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I've got the following power circuit set up. Basically I'm trying to get a 7812 (I Need 12 volts), and a 7805 (I Need 5 Volts) working with a On/Off Switch.
Everything seems good with the voltages, I'm getting both, but my issue is that I get the voltages wether or not the 'on' is on or off? Thanks.
If you're having caps in parallel, they may hold a residual voltage. If not, your switch is probably broken. (Or the wrong type) Try pulling out the power cord. Still get a voltage?
Also, it's common practice to put the switch on the supply line, not the ground line. This so because depending on the circuit, ground may be connected to other ground in other places, and the switch may not work as it's supposed to. Maybe this is not a problem in your design, but it's always good to learn to do things the "right" way.
Could you perhaps also post a photo of what you've built so far, and maybe we'll spot what's wrong with it.
Also, I don't if you connected the voltage regulators that way on purpose (probably not) but it's probably a good thing. What it does is help distribute the the heat. Let's say the external supply line gives you 15 V and that you portable is using 500 mA on the 5 V line. If you connect the 7805 directly to the 15 V supply, the power consumed (and heat generated) by the 7805 is
(15-5 V) * 500 mA = 5 W
Whereas if you connect like you do now, the heat will be distributed between the two regulators:
(15-12 V) * 500 mA = 1.5 W
(12-5 V) * 500 mA = 3.5 W
Meaning some of the heat is moved to to the 7812. If you didn't understand a word of that, that's fine too.
nitro2k01 wrote:
Also, I don't if you connected the voltage regulators that way on purpose (probably not) but it's probably a good thing. What it does is help distribute the the heat. Let's say the external supply line gives you 15 V and that you portable is using 500 mA on the 5 V line. If you connect the 7805 directly to the 15 V supply, the power consumed (and heat generated) by the 7805 is
(15-5 V) * 500 mA = 5 W
Whereas if you connect like you do now, the heat will be distributed between the two regulators:
(15-12 V) * 500 mA = 1.5 W
(12-5 V) * 500 mA = 3.5 W
Meaning some of the heat is moved to to the 7812. If you didn't understand a word of that, that's fine too.
It's not always better. If you pull a lot of power through the 12 volt line the additional load from the 5V regulator could cause it to overheat or worse. It's great if you are drawing most of the power from the 5 volt line, however.
I don't get a voltage without the power cord. When I test the 5volt line with the power cord, I'm getting 4.99 volts. So I think that is good.
I think I understand what you are saying about the switch being on the power line. I will try this tomorrow & let you know if it makes a difference.
I would take a picture, but I don't think it will help in this situation. It would just look like a jumble of wires.
Thanks & I'll let you know how I make out.
nitro2k01 Wrote:
If you're having caps in parallel, they may hold a residual voltage. If not, your switch is probably broken. (Or the wrong type) Try pulling out the power cord. Still get a voltage?
Also, it's common practice to put the switch on the supply line, not the ground line. This so because depending on the circuit, ground may be connected to other ground in other places, and the switch may not work as it's supposed to. Maybe this is not a problem in your design, but it's always good to learn to do things the "right" way.
Could you perhaps also post a photo of what you've built so far, and maybe we'll spot what's wrong with it.
Also, I don't if you connected the voltage regulators that way on purpose (probably not) but it's probably a good thing. What it does is help distribute the the heat. Let's say the external supply line gives you 15 V and that you portable is using 500 mA on the 5 V line. If you connect the 7805 directly to the 15 V supply, the power consumed (and heat generated) by the 7805 is
(15-5 V) * 500 mA = 5 W
Whereas if you connect like you do now, the heat will be distributed between the two regulators:
(15-12 V) * 500 mA = 1.5 W
(12-5 V) * 500 mA = 3.5 W
Meaning some of the heat is moved to to the 7812. If you didn't understand a word of that, that's fine too.
XCVG wrote:It's not always better. If you pull a lot of power through the 12 volt line the additional load from the 5V regulator could cause it to overheat or worse. It's great if you are drawing most of the power from the 5 volt line, however.
Ofc, I jjst wanted to figure out if he knew what he was doing or just copying a schematic. And if not teach him/introduce him to the basic theory behind the connection he's using.
