC now would you ? We all have our own tasks/problems :-) \$\endgroup\$– BimpelrekkieSep 16, 2016 at 12:31
-
\$\begingroup\$ that's a respectable circuit they came up with, good on them \$\endgroup\$– KyranFSep 16, 2016 at 16:42
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
2
This is a classic diff amp circuit. In this case the circuit measures the voltage across R33, which is proportional to the current thru it.
The diff amp gives a single-ended voltage proportional to the difference between two input voltages. This is useful in that it allows the two input voltages to "float", meaning they can go up and down together without changing the output, since the difference between the inputs doesn't change.
For more details, ask your hardware design team or look up differential amplifier and current sense resistor.
answered Sep 16, 2016 at 12:06
-
1\$\begingroup\$ Additional functions going on here are low-pass filter and level-translator - Look up those as well. \$\endgroup\$ Sep 16, 2016 at 14:13
-
\$\begingroup\$ And would you mind telling me why that 10k resistor connected to 5v is needed on the + input? Wouldn't it work without it? Also, isn't it inserting a bit more current through R33 after all? \$\endgroup\$ Sep 16, 2016 at 23:06
\$\begingroup\$
\$\endgroup\$
3
You should ask your hardware team to provide you the transfer function (ADC voltage as a function of current). This is an entirely reasonable request and better than asking a bunch of strangers on the Internet.
R42 adds an offset of 454mV to the amplifier non-inverting input, which is multiplied by 11 (5.00V out at zero current) as seen at the op-amp output. That op-amp output voltage decreases by 1V/A of load current. The ADC sees half the op-amp output voltage.
So, my take on the transfer function is:
\$V_{ADC} = 2.5V - 0.5\cdot I_{LOAD} \;\;\; (0 \le I_{LOAD} \le 5A )\$.
Which you can easily re-arrange to find the load current given the ADC reading and ADC reference voltage.
It's a pretty straightforward (if sloppy in terms of potential accuracy) circuit but it's not your job to reverse-engineer their chum bucket of a design.
answered Sep 16, 2016 at 14:28
-
\$\begingroup\$ isn't the output divided by 2 by R44 and R45? \$\endgroup\$– Steve GSep 16, 2016 at 14:58
-
\$\begingroup\$ @SteveG Yes, hence I wrote: "The ADC sees half the op-amp output voltage." \$\endgroup\$ Sep 16, 2016 at 15:25
-