ACS712 OverCurrent Switch

eetech00

Joined Jun 8, 2013
2,658
Hi
View attachment 248746
https://www.electronics-lab.com/project/current-sensor-amplifier-current-switch/

It says: Pin 5 : OP2 Over Current Output ( Normally High-TTL, Goes Low at Over Limit)
Is it correct ???

if U2B 5pin set to 2V, U2B 6pin is 2.5V (zero current), OUT is Low.
and
Higher Current –> Lower U2B 6pin Voltage.
and
if U2B 6pin goes to lower than 2V, OUT is High.

Which is correct output logic ?
I don't see how that circuit can work.

What did you want to accomplish?
 

Ian0

Joined Aug 7, 2020
3,783
Which is correct output logic ?
You are.
They have used the application note circuit to increase the sensitivity, and don't seem to have realised that it is an inverting amplifier.
I don't see the point of the amplifier - just go straight from the ACS714 to the comparator - then their logic is correct.
They have used the least sensitive version then amplified the output, so the noise is equivalent to about 300mA, so no matter how much you amplify it, you can't detect signals <300mA because of the device's noise.
If you want more sensitivity, use the -05 version.
And do you really have an ACS712 (as per your title) or a ACS714 (as per their text) or an ACS715 (as per their parts list)?
 

Irving

Joined Jan 30, 2016
2,322
I don't see how that circuit can work.
Its pretty simple. The ACS712/4 outputs a voltage of 2.5v +/- 66mV per amp of current sensed. That's amplified by the first opamp by approx -3.3 so giving an output of 2.5v -/+218mV/A which goes down as current increases.. When the output of that amp drops below the comparator threshold set by PR1 the output goes high.

But there are significant flaws. As @Ian0 stated, noise limits the sensitivity after amplification to 350mA, but also amplifying it limits the dynamic range as the output of the first opamp hits the supply rails at around +/-11A rather than the +/-30A the sensor is capable of.

Basically a poor design; better to roll your own.
 

Ian0

Joined Aug 7, 2020
3,783
Shunts are quiet.
A 60mV/30A shunt has a resistance of 2mΩ and an output noise of 5pV/√Hz. The dominant noise source will be the amplifier at about 400nV/√Hz on the output, that's 18μV on the output compared to 2.4mV* for the Hall effect device in a 2kHz bandwidth. That's 42dB better!
Do you really need the isolation?

*Specified at 7mV p/p - but I have to guess at the crest-factor.
 

Ian0

Joined Aug 7, 2020
3,783
Alternatively. . .
Assuming that the current is DC, I'm guessing that there is a DC supply somewhere, and the DC supply is probably low voltage. (Too many guesses here, but perhaps the TS will tell us the supply voltage when he's read this collection of ideas)
INA199 shunt amplifier will work up to 26V, LM393 will work up to 32V - put the amplifier and comparator on that side of the supply and isolate the comparator output with a simple optoisolator.
 

eetech00

Joined Jun 8, 2013
2,658
Its pretty simple. The ACS712/4 outputs a voltage of 2.5v +/- 66mV per amp of current sensed. That's amplified by the first opamp by approx -3.3 so giving an output of 2.5v -/+218mV/A which goes down as current increases..
I don't understand (I understand the math).
VCC is 5V so U2A has a reference of 2.5v. The ACS has an offset of 2.5v. Since the offset isn't filtered, so no matter how much current is sensed, the output of U2A will always fall sharply when the reference voltage is exceeded.
 

Ian0

Joined Aug 7, 2020
3,783
I don't understand (I understand the math).
VCC is 5V so U2A has a reference of 2.5v. The ACS has an offset of 2.5v. Since the offset isn't filtered, so no matter how much current is sensed, the output of U2A will always fall sharply when the reference voltage is exceeded.
U2A is an inverting amplifier with a gain of -3.3. The non-inverting input is biassed to 2.5V, the same as the ACS offset.
So. . .
ACS output = 2.5V. U2A output is 2.5V
ACS output = 2.6V. U2A output is 2.17V -3.3*(2.6-2.5)+2.5V
ACS output = 2.7V. U2A output is 1.84V
etc.
 
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