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I bought 200 MOSFETs on EBay from China. Got the parts and tested the on resistance. The part IRLB3034 was marked and even shipped in tubes labeled from International Rectifier. The spec is 0.020 ohm with 4.5 volts on the gate. The parts I got were 0.064 ohm. May not sound like a big deal...
Sounds like you want to drive the LED directly to the 230 volt line with a constant current circuit. That would be very inefficient and the constant current circuit would need to dissipate a lot of heat. exp. if your LED runs at 4 volts at 500 ma. then your constant current circuit would need...
Your MCU is probably loading your divider making your calculated values incorrect. You could use a high impedance FET op-amp to buffer the divider or account for the MCU load with your calculated voltage divider.
Those super caps have a pretty high impedance for a normal 150 amp starter load. Even a 150 F size cap could not provide enough power to even get that starter motor turning.
You just need a switching power supply so the voltage applied is about 14 volts (one 12 volt battery). I have used a $20 unit that accepts up to 40 volt input and has an adjustable output voltage.
I don't use thermal relief on my two layer 2 oz. copper traces and can easily hand solder wires to a through hole. Maybe a short trace to the power plane would be better for you. Board house builds have no problems without thermal relief on power planes, it's just any rework that is hard to...
I've used an LM317 and resistor to make a two terminal constant current source or sink down to 1.7 ma. It may work lower than that but that was for a string of LEDs.
Good question. You would have to look it up on the data sheet since I don't think all are the same. Some may have a shut down pin. I would assume you are shutting off the input voltage to power down the circuit anyway.
Converters typically run 80 to 90% efficiency. So say you are drawing 1 amp on the 3.3 volt (3.3 watts). Then your 9 volt supply current would be 403 ma (3.63 watts) at 90% efficiency.
Using a linear regulator would draw 1 amp at 9 volts on the supply. It would need to dissipate 5.7 watts...
The DC to DC converter can increase your current output if your input has a higer voltage than your output. The linear regulator will just keep the same current but disipate the extra voltage as heat.
Re: problem with 74ls373N
A 4.7 K pull down is probably to high a value for LS chips. Use a lower value or use a HC or AC chip.
From memory the 373 is a latch. Why not use a 374 which is edge triggered.
You can get the hex code in a disassembled form but without comments and labels, it's a lot of work to find out what was done. Less work to write your own software.
Use a step up transformer and you won't need any voltage doubling. Then you can get more current to reduce the charge time.
Don't forget your peak voltage will be much higher that the average (rated) voltage of the transformer
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