circuit to convert 18v to 12v for pedal-board power supply

[MakerDP]

I have a pedal-board power supply that puts out 9v and 18v to power the pedals. Problem is, I am using a pedal that needs 12v! I am sick of using it's own dedicated wall-wart.

So... I know that if I were to make my own dedicated power supply for it, I would need a power transformer and a LM7812 to regulate the supply. But,  since my pedal-board power supply is theoretically already regulated, can't I just do a simple two-resistor voltage divider to get my regulated 12v supply for this one oddball pedal?

I come up with R1=50300R (47k+3k3 series) and R2 = 100kR to get me at 11.976 volts.

Or would I need to set up an LM7812 circuit to do the job feeding it from the 18v supply?

[sluckey]

Or would I need to set up an LM7812 circuit to do the job feeding it from the 18v supply?

That would be my first choice.

[MakerDP]

Thanks. That's what I figured.

The wall wart the pedal came with is rated at 300mA. You think I'll need to heat-sink the regulator?

[shooter]

an LM317 could also be used, but Like Slucky said, the 7812 is the 1st choice.

[sluckey]

The wall wart the pedal came with is rated at 300mA. You think I'll need to heat-sink the regulator?

The pedal probably doesn't need all of that 300mA. But if it did, then the 7812 would have to drop 6 volts and provide .3 amps. That's 1.8 watts. I would use a small clip-on style heat sink just to be safe.

[MakerDP]

The pedal probably doesn't need all of that 300mA. But if it did, then the 7812 would have to drop 6 volts and provide .3 amps. That's 1.8 watts. I would use a small clip-on style heat sink just to be safe.

Thanks for your patience walking me through this. So humor me here so I don't have to keep asking this question every time I want to do something like this...

Looking at this datasheet https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf on page 2 it says:

max operating temp = 150*C
thermal resistance (air) = 65*C/W

So, at 1.8W dissipation it could heat up to 117*C without a heat sink, assuming ambient temp is 25*C.

Now, knowing I could be playing on a lighted (hot) stage or outside on a summer afternoon in Central California (very hot) I should probably go ahead and use some sort of heat sink like you said - just to be safe.

Is that the correct way to calculate the need for the heat sink?

And the correct circuit to use would be Fig. 9 on Page 19 of the datasheet? When I build custom Arduino microcontroller boards, I use 100uF caps, not those tiny ones in the datasheet. Should I "go big" or are the tiny ones in the datasheet OK?

[sluckey]

Is that the correct way to calculate the need for the heat sink?

I just use the finger test. If hot, put a sink on it.

And the correct circuit to use would be Fig. 9 on Page 19 of the datasheet?

Yes. I would use .1µFs on the input and output. Also put a 470 to 1000µF on the input and output. You can omit the 470 input cap if you know for sure that the input is well regulated.

[MakerDP]

lol ok thanks for the guidance guys.

I have since read a good summary of why a simple voltage divider will not work... it is dependent upon a constant current draw. If the current changes at all, then the voltages will also change.

One of those things that make you go "well, duh" when you stop and think about it... after all E=IR and if R is fixed then a change in I will absolutely change E.

[PRR]

> I come up with R1=50300R (47k+3k3 series) and R2 = 100kR to get me at 11.976 volts.
> ...wart the pedal came with is rated at 300mA

1) "50300" seems awful fussy. What is the difference, 11.976V or 12.2448V?

1a) "50300" seems fussy, if a 10% "47K" resistor could be 51.5 (51700) Ohms.

2) Take your 50K+100K divider and put a 300mA (0.3A) load on it.

12V at 0.3A is like 40 Ohms. So your "100K" is really 39.984 Ohms, your "12V" is really 0.014V. (18V * 39.9/50339.9)

> "well, duh"

Ah, I missed that. So you already knew.

If we *knew* the pedal sucked 0.3A at 12V (acts-like a 40 Ohm resistor), you could just use a 20 Ohm series resistor to drop-off 6V.

But the "300mA" rating may just be the smallest size of wart they could buy. Like using a 2x3 to support a closet shelf-- that's the cheapest hunk of lumber (1x3s cost more) so we use it even if "too good" for the job.

And 12V * 0.3A is almost 4 Watts which is a LOT for a small box.

If the pedal realy sucks 30mA (a more likely number), then a dumb 20 Ohm resistor will only drop 0.6V, and 17.4V gets to the pedal. Not wise without further investigation.

You really should quantify the pedal's actual demand. You certainly can NOT design a voltage divider without that info.

A "regulator" saves trouble because you will for-sure get 12V, whatever the pedal sucks, up to the reg's limits (heat or current).

Do you need a heat-sink? Again, you *need* to know the pedal's demand.

If demand is truly 0.3V, and you need to lose 6V, power wasted is almost 2 Watts and the TO220 package will be right at its short-term limits.

If demand is really more like 30mA, 0.03A, power is like 0.2 Watts and the TO220 will run slightly hot but fine forever.

> it could heat up to 117*C

If you are selling to strangers with a 30-day warranty, do it.

Building for yourself, keep it cool. 50 deg rise (75d operating) is plenty hot. Heatsinks are cheaper than blown-gigs and repair work.

In particular, a steady 100+C may last a very long time, but repeated cycling works-loose and micro-cracks the seals, lets beer (or humidity) in, rots the Silicon, chip dies.

Yes, 3 times a night eight days a week is probably enough thermal cycling to ever kill the chip. But I just don't see a need to run chips hot.