jjstcool
• Member for 7 years, 5 months
• Last seen more than a month ago

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Yeah, that is because after the watchdog fires, you have to reset the corresponding flag before main starts again (there is a register where you can read out, what caused the reset) and to disable the ...

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up to where you estimated the voltage I understand what you do and it is correct. Now, the As such, we can set Vout=(1 V/μT)B and solve for the ratio of R1/R2 that solves properly calibrates the ...

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Yes you are basically correct with your observations. So I think what pole’s imaginary part indicates is the impulse response’s frequency. Is that correct? Yes. Laplace transform is a tool to ...

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This is a good idea. Theoretically you could use the pwm as is to dim the lights since an LED shines brighter the longer the on-periods are. Your concern about sending waves might be negligible as ...

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Your mistake is in the very last step: the derivative of $e^{-|w|}$ is different as you calculated. The (I believe) correct answer is as shown here Since I'm not totally sure, just expand your ...
looks like your guess is right. Imagine frequency = 0, then the inductor becomes a conductor and only $R_L$ is responsible for the impedance. on the other end of the spectrum, $\omega \rightarrow \... View answer 2 answers 1 votes 554 views Accepted answer 0 votes Miller capacitance is nothing that you can easily measure with a cap-meter. The effect comes from the voltage on one pin falling such that less voltage difference is seen on more of the area, such ... View answer 2 answers 2 votes 285 views 0 votes You messed up with the math. I don't fully follow, what you are doing, it doesn't seem totally wrong, but you certainly miss some stuff. First of all: you get a non-damped solution with your ... View answer 3 answers 0 votes 351 views Accepted answer 0 votes I found the answer to my question by accident: Just for those who are still interested in what I asked. Now everything makes sense and gets way clearer. Thanks for the help anyways! Edit : I made up ... View answer 2 answers 1 votes 698 views 0 votes Your intuition is right for the first cycle. Imagine at first there is$V_z$positive, say 2V and E is 1V, so there is a voltage drop in forward direction of 1V and the current$i_0\\$ starts to ...