How do I create a constant 5V Source?

Thread Starter

corner_boy

Joined Jun 24, 2008
36
The chips I have need a 5 volt power source. I originally wanted to use batteries, but i need to know if theres other ways to make a power supply for this. I only ask because the person I'm making this for may like the fact that a change of batteries wont be needed. Set it and forget, in a way.

I got a 4-AAA battery source with a switch i got from radioshack for now.

P.S.
I have a battery charger for a motorlla cell phone i GOT. It says:
--Input: 100-240V~50/60Hz 0.2 A
--Output: 5.0V----- 550 mA
(What do you guys think??)

edit---- where can i buy a charger plug like the one mentioned on the PS, if that might be the best solution?
 
Last edited:

SgtWookie

Joined Jul 17, 2007
22,201
The charger for the phone won't be regulated - the charging circuit was inside the phone itself. 5v unregulated won't give you enough "headroom" to regulate the voltage.

You could get something like this wall-wart:
http://www.mpja.com/prodinfo.asp?number=17419+PD

and a 7805 regulator IC, a heatsink for it, a couple of capacitors, one for the input, one for the output, and a small box to put it in.
 

Thread Starter

corner_boy

Joined Jun 24, 2008
36
The charger for the phone won't be regulated - the charging circuit was inside the phone itself. 5v unregulated won't give you enough "headroom" to regulate the voltage.

You could get something like this wall-wart:
http://www.mpja.com/prodinfo.asp?number=17419+PD

and a 7805 regulator IC, a heatsink for it, a couple of capacitors, one for the input, one for the output, and a small box to put it in.
wow, thanx. i almost made a mistake assuming that the charger would have been fine.

---questions---
What are the capacitors needed for?

Since a heatsink seems needed to prevent overheating, which ones would be the best for a circuit that uses about 5 chips (ISD, mt8870, 2 potentiometers, and 2 LM386 amplifiers)?
http://search.digikey.com/scripts/DkSearch/dksus.dll?lang=en&WT.campaign=1136&WT.medium=cpc&WT.term=heatsink&site=US&y=15&KeyWords=lambda+heat+sink&x=18&WT.content=text&WT.source=google&cshift_ck=1119267935cs501936302&WT.srch=1
 

eblc1388

Joined Nov 28, 2008
1,543
wow, thanx. i almost made a mistake assuming that the charger would have been fine.
Well that depends on whether your charger is "heavy" or light. A heavy(relatively speaking) one contains an iron transformer and would not be suitable. However, most chargers supplied by phone companies are now of another type.

A light one, however, means it contains a switching regulator and that might be usable as a power source for your experiment.

To check if it is suitable, connect a voltmeter to the charger output and reads the voltage. Then place a resistor of 47 ohm across the charger output and reads the voltage again. If there is only a drop of perhaps 0.2~ 0.5V, then the charger is usable.
 

Thread Starter

corner_boy

Joined Jun 24, 2008
36
Well that depends on whether your charger is "heavy" or light. A heavy(relatively speaking) one contains an iron transformer and would not be suitable. However, most chargers supplied by phone companies are now of another type.

A light one, however, means it contains a switching regulator and that might be usable as a power source for your experiment.

To check if it is suitable, connect a voltmeter to the charger output and reads the voltage. Then place a resistor of 47 ohm across the charger output and reads the voltage again. If there is only a drop of perhaps 0.2~ 0.5V, then the charger is usable.
using the charger as a source to a board i got the 2 values from a multimeter.

1) connected to the wall - 5.083 volts
2) with a 47ohm resistor - 5.1 volts
 

kahafeez

Joined Dec 2, 2008
150
dude just use a simple 7805. dont make it complicated. just use an adapter and a 7805. in my country we've adapter which have a switch on it with which u can select its output. i keep the adapter to 9v. it gives me 9v dc and then i put a 7805 which turns the 9v dc to a 5v dc. simple. no capacitors. no nothing. easy. but if u need more current u'll have to put some extra stuff. if u want to run an ic or something of that sort its ok. i've been running an 8051 using the technique i mentioned. it works.
 

eblc1388

Joined Nov 28, 2008
1,543
using the charger as a source to a board i got the 2 values from a multimeter.

1) connected to the wall - 5.083 volts
2) with a 47ohm resistor - 5.1 volts
The voltage seems perfect. You can use it then.

kahafeez said:
it gives me 9v dc and then i put a 7805 which turns the 9v dc to a 5v dc. simple. no capacitors. no nothing. easy.
Do you know a 0.1uF capacitor is needed on the output of the 7805? I bet you don't.
 
Last edited:

bertus

Joined Apr 5, 2008
19,939
Hello,

The capacitors at the inout and output of the regulator are there for better stability.
Without the capacitors there is a possibility of oscillation, wich can cause the regulator to become instable.
Sometimes the osciilation heats up the regulator so it will go in thermal shutdown.

Greetings,
Bertus
 

millwood

Joined Dec 31, 1969
0
The charger for the phone won't be regulated - the charging circuit was inside the phone itself. 5v unregulated won't give you enough "headroom" to regulate the voltage.
most (all?) of phone chargers are regulated smps.

yeah, you can get one of those from walmart for next to nothing, or to get it from friends who have unused phones.
 
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