24V from a 48V switched-mode PSU

I've built a  24V LM317 supply for powering two  (single stage) BA283 mic pres. It works but gets too hot to touch, the heatsink is a piece of aluminium L-profile which should be enough. So from where is the heat coming from, is the 24V difference too much? AFAIK it's not enough to kill the internal opamp (comparator) which is powered by the voltage difference (so the max. diff is the 36-37V which the same as what most opamps take). Or does the amp have some oscillation somewhere? Or the LM317?

So should I build another LM317 stage for the other channel, or use it as pre-regulator (like 48V -> 36V for the first and 36V -> 24V for the second one). Third option is to parallel the LM317 with another similar one (is it possible?)

Datasheet contains interesting ideas like tracking preregulator (8.3.4) and slow turn-on (8.3.8):

http://www.ti.com/lit/ds/symlink/lm317.pdf

As the input is already regulated I guess a regular 3055 single transistor circuit with a zener or just a resistor divider used for biasing could do the job, right?

The PSU I use is a one for Cisco IP phones from Ebay (380mA output).
http://www.ebay.co.uk/itm/Genuine-CISCO-48V-Mains-AC-DC-Adapter-Power-Supply-for-Cisco-1142N-Aironet-1140-/120994607996

mhelin said:

............. the heatsink is a piece of aluminium L-profile which should be enough. ...............

Click to expand...

Did you calculate?

The power wasted on the regulator is the voltage drop over it times the current drawn. 

Increase your heatsink size and you'll be fine.

sahib said:

Did you calculate?

The power wasted on the regulator is the voltage drop over it times the current drawn. 

Increase your heatsink size and you'll be fine.

Click to expand...

Right.  Popular misconception seems to be that heatsink should be sized for total current drawn times regulated power supply rail.  Not true.  Size for voltage dropped from regulator input to output.

If your loading on the regulator is somewhat constant, then you can drop some of the input-to-output voltage across the regulator by inserting a power resistor between input supply and regulator input.  The resistor is sized to drop say 50% of voltage, so your regulator then dissipates only 50% of what it was.

without knowing the current draw its hard to calculate, but with a 24v drop we can assume there's going to be a fair amount of heat.

Lets assume 100ma per preamp? 200ma total = ~4.8w

The quick and dirty way to start looking at suitable heatsinks would use the following maths/calcs to determine the value of thermal resistance required.

max_IC_temp(60c) - ambient_temp(25c) = 35c

35c/4.8w = ~7.3c/w

Give yourself some "headroom" so I suggest heatsinks with a thermal resistance of 5c/w or lower.

Thanks a lot for the ideas. I will try some resistor like 47 ohms or so before the regulator to drop couple of volts. Being a class-A amplifier the current should be quite constant. Could also use the case (19" 1U rack) as a heatsink though not sure if it is a good idea because of the thermal resistance of steel vs. aluminium. There isn't too much space  for a proper heatsink anyway.

The 47 ohms resistor drops the voltage just about 4.7V so the current taken by  the two preamps (plus the Bo Hansen DI circuit on both pres) is just 100 mA. Shouldn't it be at least 60 mA per channel (output alone, 100 mA for 1272 / channel)? BA283 output is strange, DC at collector is quite high ~22.2V so there is little current going through the  transformer primaries.  For 200 mA total current draw the 12V drop means I need about 60 ohms resistor.