Hello E2E Experts,
Good day.
I am using TMCS1108-Q1 current sensor IC. According to the datasheet the Vout equation I can't get proper output voltage at the Vout pin.
In this condition when my led strip, which I have connected to p1 and P2, the Vout=0.28v, and in the off condition I get 0.333V.
Please give me a one-solved example of how to calculate the Vout from the given equation.
Regards,
CSC
Thank you for your response.
My actual question is if I want measure current through this given equation and also through software using this equation how to i find it?
if possible give one solved example use of Equation which is given in screen shot.
Also i attached my 2nd schematic i have to also find current and voltage using this equation.
Damini,
This is going to be dependent on the frequency of the AC signal, and the bandwidth of the device. Check out the specs of the noise floor from the datasheet:
To determine the peak to peak noise of the device, you would need to multiply the above number for the appropriate device by the square root of the desired bandwidth, and then multiply that result by 6 (for a 3 sigma two tailed approximation of 99%). This will give you your peak to peak noise referred to the output, and you can divide that by the sensitivity to refer to the input. To reduce the noise level, you can filter the output which will help you achieve better resolution on the AC signal, but this will be at the expense of bandwidth.
Damini,
The above is the correct answer here. If you are looking to measure an AC signal, the AC signal strength must be greater than the magnitude of the noise floor to be detectable (this is called signal to noise ratio).
From the datasheet snippet I posted above, you can calculate for the A2 variant from my instructions above that the RMS noise is (330uA/sqrt(Hz))*(sqrt(80kHz) = 93.3mArms, peak to peak noise is (2.95mVrms)*6 = 560mApp, referred to the input. What this means is that at full bandwidth, a signal with an amplitude of less than 560mA will at worst be undetectable, and at best, highly distorted due to the noise. If you filter the output and reduce the BW as a result, this will also lower the noise floor of the device, and therefore your SNR will improve, but you also will lost the ability to detect higher frequencies. Based on my results, I would say that the textbook answer to your question "What is the minimum amount i am measuring AC current," for full bandwidth, would be 560mApp, but higher SNR results in less distortion.
With the understanding you are looking for 200uA resolution, not 200mA, I do not think this is going to be achievable with any Hall effect sensor, let alone the TMCS110x family. They are simply too noisy to achieve this level of resolution.
Please check out this TI Precision labs video for more instruction on this topic.
