Regulators how do they - Page 1
Forum section: electronics / beginners

author: M0BSW
author_rank: Frequent Contributor
timestamp: December 19, 2013, 06:14:04 pm
content: Here's a question I have a 5 volt output regulator with and input of 12 volts, so where does the 7 volts go to, how do they work !!!!!!

author: Phaedrus
author_rank: Frequent Contributor
timestamp: December 19, 2013, 06:25:05 pm
content: The extra energy, in a linear regulator anyway, is converted to heat and dissipated. This is why linear regulators are only used for low power, because otherwise they would be too hot.In a switching regulator, the voltage is exchanged for higher current.

author: PA0PBZ
author_rank: Super Contributor
timestamp: December 19, 2013, 06:25:40 pm
content: Used to warm your house...

author: c4757p
author_rank: Super Contributor
timestamp: December 19, 2013, 06:30:34 pm
content: Voltage is just the relative difference in potential energy per electron (actually per Coulomb) between two points. By giving off heat, they have less energy, so a lower energy per electron relative to ground.

author: M0BSW
author_rank: Frequent Contributor
timestamp: December 19, 2013, 06:32:22 pm
content: OK I can understand that, so I'm about to use one for the first time 7805 5 volts output, which will be powered by my 12 volt supply, would it help if I put a heat sink on it, although the diagram doesn't show one.

author: Phaedrus
author_rank: Frequent Contributor
timestamp: December 19, 2013, 06:40:43 pm
content: How much current are you pulling? If it's less than an amp you'll probably be ok. An amp or more, find a small heatsink.Cheap ATX PSUs will sometimes have small floating heatsinks on their bridge rectifiers, with convenient mounting screw holes. I find them useful for minor heatsinking needs. Of course, I have access to a near-endless supply of power supplies, so it's cheaper for me to grab one than to buy a heatsink...

author: Monkeh
author_rank: Super Contributor
timestamp: December 19, 2013, 06:51:52 pm
content: Quote from: Phaedrus on December 19, 2013, 06:40:43 pmWith 7V dropout? You'll want a heatsink at anything over 150mA..

author: PA0PBZ
author_rank: Super Contributor
timestamp: December 19, 2013, 06:52:24 pm
content: You can use it without a heatsink if you are not dissipating more than about 1.25 Watt, so in your case a current of 1.25/(12-5), about 180 mA.

author: mariush
author_rank: Super Contributor
timestamp: December 19, 2013, 07:02:13 pm
content: The 7805 can handle a certain amount of power dissipated through that metal tab, without any heatsink attached to it. Usually, that's about 1-2 watts. How much your regulator will dissipate you can determine using formula Power dissipated = (Vin - Vout) x Current.The datasheets usually have formulas inside that tell you how to determine the proper heatsink (if necessary) for the amount of current you estimate the device is going to use.There is an online calculator that will help you with the math and also explains the formula towards the bottom of the page:http://www.ef-uk.net/data/heatsinking.htmThe calculator works with all the regulators in the 7800 family.

author: ivan747
author_rank: Super Contributor
timestamp: December 19, 2013, 08:07:18 pm
content: All about heatsinks:

author: M0BSW
author_rank: Frequent Contributor
timestamp: December 19, 2013, 08:12:56 pm
content: Quote from: mariush on December 19, 2013, 07:02:13 pmThank you for the formula, the project is the constant current load that Martin Lorton made, which is a long the lines of daves.

author: mariush
author_rank: Super Contributor
timestamp: December 19, 2013, 09:14:49 pm
content: Then the 5v is basically used to power the operational amplifier and maybe a power on led, so that's only a few mA of power. A heatsink is not required for that.The n-channel mosfet will however require a good heatsink to dissipate the power it uses.