Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
wyrebiker said:I wish to connect a permanant lead to my motorcycle battery (12 Volt DC) reducing output to 5 volt DC. Cheapest and easiest method please.
horzcat said:Like most of the guys said, use a 7805 IC. But what they didn't say is, you must use a heat sink with the IC. Otherwise, the IC won't last more than a few months. Both the IC and the heatsink are cheap. You can find them in any radioshack.
manish12 said:a series of 10 si diode give more current than 7805 , and it is chep too !
what do you think
12-5 = 7
number of diode = 7/0.7 = 10
and you have a option 5v, 5.7v , 6.4v , 7.1v , 7.8v etc .
manish12 said:a series of 10 si diode give more current than 7805 , and it is chep too !
what do you think
12-5 = 7
number of diode = 7/0.7 = 10
and you have a option 5v, 5.7v , 6.4v , 7.1v , 7.8v etc .
Audioguru said:Your booster circuit has a diode in the ground connection of the 78L05, so its voltage regulation will be poor.
It has an emitter-follower on its outpput so its voltage regulation will be poor again.
Two poors equals a very lousy regulator.