Low side current sense application using Op-amp

Thread Starter

Ajay_P

Joined Nov 26, 2023
7
Hi,
Can anyone please help me for derriving the gain formula for Low side current sense circuit which is a differential amplifier used for overcurrent protection for BLDC motor controller? I will attach a screenshot of the circuit.




1701060517339.png
1701060517339.png
 

Sensacell

Joined Jun 19, 2012
3,398
Are you fixing this? Designing this?
Low-side current sensing can be done without a differential amp, if done correctly there is no common-mode voltage to contend with.
 

dl324

Joined Mar 30, 2015
16,677
Welcome to AAC!
Can anyone please help me for derriving the gain formula for Low side current sense circuit which is a differential amplifier used for overcurrent protection for BLDC motor controller?
I find your schematic very tedious to read with it's meaningless colors, scenic routing, unnecessary whitespace, wire bends, and crossings.

Crop it to the area where you want to insert your current sense. What are you going to do with the "sensed" current? Or do you intend to limit current?
 

crutschow

Joined Mar 14, 2008
34,044
As dl324 noted, yours is a good example of how not to draw an easy-to-read schematic.
If I have to scroll to view the schematic, then it has too much white space, and I stop looking.

Sensing low side-current can be readily done with a single, non-differential amplifier.
What are you exact requirements (circuit voltage, max current, desired output, etc.)?
 

Ian0

Joined Aug 7, 2020
9,497
Unless you can be absolutely certain that your amplifier connects to the same ground as the power, and that there is no track resistance between that ground and your current sense resistor, then I would no recommend trying to use a single-ended amplifier. Differential is the way to go. It allows you to put the current sense resistor where you please.
The gain would be R9/R8 if you got the topology correct.
R7 is superfluous, and R5 should be the same value as R9 and should go to your reference point.
If you are measuring current in one direction only, use ground as the reference.
If you are measuring current which can flow either way, the reference should be half the supply and that can be achieved by doubling the value of R5 and connecting a resistor of identical value to the positive supply.
Diode, preferably schottky, clamps to the negative supply from the op-amp inputs are a very good idea.
Make sure that you don't exceed the common mode range of the op amp.
Better still, use an AD8418/INA180/INA181
[edit] SGS-Thomson's TSC2010 is a version of AD8418 with better common mode range. I discovered it during the post-Covid component shortage when AD8418 were on a year's leadtime.
 
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Thread Starter

Ajay_P

Joined Nov 26, 2023
7
Welcome to AAC!
I find your schematic very tedious to read with it's meaningless colors, scenic routing, unnecessary whitespace, wire bends, and crossings.

Crop it to the area where you want to insert your current sense. What are you going to do with the "sensed" current? Or do you intend to limit current?
I intend to limit the current
 

Thread Starter

Ajay_P

Joined Nov 26, 2023
7
As dl324 noted, yours is a good example of how not to draw an easy-to-read schematic.
If I have to scroll to view the schematic, then it has too much white space, and I stop looking.

Sensing low side-current can be readily done with a single, non-differential amplifier.
What are you exact requirements (circuit voltage, max current, desired output, etc.)?
1701075742822.png



See, the image attached is same as before, here the input of the differential amplifier is the drop across RS3.
 

Ian0

Joined Aug 7, 2020
9,497
I did not understand, can you explain clearly or provide some reference notes for understanding the same.
This is a differential amplifier circuit. Yours doesn‘t look like this.
BEE79FF8-6DC1-4399-87D6-7A522B87B3EE.jpegIf you redraw your circuit with the differential amplifier correct, then the gain is easy.
R1=R2,
Rf=Rg
gain is Rg/R1
 

Thread Starter

Ajay_P

Joined Nov 26, 2023
7
This is a differential amplifier circuit. Yours doesn‘t look like this.
View attachment 308531If you redraw your circuit with the differential amplifier correct, then the gain is easy.
R1=R2,
Rf=Rg
gain is Rg/R1
Can you please tell me which configuration have been used? Is it differentiator because of the presence of capacitor?
 

Ian0

Joined Aug 7, 2020
9,497
Can you please tell me which configuration have been used? Is it differentiator because of the presence of capacitor?
It’s not a differentiator at all. You don’t need a differentiator.
You need a differential amplifier.
The capacitor is just for filtering high frequency noise. There should be two capacitors, one across Rf and one across Rg. Ignore them for now until you have the amplifier configured correctly.
Are you measuring current in one direction only, or is it a bidirectional measurement?
 
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