lowering the voltage

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Rewind the transformer...

or employ a regulator chip (type required entirely dependent on your current draw requirements). In some cases, the regulator circuit will only add 3 additional parts. The voltage difference in your case does not even mandate a Low Drop Out regulator.

Lots of options when it comes to regulators.

pw
 
what i was wondering is .....what specifically should i do

what type of resistor or anything should i use.....

i'm not too technical savy when it comes to internals.....

i can solder some and once told what will do the job i can find it and put it in......i just need to know

is there a page anywhere that shows the amount of volt drop for differant types of resistors....or regulators?

i don't know what the 1/4w, 1/2w, and so on differance is......

i wanna drop the volts......that's all i know

this is all AC by the way.....no DC
 
If we know precisely both the voltage *and* constant current delivered to the load, you are correct... we can simply connect a resistor in series with the power source. This value is calculated in accordance with Ohm's Law. This resistor will drop some of the source voltage, leaving the right amount for the actual load.
Again... the voltage drop has a direct relationship to the load current.

Usually, however, this approach is not ideal. The required value of the series dropping resistor will almost never be a standard value so you right away have to start fudging.


While a resistor MIGHT cut it if your load were constant... chances are your load varies. When the Load varies, the mathematics OHMS LAW is in full effect. Vary the current... vary the voltage drop.

You'd be better off with a carefully calculatated resistor and ZENER DIODE with a voltage rating of 12V. You also must know the current draw of the load. This will also help determine the power dissipation of the resistor and the cost. A high current resistor can cost as much as a run of the mill 3 terminal regulator.

So, in in effort to help you help yourself... set your browser to www.google.com and ask about "ohms law".
 
You could use the series resistor. I am not aware of your circuit but generally speaking, you could run that resistor to a capacitor whose negative lead goes to ground so the cap is across your load (circuit).

The resistor could be a value that drops 3v when the circuit is idle, and the capacitor should be big enough to effectively absorb the changes in load when the circuit is active.

If unsure, start with a large value of resistance and a large capacitor. Lower the resistance until you get 12v.

Then - divide the voltage across the resistor (3v) by its value in ohms to get the idle current of the circuit. Perform 12(v) divided by the idle current to get the circuit impedance (effectively its resistance). Then do 1/(2xPIx2xthe impedance).

Not knowing your circuit, this is probably a good minimum capacitance value. Don't go too far overboard here (maybe a little).
 
Hi,

what's all this regulators, LDOs, Zeners, etc?
this is all AC by the way.....no DC

The poster says he is talking AC.

The only answers that come to mind are
1. buy a 12Vac transformer.
2. attach a variac to the primary side of the 15Vac transformer.
3. load up the 15Vac transformer until it's voltage is down to 12Vac. Then wait till it overheats or not.

The last option may not be daft. It depends on the regulaton of the existing transformer.
 
AndrewT said:

what's all this regulators, LDOs, Zeners, etc?

The poster says he is talking AC.

Thanks andrew, must have been asleep.

Actually, I was assuming Lowjacker was going to rectify first. Even if the device calls for AC, there is almost always rectification within, and the answer moves inside the unit.

Otherwise I tend to agree with Andrew T.

BTW, most plugpacks giving AC tend to drop from a value like 15V to a value like 12V under their standard load.
 
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