# I want to convert 12v DC into 5v DC

I know there are lot of examples are given on internet. I have searched and I found good one. I am not professional in electronics engineering. I am Software Developer and I am trying to create a module that converts 12v into 5v. Actually I want to charge my usb device(mobile/mp3 player) from my motor cycle battery. I have found good tutorial on electroschematics site. but, I am confused in selecting capacitor for C1 and C2 defined in diagram. I dont know what is the voltage for those capacitor?

• If you've found one that uses a 7805 then you haven't found a good one, sorry. Mar 21, 2015 at 3:34

This is a terrible circuit and will not give you anywhere near the 5V@330mA specified.

Simple envelope calculation to prove this: just use Ohm's law with the full 12V across the input 100 ohm resistor. The maximum current which can flow through this is 120mA. More practically, I'd expect you to get only 60mA max at the output assuming you had a perfect linear regulator, and more likely only 40mA max at the output.

Other things I've noted:

1. C1 and C2's values are massive. 330mF on the input and 100mF on the output? really? More typical values are in the 100's of microFarads. In theory having large caps doesn't hurt, but ones this large either have very low voltage ratings and/or are much more expensive.
2. The two resistors really just shouldn't be there.
3. C3 and C2 are redundant. Get rid of one of them. Note that if you remove the 100 ohm resistor you need to remove the zener diode as well!
4. Some mobile phones won't charge just because you give them 5V to a USB port. You need to connect the data lines as well. How to do this is beyond the scope of this answer.
5. This is really pushing a linear regulator to its limit (probably beyond its limit). In the best case you're going to be dissipating 660mW in the regulator, and at the worst you'll be dissipating 3.1W. That's a lot of heat. You'll certainly need some sort of heat sinking in place.

As far as the absolute minimum voltage ratings requires, C1 must have a voltage rating of at least 7.2V in this circuit and C2 must have a minimum voltage rating of 5V. However, you'd have to intentionally go out of your way to find ones rated this low, and it's a bad idea to run caps so close to their max voltage. It's very common to find ones rated at 16V, 25V, 50V, or even higher voltages for pennies. Use these instead.

I would highly suggest you get some pre-made module (preferably one with a DC-DC buck converter rather than a linear regulator). One example is this module. This doesn't solve the data-line issue (4), but it does solve all of the other issues I listed.

• I would read C1 and C2 as 0.33 uF and 0.1 uF, not 0.33 Farad and 0.1 Farad. It is common to assume uF if no unit is shown, and, if I recall correctly, these are the recommended values from datasheet example circuits. Mar 21, 2015 at 6:06
• True, except the schematic has a mix of components with uF and without. Also, as the OP indicated they aren't experienced with electronics so I was being explicitly clear. Mar 21, 2015 at 7:30

The Voltage of the C1 and C2 should be at least double of the expected voltage in this circumstance. Typically Capacitors voltages in the range 0.1uf to 0.3uf are around 100 VDC not really too important again as long as it is above the expected Voltage. I do not know why the a 7805 IC will not work. It is rated at 1 amp with a maximum input voltage 35 volts. A heat sink is required to dissipate excess heat. It is very simple, 3 pins Input voltage, ground, and output. The only other components you will need are the R1 and C3. Since their is no AC component to this circuit, the other capacitors really are not necessary.---------- After reading the above answer you definitly need the R1 to limit the charging current of the phone battery, It would more than likely overheat otherwise.

I'd go with 25V or more for C1 and 10V or more for C2 and C3. Basically, the capacitor voltage rating is the limit up to which it's guaranteed by the manufacturer to work properly and not leak more than the specified value and, most importantly in your case, not break down. Generally it's a good idea to leave some margin, especially if you expect temperature variations, but it doesn't usually make sense to go higher than twice the expected voltage. It won't hurt, but the part may be more expensive and larger.

I don't see why a 7805 shouldn't work in the correct setup. This is a long shot and a bit cheap and I don't know whether or not it is helpful. You could consider running a line from the bike's 12 volts to a cigarette lighter socket kit and plugging an automotive USB adapter into it to give you a usb port (or two) with the needed 5 volts. The wire could be run hidden and the socket could be placed in the saddle bag out sight or up front with the speedo/tach. A quick and inexpensive approach. I hope that this is helpful.