I want to measure the coil voltage (coil+ to coil-) of the following circuit with my microcontroller, but I don't know how. Do you have some hints or suggestions for me?
Thanks!
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closed as unclear what you're asking by Marcus Müller, laptop2d, Elliot Alderson, Edgar Brown, Dwayne Reid Feb 1 at 5:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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\$\begingroup\$ So you have a circuit, you already know at which points you want to measure. What is it that you then still need help with? \$\endgroup\$ – Marcus Müller Jan 28 at 14:30
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\$\begingroup\$ I don't want to measure the coil- and coil+ voltage relative to ground. Is there a way to measure the potential difference across the coil directly? I don't know but maybe by using a differential ADC or something like that? \$\endgroup\$ – Michael H Jan 28 at 14:40
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\$\begingroup\$ I don't understand the question, you have a less than 0.5us/sample ADC in your ucontroller to do that ? You need only peak to peak value? \$\endgroup\$ – Dorian Jan 28 at 14:54
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\$\begingroup\$ The ADC requires at least 1us/sample. Yes peak to peak value would work. \$\endgroup\$ – Michael H Jan 28 at 15:02
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\$\begingroup\$ Since the driving waveforms are symmetrical, so are \$ V_{coil+}, V_{coil-} \$. A single-ended ADC could monitor one or the other, provided that its input impedance is high compared with that 48 ohm source Z. Just double changes of the single-ended result. A very fast ADC would be required to sample that 1 MHz waveform. \$\endgroup\$ – glen_geek Jan 28 at 15:59
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Of course you could use an isolation amplifier if you had a power supply on the isolated side so I'm guessing you don't. For a passive measurement, you can get a measurement of the output coil by winding it bifilar with a sense coil. The sense coil can be a different, lighter gauge, since it is not going to have any current. This sense coil would have the same voltage as the secondary coil, were it not for the slight voltage drop due to the resistance of the secondary coil under load. This means you may have to subtract the value of the resistive current loss, if it is significant. You could do this by means of a table if you have a fixed load or possibly you can infer it from the primary current if you have a varying load.
