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simple 5v regulator? Answered

today i made a shaking generator for a shaking flashlight.i am using a 5.5v 1F capacitor as storage device.but output of  the generator is 3v~6v DC.how i can regulate the output ≤5v?.i am looking for most simplest and cheapest method.
can i use a resistor?

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-max-vina1991

Answer 3 years ago

A zener diode will not conduct current in the reverse direction until you exceed the zener breakdown voltage. Once you do, it starts acting like a more like a short circuit, conducting potentially a lot of current (depending how far you exceed the breakdown voltage).

So if you add a zener diode and a resistor across a unregulated power source, the voltage across the zener diode will be approximately the breakdown voltage. The resistor limits current so that the power supply does not force too high a voltage across the little zener diode. The voltage dropped across the zener diode, times the current through it will give you the watts of power it is dissipating, converting into heat. (That's true for everything*) Most small zeners are rated less than 1W, so 200mA across a 5V zener will push it to it's limits.

Because the zener diodes are not rated for much power, they do not make great power supplies. Allso they are very inefficient. at best, the efficiency will be the voltage out divided by the voltage in, and at worst, the efficiency will be zero. You have to make sure that the resistor used doesn't allow too much current to flow through the zener diode. if you have a 12V power supply, and a 5V zener diode, then the voltage dropped by the resistor will be about 7V, and 7V, with a maximum of 200mA for the zener diode, means that you need a 35 ohm, 1.4 watt resistor. This 'shunt' regulator will only be able to deliver up to 200mA at 5V. Once the load is connected and is drawing the maximum of 200mA, all the current from the resistor in series with the power supply is going into the load and not through the zener diode. That means for any current higher than that, the voltage on the output will depend on that resistor being in series with the power supply, and the voltage will sag to less than 5V. (also the resistor might overheat with more voltage drop across it. Use a 5W resistor in this case.)

The schematic below was for a tesla coil design, and at the time of developing the circuit, I burned out my 7812 linear voltage regulator, so I made a "ghetto" zener shunt regulator. Unfortunately the resistors were under-rated wattage-wise and overheat if left on for too long.

Solid State Tesla Coil Slayer Exciter rev 2 with regulator.png
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-max--max-

Answer 3 years ago

LEDs make good shunt regulators too, I used a white one as both a 3.3V voltage source and as indicator lamp!

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Downunder35mvina1991

Answer 3 years ago

Same way it is shown to be used for voltage regulation in a simple Google search.
Too lazy right now to draw it up.

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Yonatan24Downunder35m

Answer 3 years ago

Is it better than an LM7805?

I didn't know that Zener Diodes can lower the voltage like voltage regulators...

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steveastroukYonatan24

Answer 3 years ago

Yes, but they are not as useful. They are really used as reference voltage sources. Its hard to get Zener diodes in practical ratings compared to a regulator.

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Downunder35msteveastrouk

Answer 3 years ago

Considering the low power involved the losses are far less than with a regulator.
The zener basically just cuts off everything above 5.1V and puts it to a waste.
So only what's save will enter the storage cap.

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-max-

3 years ago

If you want efficacy, get a buck-boost converter. a zener shunt regulator would be the easiest method, requiring nothing more than a zener diode placed across the hand crank generator (assuming that it is only one of those small toy motors)

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steveastrouk

3 years ago

If you are using an LED which wants nice clean, constant current AND you want to maximise the power from your shaking, you really need to build a constant current supply.