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I would say no. The effects of temperature coefficient can be expressed as $$ R= R_0 \left[1 + \alpha\left(T_0-T_{\text{ref}}\right) \right]$$ So, only the change in resistance (not the resistance) will be proportional to the temperature difference, and that holds best for small temp changes about a working point. To be even more specific, $$ R=\frac{\...


3

The cylindrical field around a wire is fairly weak, so Hall effect current sensors use a C-shaped core to reduce the reluctance and direct the flux so that you get a decent signal from a Hall sensor placed in the gap. picture from here It is possible to buy these sensors as ready made, with a 2-part core, that then snap over the wire.


2

A DC analog ammeter is essentially a voltmeter in parallel with a shunt resistor. It's a bit more complex than that, because a typical (d'Arsonval moving coil with taut band suspension) meter movement actually measures current, and has a resistance that is temperature sensitive since it's typically wound with copper wire rather than an alloy like ...


2

No, it is not correct in general. There are many ways that "quanity \$y\$ increases with quantity \$x\$", e.g.: \$y=ax+b\$ (with \$a > 0\$) or \$y=x^2\$ or \$y=\sqrt x\$, etc. If you say, however, "\$y\$ is proportional to \$x\$" you especially mean that the relation between the quantities is \$y= cx\$, where \$c\$ (slope) is a constant. As you can ...


1

R3 is way too large. Its voltage drop will hit 3.3 V at a current of just 3.3 V/50 kΩ = 66 µA, which corresponds to a load current of 100 Ω/0.1 Ω × 66 µA = 66 mA. If you want to measure currents up to 10 A, which would be 10 mA through R3, its value can be no more than 3.3 V/10 mA = 330 Ω. In fact, it will have to ...


1

I'm not sure if the machine you're observing already exists or has already been designed, but... While not exactly an answer to the question, a coil energization speed-up circuit will improve pull-in time and potentially bring any of your solenoids that haven't outright failed into the desired speed range. This approach is commonly used in industrial ...


1

If you assumed that the current was really peaking at 7.8 amps AND, that the inductor is 68 uH (as per your comment) AND, that it is used in a boost converter as the energy storage inductor then, you could calculate what the current was (providing you know the power supply voltage). But, assuming that the current is 7.8 amps and that it rises from zero amps ...


1

The biasing approach is what you need to do - offset the signal such that a zero input is somewhere near to the mid-range of the A-D scale. You then subtract the reading you get with a zero input from the A-D during measurement, and have a signed value resulting. You can use one of your op-amps to introduce the offset. You have a shorting relay across the ...


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I'm using and ESP32 which has a built-in ADC with a reference voltage of 3.3V. Hence, the signal will be alternating from 1.5V to 3.5V which will then be processed by the MCU accordingly. I understand that I can simply use a voltage divider to step down the voltage but that will lose some precision. How much precision will you lose? To reduce 3.5V ...


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I did not understand how does this ciruict works! *Why have we used R24=1.2M? Especially in case we remove R21 (As it is noted by DNP) Recall, that this is circuit is both a battery charger and power supply, so the battery current is bi-directional. When operating as a power supply the battery is discharging and the current-shunt output potential is below ...


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My first question is if we consider U4:A as an error amplifier (As it appears), what is the purpose to connect the inverting input of this OPAMP to the output of the 3rd order low pass filter and the non-inverting input to current sensing circuit of the battery? The IREF input (the input of the 2nd order LPF as you've said) comes from the MCU's RA6/AN5/PGC ...


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