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It will be OK
Differential input voltage ±Supply voltage
The common mode voltage is to meet the common mode rejection.
your output will be slightly less then the 5V rail.
Is this for some kind of current limit on a load connected to the output?
Not sure why you would buffer the rail voltage.
This could be a loose connection on the circuit breaker.
After years the connection can become loose and you get some voltage drop with high current loads.
Outdoor boxes are prone to this because of hot to cold cycling.
One question, Do some lights get brighter ?
That would be a open neutral...
It appears that you have a input offset voltage of about 1.4 mv , with a gain of 100, you get 141 mv when you should get zero (room temperature)
And has far as noise in your measurement, getting two sensors to couple together is problematic. Best if they are both submerged in a oil bath (high...
TC voltages are relative to the temperature at the connecting end.
Your calibration would drift with temperature of the junction, where you connect the wire to your amplifier.
This may be OK if you are trying to measure something very hot and you don't care about a few degrees in error.
I like your approach, but something seems off.
so now we take an integral from t=0 to t=10us (di/dt)
6.7k/8.7k *(24-i1(t) *2k) dt
100*(6.7/8.7) * 24t-1K*i1(t)^2 from t=0 to t=10us
I don't see the exponential expression?
I would start by, removing the 8 K, like they say above, add 3 ma (24 volts/8K ohms) back in later.
add the 2K and the 4.7K to give you a 6.7K
Now look at the voltage divider you have with the 2 K and the 6.7K resistor on 24V
V= 24V * 6.7K/(2K+6.7K) = 18.483..V
calculate the impedance of...
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