Convert 12 V to 5 Volt?

[Sanghera]

Hi all, just wondering how to convert 12 V into 5 Volts, do you need a DC to DC converter or can you use something else?
Thanks.

[abasir]

Do you want to convert 12V power supply _or_ 12V signal to 5V?

If power supply, you'll need DC/DC convertor (expensive) or voltage regulator IC (cheap) depending on amperage & efficiency you required. For low amp (about 1A) the standard here is LM7805. If you need about 3A, you can use LM123. For extended use at rated amp, heatsink is required.

If signal, just use a voltage divider, i.e., two resistors.

[Sanghera]

So what value of resistor for 12 volt down to 5 volt for signal? Can you get 12V from the computer power supply just as easy as 5 Volts?
Thanks.

[abasir]

Originally Posted by Sanghera So what value of resistor for 12 volt down to 5 volt for signal? Can you get 12V from the computer power supply just as easy as 5 Volts? Thanks.

I would use 6.8k & 4.7k or something near that.
All desktop computer power supply I came across have +12V (yellow wire), +5V (red wire), -5V, and -12V. Typically the +12V and +5V can supply several amps.

[Sanghera]

Cool.

[arvidb]

Sanghera, just to clarify the use of the voltage divider (in case you don't know this already):

Connect the two resistors in series. Connect the 4k7 end to GND and the 6k8 one to the 12 V signal. You can then pick the 5 V signal between the junction of the resistors and GND.

Arvid

[Sanghera]

How did you do the math for this?
I mean, say you have a 1.5 Volt battery and want 1 Volts out of it? Not that I need this info. just as an example. How would you figure how many ohm resistor and what not? Ohm's law eh? You would have to know the amps of the 1.5 Volt battery wouldn't you? If you do, what would be the proper way with a multimeter?
Thanks.

[Sanghera]

Scratch this post.

[Al_The_Man]

Here is a program I wrote to run under Qbasic for the three terminal voltage regulator LM317 The value of R1 is typically around 280 ohms and is connected between the output pin and the Adj pin, R2 is connected from the Adj pin and ground. This gives you the value of R2 for a given output and and the current through it.

REM PROGRAM TO CALCULATE LM317 REGULATOR RESISTOR
CLS
PRINT "R1 = ";
INPUT R1
PRINT "Voltout = ";
INPUT Voltout
LET I1 = 1.25 / R1
LET R2 = (Voltout - 1.25) / I1
PRINT "Resistor Divider Current ="; I1
PRINT "R2 ="; R2
END

Al

[arvidb]

Originally Posted by Sanghera How did you do the math for this? I mean, say you have a 1.5 Volt battery and want 1 Volts out of it? Not that I need this info. just as an example. How would you figure how many ohm resistor and what not? Ohm's law eh? You would have to know the amps of the 1.5 Volt battery wouldn't you? If you do, what would be the proper way with a multimeter? Thanks.

You calculate the current using Ohm's law, with R = resistance of the two resistors in series and U = battery voltage of 1.5 V. When you know the current through the resistors, you can calculate the voltage drop over each resistor.

If the resistors are R1 and R2 their series resistance is R1 + R2, so the current through them by Ohm's law becomes I = U/(R1 + R2). The voltage over R1 is then, also by Ohm's law, U1 = R1/(R1 + R2)*U, and the voltage U2 over R2 = R2/(R1 + R2)*U.

In the original question, U = 12 V, and with R1 = 4k7 and R2 = 6k8, U1 = 4k7/(4k7 + 6k8)*12 = 4.9 V, which is good enough.

Observe also that U1 + U2 is always equal to U, as it must. Also notice that if you draw current from the point where the resistors are connected, that current will be drawn from the positive, so will pass only through the resistor not connected to ground. This will increase the voltage over that resistor (again, Ohm's law), meaning less voltage and current through he "lower" one, and so a lower voltage. That's why voltage dividers like this only work for "signal" levels (where the current draw of whatever is connected between the resistors is insignificant).

Arvid

[Sanghera]

So, this would not be a practical way to do for a PS? Or does current not even come into play here?
Thanks.

[abasir]

Voltage divider above is not for PS. Current flow is only about 1mA (12V/10k Ohm). If you connect another significant load (taking more than 0.1mA) to the divider, the voltage will change as the total resistance is changed (taking into account the load resistance).

As I indicated earlier, if you need PS conversion, get a voltage regulator -- LM7805 with heatsink will give you 1A continuous.