Current sampling

QWE4562009

Current sampling [Copy link]

Current sampling

The load current is instantaneous, ≤90mA, with a frequency of 2-100HZ, which is a low-frequency pulse current. The microcontroller is powered by 3.3V.

Question: What is the appropriate value for the current sampling resistor R9 and the amplification ratio (1+R12/R7)?

lugl4313820

While waiting for the big guys to show up, I will also learn from them.

vincentc543

If the power supply is 3.3V, the load current is <90mA

My method is to amplify the maximum scale to about 3V at 90mA. The accuracy will be higher than the maximum scale of 1.665V.

3/(0.09x0.5)=67 Amplified 67 times, close to 3V

Substitute the magnification into the formula:

R7=Default 1K

67=1+(R12/R7) ,R7=1K, R12=66K

QWE4562009

Sampling resistor R16

fjdeepblue

This is also related to the reference voltage of the microcontroller ADC. The reference voltage is not necessarily the power supply voltage.

士也土

A current sampling resistor of several tens of ohms is enough

QWE4562009

vincentc543 posted on 2022-6-6 22:15 If the power supply is 3.3V, and the load current is <90mA, my approach is to amplify the maximum scale to about 3V when the current is 90mA. The accuracy will be better than the maximum scale of 1.6 ...

R7 = 1K by default ---- What is the basis for this?

vincentc543

The magnification is a formula for the resistance ratio. 1K only falls on the mA current level, which is considered a medium matching method.

Of course, 10K ohm and 100Kohm can also be used, but the load effect must be avoided, as impedance distortion will cause amplification distortion.

QWE4562009

vincentc543 posted on 2022-6-6 22:15 If the power supply is 3.3V, and the load current is <90mA, my approach is to amplify the maximum scale to about 3V when the current is 90mA. The accuracy will be better than the maximum scale of 1.6 ...

This is a master

QWE4562009

vincentc543 posted on 2024-1-30 19:52 The magnification is a formula for the resistance ratio. 1K only falls on the mA current level, which is considered a medium matching method. Of course, 10K ohm, 100Kohm can also be used, but...

How to measure load effect?

vincentc543

QWE4562009 posted on 2024-2-19 16:13 How to measure the load effect

You need to know the line impedance first. If the original line is 10K, and the impedance you add is also 10K ohm, then 10K in parallel with 10K will equal 5K, affecting 50% impedance.

If the original line is 10K, and the impedance you add is 200K ohm, then 10K in parallel with 200K equals 9.52K.

Therefore, to reduce the load effect, an impedance greater than 20 times must be used to control the error within 5%.

This is the parallel impedance load effect. The same is true for the series impedance load effect. The algorithm uses the series method.