# Tag Info

4

Decide on what window you want the AC RMS value over. Determine the mean of the samples in the window. For each sample, subtract the mean. Square each sample and sum them together. Divide the sum of the squares by the number of samples. This is the mean squared value. Take the square root of that result to obtain RMS. Note that it will rarely be zero. ...

4

I believe this whole question is based on the misinterpretation of the oscilloscope screenshot shown first, before adding a capacitor & resistor to the LM393 output. You said: I want to change my square wave signal [...] 5 volts to 0 volts This is what is shown in the first image anyway! In the first picture, it is my original square wave coming from ...

2

The LM324/LM358 can only sink a few tens of uA down to (near) the negative rail. Try increasing your resistors and maybe add a pull-down resistor. From the datasheet: Note that the 50uA is nominal, only 12uA is guaranteed, and that only at 25°C. I would suspect your SPICE model is not very accurate if the simulation does not reflect actual behavior ...

2

You need another resistor (from the non-inverting input to ground) if attenuation is required. Call that Rs. Let's pick R1||R2||Rs = 10K Then we can immediately write: (12V/R1)*10K = 0.8V so R1 = 150K The output will be zero when the input is -(1/8)*12V or -1.5V. So Vref/R2 = 1.5/150K by KCL. Say Vref is 5V then R2 = 500K (499K is closest E96 value) And ...

1

With an input signal amplitude of 200 mV p-p, the diode is ineffective at clamping because it just won't conduct any significant amount at these voltages. Here's a picture from another post that shows how much current a 1N4148 can be expected to take at various low voltages: - The Y axis is the diode forward voltage and the X axis is the current it takes. ...

1

During the negative cycles, the capacitor needs to charge so that a positive potential develops on its right side plate. This voltage will add to the waveform during rest of the operation of the circuit. To charge the capacitor during negative half cycles only, the diode needs to be conducting during negative half cycles. -100mV amplitude is not enough to ...

1

It should function, and you probably want two op-amps if you have to have a high impedance input. I do have some comments: Okay, starting at the input. Why so low for R6/R7? You're drawing 10mA from U1 at 23V in. Since you have a 100K resistor after it you could use 220K and 20K with 82K instead of 100K. C2 may be a source of noise, from microphonics and ...

1

You have a symmetrical power supply, so you can make use of that instead of squeezing the input signal only in the positive voltage range. Change your voltage divider to get output values between 0 and 5V instead of 0 and 2V. Then use a summing amplifier to add -2.5V to the signal. This will give you a $\pm 2.5V$ signal. I changed a few details in your ...

1

You have a DC gain of 252 in the first stage and a gain of 1 in the following stages. The maximum Vos of the INA129 is at room temperature is 52uV + temperature drift so you could see 13mV from the front end. The LM358s have an offset of +/-3mV but also a bias current of as much as 35nA at room temperature, so you could get another 96mV from the unbalanced ...

1

At least two ways come to mind. 1) Put a summing amplifier after your DAC adding in a -23 mV offset to the signal. Need to be careful and use precision resistors so that you don't introduce more error than the offset you're trying to correct for. 2) Do it in software, which was suggested by the data sheet and Elliott Alderson. Just subtract the digital ...

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In software, you could subtract 23 from the digital value before sending it to the DAC. If I am reading the datasheet correctly then this offset value is within the specified maximum $V_{OS}$ for the converter, which is 0.75% of full scale. If the full-scale output is 4.096V then the maximum offset should be less than $\pm 30.72\,\text{mV}$. However, ...

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The electret microphone has a FET buffer inside of it and has a DC offset. The capacitor blocks the offset, however you need a path for the bias current of the op-amp. Bias current is a current flowing into or out of the op-amp input. The capacitor also blocks that voltage, so it will tend to float up towards one rail or the other rail depending on the op-...

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The proper way to have a consistent offset over most load conditions would be to have 2 power supplies in series, This way no matter how the load changed, it will be a consistent offset As your device normally runs off 12V, you would want atleast 1 floating supply, in a pinch this could be a battery that only your circuit is connected to To the negative of ...

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Many years later and spending time using these sensors in the field, I'd thought it would be good to share some lessons learned with these BME280 units. To answer the original question: the BME280 does not appear to be orientation dependent. This was tested positive/negative along xyz. Can't say anything about the acceleration dependence, these were paired ...

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