/diy/ - Do It Yourself

>>1181641
Most definitely. If you didn't already know, Zener diodes have a reverse breakdown voltage at a precisely fixed level, meaning if you put more voltage then that voltage, the current through the Zener increases more than exponentially. If it decreases, the current through the Zener goes very low. This means that when we put a Zener diode in series with a resistor, regardless of the value of the resistor and voltage of the source (up to a certain point), the voltage across the zener will remain constant.

Note, if the resistance is too high, not enough current will flow through the Zener and the voltage across it will decrease to the point of no longer being in the Zener region, thus decreasing the voltage across it sharply. If the resistance is too low, the Zener diode will be conducting too much current, and will therefore be dissipating too much power. Always keep in mind the power rating for your Zener diode, and make the maximum current through your diode enough to get to half this power value for safety's sake.

When designing a Zener power supply such as pic related, your main concern will be with the power dissipated by the diode. If it's a 5W zener diode, then you limit the power through it to be 2.5W, which through P=I*V gives you a maximum current of 2.5/3.3 = 0.76A, or 760mA. Since the voltage across the series resistor will by the law of voltage dividers be the value of the Zener subtract the voltage of the power supply, or 5-3.3 = 1.7V, we can calculate its resistance through R=V/I. Now here's where you decide whether your load is running 24/7. Since it's a µC and can draw a variety of currents, let's say it isn't on 24/7, and take the series resistor current to be the only current when it's off. If you were using an LED or something you would add the current through the LED to the current through the Zener to get the current through the series resistor, but we won't do so here because the Zener diode would dissipate more power.

cont.

>>1176243

this is the simplest way to do what you want. you would probably want the zener to conduct (at least) twice as much current as the load, so do your calculations accordingly.

i've seen this exact circuit used to generate 5Vdc from 120Vac inside a commercial product. for good form, you'd add a fat electrolytic cap across the zener. at least 1uF per mA of expected load current.

>>683421

Assuming shit for brains level of electrical knowledge

Ok start with using the multimeter on the internal battery to determine the polarity of said battery. And mark the positive pole (because you forget that shit)

Solder one end of the resistor to said pole and the other end to the end of the diode with the black mark. This is your 5.1v point
solder the other end of the diode to the negative pole of the battery.

then strip your usb cable. it has 4 wires, red,black, green and white. the last two are for data but we dont use em here so fuck em. Solder the black one to the negative and the red one to the positive.

slap some tape around to prevent shorts and try to cram all he stuff back in the case.

that should do the trick