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OK. You say that it was a single jolt, and you pulled your hand away immediately. It was one of two things -- either it was static discharge and a true shock, but exactly what you would have gotten from a doorknob, you burnt yourself, or it was an actual shock and I can't count. You're running 5V through a \$10\Omega\$ resistor, which means that it's ...


3

Voh and Vol are provided, just look at the datasheet for GPIO DC specifications. They are just based on the pin drive strength and at the rated current for that pin strength. At the rated current the drop is 0.8V. With less current, the drop is less. With no or very low DC load, the drop is almost nothing, 0V. Since the LED driver is a high impedance input, ...


3

I've done a half dozen simple but moderately high frequency (50-100 MHz) opamps on two layer boards with no dedicated power planes and never had a serious problem that a power plane would have fixed. Provided you route the power lines carefully, that you keep as solid as possible a ground under the power and signal traces and the board is simple enough you ...


3

The 1.8 V LDO I intend to use has an output voltage accuracy of +/- 2% which means the voltage can vary by +/- 36 mV, which translates to 72 mVpp voltage. That's accuracy, not ripple. It includes effects like part-to-part variation and thermal drift, but unless you look at it on a very slow timescale, drift is not ripple. If your sensor is ratiometric (like ...


2

My guess is that it is an induction furnace. It doesn't heat the metal in the furnace (oven) by the heat dissipated in the water cooled copper coil, but instead the magnetic field created by the coil creates eddy currents in the metal in the furnace that needs to be heated. These currents in the metal cause heat dissipation in the metal, so they heat the ...


2

The following circuit will select the higher of the two input voltages to output. The circuit will work if at least one of the batteries has 3.7V. simulate this circuit – Schematic created using CircuitLab Diodes DM1, 2, 3 and 4 are the body diodes of the M1, 2, 3 and 4 respectively, and are not actual discrete diodes in the circuit. The purpose of ...


2

If you're going to use a series resistor then it has to drop 24 - 6.4 = 17.6 V. If you want 20 mA then \$ R = \frac V I = \frac {17.6}{20m} = 0.88\ \mathrm{k\Omega} \$ or 820 Ω for the nearest E12 value. Recalculating for that gives us \$ I = \frac V R = \frac {17.6}{820} = 21.5 \ \text{mA}\$. The voltage will remain fairly constant across the LEDs over a ...


2

It was either just a static discharge, or it can't just shock you, you must have also been in contact with some other metallic structure, such as a radiator. If it was a short sharp pain, then it was just static discharge. You get charged up moving in your chair wearing plastic clothing. Then you discharge when you touch something, like another person or ...


1

As mentioned from user Pat this is possible by using three phase bridge rectifier. However the diodes conduct only if phase voltage is higher than the other phase voltage and of course together higher that output capacitor or battery voltage. The result is a spiking current, not a contionous current. Source if the image You can see the waveform of a phase ...


1

Yes, it is possible to convert a 3-phase AC supply to a DC supply by using a full-wave 3-phase bridge rectifier. Assuming a constant DC load, each phase of the AC supply will be loaded equally. The resulting DC supply could then be converted to a single phase AC supply using pulse width modulation. Again, given a constant downstream single phase AC load, ...


1

In a cable tray, the electrons don't care if you parallel ten 1000 kcmil aluminum wires or seven 1000 kcmil copper wires. However the person paying for the wire cares, so you use aluminum because it's 1/3 the price. Different deal in a winding where you're all about making a lot of magnetic flux in a small space. There, density is so critical that you ...


1

@DKNguyen was spot on. Thank you very much for taking the time to point out the mistake. L1 above should have been connected to SW but was connected to the other pad of Cboot1, BOOT. The whole time the entire circuit was running off the bootstrap voltage.


1

It might differ outside of Europe, but the following could be used as some basic overview of how to evaluate and select the standards needed, before comercializing a power supply. Before starting with the design process you should define the following: Customer / Field (e.g. medical, industrial, etc) Location to be used (e.g. country) Standards Depending ...


1

Sometimes the USB port is enabled to deliver the 5 V voltage by a correct biasing of the d+ and d- contacts. Try to connect a cellular phone to the USB port and let your motor soldered to the pc board as in your last photo. It should spin correctly.


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