About adjustable voltage regulators (LM317)

Hi,

I'm currently working on designing a PSU for my guitar stompboxes. I'd like to use LM317, because of their very clean regulation and also because of their adjustability.

While reading the datasheet, I can't find any clear info about the relationship between the input current and the output current. In other words : for 1mA at the output, how much current consumed at the input ? Is it a function of (Vin-Vout) ? t the "Heatsink requirement" paragraph I can see that the power dissipated by the component is ~ (Vin-Vout) x I(load).

I'd like 7 isolated outputs, 4x 120mA 9V/12V, 2x 500mA 9V/12V and 1x 150mA 9V/12V/18V. The current rating will depend on the power trafos used, as the 317 can handle a lot more current than that. So, will my output's current capability differ depending on the output voltage ? The input voltage stays constant.

LM317's datasheet : www.national.com/ds/LM/LM117.pdf

Best regards.

Eric

ricothetroll said:

Hi,

I'm currently working on designing a PSU for my guitar stompboxes. I'd like to use LM317, because of their very clean regulation and also because of their adjustability.

While reading the datasheet, I can't find any clear info about the relationship between the input current and the output current. In other words : for 1mA at the output, how much current consumed at the input ?

Click to expand...

the input current is equal to the output current plus the voltage divider current.
The voltage divider current is set by the value of the resistor located between the output and adjustment input. Typically 240ohms gives 0.5mA. 

Is it a function of (Vin-Vout) ? 

Click to expand...

No. 

  the "Heatsink requirement" paragraph I can see that the power dissipated by the component is ~ (Vin-Vout) x I(load).
I'd like 7 isolated outputs, 4x 120mA 9V/12V, 2x 500mA 9V/12V and 1x 150mA 9V/12V/18V. The current rating will depend on the power trafos used, as the 317 can handle a lot more current than that. So, will my output's current capability differ depending on the output voltage ? The input voltage stays constant. 

Click to expand...

It will depend on the input voltage too. You must choose your transformer and smoothing caps for at least 3V more than the regulated voltage at the input in worst conditions (low mains voltage, maximum current demand), but you must make sure that the input voltage is not too high when there is no current drawn and mains is high.
Since you want one 18V-capable output, you would need an 18V/3amp-rated xfmr, which would work out at 24V/2amp DC nominal. In case of low mains (-15%), you would have the required 21V pre regulation.
In case of high mains (+15%) you would have 29V pre reg.
Now the dissipation on the 500mA regs would be (29-9)x0.5 =>10W! You would need a hefty heatsink and probably the larger TO3 package to dissipate all that power.
I don't think you have the option of having a custom-made xfmr, so my suggestion would be to use two xfmrs, one for the 9/12V rail and one for the 9/12/18V rail.
Even then, you would have to seriously watch the thermal dissipation.
With a 12V 2.5A xfmr (sticking to commonly available types), you would be ok for low mains at -10%. Worst-case dissipation (full-current, 9v output at 0.5A, high mains at +15%) would be about 5W. It still requires a relatively large heatsink.
Sorry to complicate things for you, but that's the way it is...

Hi,

Thanx a lot for your answer ! Don't be sorry, that's exactly the kind of information I was looking for 
I actually though of one 2x12Vx500mA, two 2x12Vx133mA and one 1x18Vx150mA.
I took an ambiant temp of 55°C that gave me a maximum authorized elevation of 70°.

- For the first trafo, I have a mawimal allowed DC current of 500mA / sqrt(2) = 357mA
    => Pd = ((12 x sqrt(2) x 1.15) - 9) x 0.357 = 3,75W
    => PHI(ja) = 70 / 3,75 = 18,7 °C/W
    => PHI(ja) - (PHI(ch) + PHI(jc) = 18,7 - (0,5+4) = 14,16°C/W
    => that should be OK with this heatsink : http://be.farnell.com/jsp/search/productdetail.jsp?SKU=1710638

- For the second trafo, I have a mawimal allowed DC current of 133mA / sqrt(2) = 95mA
    => Pd = ((12 x sqrt(2) x 1.15) - 9) x 0.095 = 1W
    => PHI(ja) = 70 / 1 = 70 °C/W
    => PHI(ja) - (PHI(ch) + PHI(jc) = 70 - (0,5+4) = 65,5°C/W
    => that should be OK without heatsink (thermal resistance without heatsink is 50°C/W)

- For the third trafo, I have a mawimal allowed DC current of 150mA / sqrt(2) = 107mA
    => Pd = ((12 x sqrt(2) x 1.15) - 9) x 0.107 = 2,17W
    => PHI(ja) = 70 / 2,17 = 32,27 °C/W
    => PHI(ja) - (PHI(ch) + PHI(jc) = 32,27 - (0,5+4) = 27,7°C/W
    => that should be OK with the same heatsink as for the first trafo

Do you think that could all run safely in a closed box like this one : http://be.farnell.com/jsp/search/productdetail.jsp?SKU=1877077 ? Should I drill holes in it for a better ventilation ?

Best regards.

Eric

ricothetroll said:

Do you think that could all run safely in a closed box like this one : http://be.farnell.com/jsp/search/productdetail.jsp?SKU=1877077 ? Should I drill holes in it for a better ventilation ?

Click to expand...

Your dissipation calacuklations seem correct (I haven't checked the actual figures, but it looks ok), but the rated thermal resistance of heatsinks is for a thermal exchange with the ambient air, supposed to be circulating at such a rate that it is at constant 25°C.
This is not really the case when they are inside a box. So you must either manage to thermally link the heatsinks to the box (if it is capable of dissipating), or redo your calculations, based on a much higher internal temperature.
In the absence of numerical value for the box thermal dissipation, you would have to do a calculation based on the exposed area.

My enclosure is made of 2mm thick unfinished aluminium. Its size is WxLxH 200x120x64mm. According to the "quick & dirty" formula found on ESP's site, the thermal resistance would be 50/sqrt(20x12 + 2x20x6,4 + 2x12x6,4) = 1,96°C/W (I neglected the bottom plate as it's contact with the rest of the chassis is not "thermally" good)
http://sound.westhost.com/heatsinks.htm#s15

That value seems ok to cool all 7 regulators. I assumed that all thermal resistances are put in parallel :
1/Rtot = (2/14,16) + (4/65.5) + (1/27,7) => Rtot = 4,2°C/W > 1,96°C/W.
Is my reasoning good ? I'm still considering an ambiant temp of 55°C.

I would put the regulators as far away as possible from each other to reduce thermal interraction.

> 4x 120mA 9V/12V, 2x 500mA 9V/12V and 1x 150mA 9V/12V/18V

4*120= 480mA, 2X500= 1000mA, and 150mA is 1.63 Amps.

As said, your 18V reg suggests a 25V raw supply. If all outputs are set for 9V we have 16V drop at 1.6A is 26 Watts total dissipation.

Take a 26 Watts incandescent lamp and put it inside the Hammond 1590DE case. It will get VERY hot.

In my units the 1590DE has ~~100 square inches of surface area top and 4 sides. A rough rule is 100 deg C per Watt per square inch. So 100 sq.in. is 1 deg per Watt, and 26 Watts makes 26 deg C heat rise. That would be OK.

However if you do a device/sink to air transfer inside the box, air to box transfer, and box to air transfer through the sealed box, you are likely to be more like 78 deg C rise at the device. That's awful hot.

You can get regulators in "totally insulated" packages now. The heat-tab is plastic. Bolt these directly to the inside of the 1590DE. Be sure the inner surface is flat. Use grease.

The "26 Watts" number assumes all outputs at max current and lowest voltage. Real use may not be so bad. But I do think you are trying to stuff up-to 10 pounds of potatoes in a 5 pound bag.

18V raw supply makes a lot more sense. If I can guess what you are doing: your "18V" need is not 18V exactly, it is "lots more than old 9V designs", i.e. a pedal hot-rodded with two batteries. The 2*9V was convenient but isn't strictly needed. Also such designs are typically LOW current (from battery tradition).

So then you can regulator-drop ~~18V to 12V or 9V with more like 15 Watts dissipation, and brute-force R-C filter your raw suply for a "two battery" equivalent.

And 18VDC is a simple 12VAC winding to a FWB and large cap. This may want to be 2,000uFd or more. And the cap is more sensitive to heat than the semiconductors: it won't melt today but a hot cap's life can be shortened from decades to months. (But if the box runs hot enough to kill a cap in a year, it will also burn the wax on the floor and your fingers after the show.)

make sure you have a load resistor that drags a few ma, some of those regs do weird things when completely unloaded, like shut down,

there should be a minimum load spec on the data sheet,

as long as there is at least one stomp box loading the reg, you are ok,

but if you kick in a box on an unloaded reg, i do not know what will happen.

why use a 317 if all you need are standard voltages, 9/12/18?

more resistors needed with the 317 divider,

oh, now i know, you want the hendrix dead battery fuzz face sound, so you can dial the 9 volts into 7 for max sizzle my bizzle?  ;D

CJ said:

oh, now i know, you want the hendrix dead battery fuzz face sound, so you can dial the 9 volts into 7 for max sizzle my bizzle?  ;D

Click to expand...

That's exactly it, brutha...

If you stick yer tongue to the reg and it gives you a zip, you know it's not dead enough yet.

Disclaimer: please do not actually put your tongue on a voltage regulator...

unless you want to regulate your taste buds like warren G,

how full is fern ridge lake?

do i have to pull my sailboat out yet?

frickin corps, their pulldown schedule is so whacked,  :-X

Hi,

Thanx a lot for your answers ! Here is my schematic :
http://www.wuala.com/ricothetroll/public/Guitar_PSU_Schem.png/

I actually though of using 3x 12VAC trafos for 9/12VDC out, and one 18VAC trafo for 18VDC out. So there will be only one 18VDC capable output. I wanted 4 isolated ~100mA outputs, and two beefy >350mA outputs to be able to power current eater pedals like Moogerfooger Analog Delay (300mA!!). I planned to use LM317s because for what I know they provide the best regulation and are also adjustable, what is nice for my use. Some stompboxes are pretty sensible to PSU noise, like heavy distortions.

Possible trafos :
- 2x 12VAC / 12VA for the 2x500mA 9/12VDC output (actually at first I didn't think of the AC to DC ratio, so that's actually going to be 357mA, that's still enough)
http://be.farnell.com/jsp/search/productdetail.jsp?SKU=1712709
- 2x 12VAC / 3,2VA for the 2x2x95mA 9/12VDC output
http://be.farnell.com/jsp/search/productdetail.jsp?SKU=1131637
- 1x18V / 2,8VA for the 18VDC / 107mA output
http://be.farnell.com/jsp/search/productdetail.jsp?SKU=1689072

You can get regulators in "totally insulated" packages now. The heat-tab is plastic. Bolt these directly to the inside of the 1590DE. Be sure the inner surface is flat. Use grease.

Click to expand...

Thanks for the tip ! I actually didn't know this exists  :-[ I think I'm gonna buy some of those :
http://be.farnell.com/stmicroelectronics/lm317p/ic-voltage-regulator/dp/1703358

make sure you have a load resistor that drags a few ma, some of those regs do weird things when completely unloaded, like shut down,

Click to expand...

I think that with 317s the voltage divider is already providing that load, isn't it ? With 220+680+680R / 9V I have 5,7mA, same for 1800+91R / 12V and 1500+1500R / 18V.

Disclaimer: please do not actually put your tongue on a voltage regulator...

Click to expand...

unless you want to regulate your taste buds like warren G,

Click to expand...

;D

ricothetroll said:

I think that with 317s the voltage divider is already providing that load, isn't it ?

Click to expand...

Look twice at the datasheet of your voltage regulator. From the NationalSemi link you might have looked at pg.2 for the LM117 part. The LM317 is on pg.3. Other manufacturers (Vishay, SGS-Thomson, TI, Fairchild, ..) come with up to 14mA min.load requirement. Get this R1 value from (min)reference voltage/(max)min.current load for your part.

Look twice at the datasheet of your voltage regulator. From the NationalSemi link you might have looked at pg.2 for the LM117 part. The LM317 is on pg.3. Other manufacturers (Vishay, SGS-Thomson, TI, Fairchild, ..) come with up to 14mA min.load requirement. Get this R1 value from (min)reference voltage/(max)min.current load for your part.

Click to expand...

You're right !  :-[
My bad...