I have a power supply 15VAC and i want to ajust the output at 5-6 VDC.I use one part that is for turn the AC to DC (i don’t know its name in English) and one resistor 220 ohm.But when i use the voltmeter i always take the same readings regardless if i use 220 ohm or 2000 ohm or with no resistor at 7 VDC.
What i do wrong and is there a way to do this?
It’s called a “Rectifier”.
Try putting a 16v light bulb on the output of your circuit and then reading the voltage again. The resistor doesn’t take effect without a bit of a load on the circuit.
You would have a much more accurate circuit by using a voltage regulator instead of a resistor.
Here are a couple of good sources for circuit diagrams:
http://home.cogeco.ca/~rpaisley4/CircuitIndex.html
http://www.mrollins.com/circuit.html
Good luck…
It’s called a “Rectifier”.
Try putting a 16v light bulb on the output of your circuit and then reading the voltage again. The resistor doesn’t take effect without a bit of a load on the circuit.
You would have a much more accurate circuit by using a voltage regulator instead of a resistor.
Here are a couple of good sources for circuit diagrams:
http://home.cogeco.ca/~rpaisley4/CircuitIndex.html
http://www.mrollins.com/circuit.html
Good luck…
Can i use for voltage regulator an old AC tranformer from marklin with a Rectifier?
The part is called a rectifier.
You cannot change the voltage with just a resistor because the ouput voltage will change according to the load (the current that is drawn). A modern voltmeter has in input impedance of many MegOhms and will not load the circuit so even at 220,000 ohms you will still see a constant voltage.
You need to install a “voltage regulator”.
AC supply feeds a rectifier which feeds a small resistor (47-220 ohms at 5 watts - not critical and can be omitted). Place a capacitor ± 100uFD from hot to ground and connect input of regulator to the hot lead. Regulator ground goes to ground, output gets another capacitor (about 1000 uFd - bigger makes cleaner DC).
If you use an LM7805 regulator the output will be 5vDC. to make 5.6 to 5.7 volts place a single rectifier diode in series with the ground lead of the regulator.
I’m writing this quickly so I hope that it makes sense.
A similar circuit is below. He uses smaller capacitors. R is not needed as your load will take its place, C2 adds stability and may not be needed.
Go here for more in depth information.
Good luck,
Karl
If you mean with the regulator, Yes.
If you mean instead of the regulator, maybe. You might be able to adjust the output control to what you need. Just be careful to not draw too much current.
Thanks you all for your replies.
I have a pc power supply ( a good one, Enermax 350W) and made a test.I turn it on by just shorting with a cable the Green cable and the Ground.If i remove the shorting cable it just stops. After i made some readings and i have 5,14 v on 5 and 12,16 v on 12.I believe it’s ok.
Now i am thinking to use the 12v line for swithing, 5v for DPDT switch, and 3,3v for Leds, and the 16V AC by marklin for my 130’ turntable.
What do you think about my thoughts?
If you are wanting to use an old PC power supply, I suggest you check out this site …
[link]http://web2.murraystate.edu/andy.batts/ps/POWERSUPPLY.HTM[/link]
PC supplies need a couple modifications in order to work properly without damaging.
Mark.
I read the article and i have a question about “static load to function”. My power supply turn on and stays on just shorted with a wire,without static load needed.So do i need the static load?
This is the way I understand it (somebody correct me if I’m wrong) -
These power supplies are DESIGNED for use by a computer and not as a bench top power supply. As such, they have numerous built in protection devices. Granted, the unit WILL turn on, but unless certain conditions are met, it will not provide the rated output. The resistor “fools” the supply into thinking it is properly supplying the mother board in a computer. If this load is not recognized, certain aspects of the supply will not power up to their potential to prevent damaging other critical components.
As I’ve already said, it will appear to turn on, and you will be able to get voltage readings, but those voltages aren’t at their full potential due to the supply not recognizing a critical load. You will be greatly disappointed at how little it will run without this modification.
DISCLAIMER - I’m not a computer geek nor an electronic whiz - these are just my findings, correct or otherwise. [;)]
Mark.