How to Calculate Resistor Value to Drop 12V Supply Down to 6V for a Circuit?

I am a beginner: how to calculate what kind of resistor I need? if for example I have a 12V power supply and I want to downsize to 6V how to calculate what kind of resistor I need????

it depends on what current will flow through it, and it is calculated from the formula R=U/I. In this case, U=6V because that's how much you want to lower the voltage. you need to know the current to select the resistor, and do not forget about the power of loss on the resistor P=U*U/R or P=I*I*R so as not to burn it out

so it's 12V/3A

If you want to reduce the voltage from 12V to 6V then you need two resistors, make them a voltage divider - pop to this page, everything is nicely described there: http://www.eres.alpha.pl/index.php?text=21

if it is 3A then you need a 2 ohm resistor but the problem may be with the power because it will give off about 18W on the resistor and that is quite a lot of power

... or more precisely ΔU/Io =>> (Ui-Uo)/Io
Where:
Ui=12V Input voltage
Uo=6V Output voltage
Io= .... [A] Load current (you must, unfortunately, approximately know the current you intend to draw with this Uo=6V !)
ΔU=Ui-Uo Voltage drop across the resistor

From the above you get =>> ΔU=(12-6)V=6V
Dividing this by the current Io you get the resistance of the resistor needed.
His power (loss power) will be ΔU=6V (voltage shunt on the resistor) * Io, which will flow through it.
For example - assuming Io=1A

The value of the resistor will be: R=Ur/Io=6V/1A=6Ω (Ohms)
The power of the resistor Pr you will calculate like this: Pr=Ur*Io=6V*1A=6W (Watts)
That is, the resistor will have a value of 6Ω and a power of 6W.


.... you're welcome

Greetings ...!
"djvu" .

thanks helped!


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Bicuś

Quote:

it depends what current will flow through it, and is calculated from the formula R=U/I. in this case U=6V because that's how much you want to lower the voltage. you need to know the current to select the resistor, and don't forget the power loss on the resistor P=U*U/R or P=I*I*R to not burn it

This is true, but in this case you don't need to calculate anything from Ohm's law - the resistor must have the same value and power as the load.

Quote:

Hence it is 12V/3A

Then why should it work on 6V? What is it even
If it draws 3A on 12V, it will probably draw about 1.5A on 6V (still depends on what the receiver is).
The limiting resistor will be 6V/1.5A=4Ω. Its minimum power: 6V*1.5A=9W.

Quote:

If you want to reduce the voltage from 12V to 6V then you need two resistors, make them a voltage divider - pop to this page, there is everything nicely described: http://www.eres.alpha.pl/index.php?text=21

You are wrong. That second resistor in the divider is the receiver.

Not true !!! - the second resistor is superfluous, because it is the receiver itself, whose equivalent resistance can be determined in the same way as the series resistor.

Otherwise, Ohm's law has no substitutes - and quite rightly, because isn't P=U²/R the same as Ohm's law ??? - admittedly after a slight transformation, but still ! ( P=U*I where I=U/R !!!)
You can just as well write (depending on your needs) P=I²*R, which, however, will in no way change the result but only the way it is calculated

Using two resistors as a divider, unfortunately the power loss will be greater than 50%, as much as 75%
=>50% on the upper resistor and 25% on the lower and only 25% on the receiver !!!
.... which is also easy to calculate

Therefore, for obvious reasons, it is not worth using a potentiometric divider !!! (especially at high power ...)


"djvu" .

Quote:

Otherwise Ohm's law has no substitutes - and quite rightly, because isn't P=U²/R the same as Ohm's law ???? - admittedly after a slight transformation, but still ! (P=U*I where I=U/R !!!)

Is this statement referring to the page that marcinj12 provided, or to my statement?
Neither way - I disagree. It's as if in mathematics you want to count the area of a square from the formula for the area of a trapezoid. Count you will count. Without knowing the formula for the area of a square - you will derive it without a problem. Just why make life more difficult for yourself????

According to what I wrote earlier"
Resistance of resistor R =ΔU/I=6V/3A= 2Ω (Ohm)
and its minimum power P =ΔU*I=6V*3A= 18W
I recommend, however, to use a higher-power resistor, because in case of a short circuit, which may happen one day, the resistor may burn out immediately.
More power will significantly extend its life !

Super would be a resistor with a heatsink, e.g. of the type " AX50WR " prod. Vishay (here link to the data page =>> http://zefiryn.tme.pl/arts/pl/Rezystory/ax10wr_0r1.html )
available e.g. from TME ( http://www.tme.pl )

Unless there is a value of 2,0Ω in the series, so they remain to choose: 1,5Ω or 2,2Ω .
If the voltage can be about 10% lower - take this 2,2Ω
.... and if it can't but can be raised by about 10% - take the 1.5Ω

In the type" AX25W " there is 1.0Ω but they have 25W each.
As you put the two serially together it comes out exactly 2.0Ω


I consider the topic closed for me (completed)


If you have any questions or concerns, write to me on PW


Good luck ..!!
"djvu" .