Hello, I'm fairly new to building psu's and I encountered a problem - thermal resistance/power dissipation to be exact.
I was trying to build a lm338 psu, that would output from 1.25v to 30v dc (I would mainly use 24v as I want to use this psu for an audio amplifier) and I thought a small 35x35x10mm fin plate radiator would be enough (oh boy was I wrong). Luckily, I decided to read up on the thermal problems before building it. So, I have few questions, but first I'll give you some specs:

I have a 230v-24v toroidal transformer that can output around 6 amps, some p1000 diodes,
1x 50v6800uF, 2x50v 4700uF, lot of 100nF caps, lm338 (I also have some lm317t) and other basic components like resistors.

I tried to calculate power loss on lm338 and picked the worst case scenario values, and here is my first question:
When calculating power loss what value should I put in for Vin?
Should it be the rms value - 24v, 33v - peak amplitude voltage 24*√2, or somewhere in between? (my guess is the 3rd option as the capacitors wont charge and discharge at exactly 33v but im not 100% sure)
I hope it makes sense

I used 33v as the worst case scenario, just in case.
Here are my calculations: (33-1.25)*6A = 190,5W
I thought to myself, wow, there is something wrong here, but I decided to calculate maximum power dissipation of lm338 anyway.

125°C - Maximum operating temperature
30°C - Ambient temperature
0,7°C/W - Junction to case thermal resistance (wasnt sure if i used the right one)
2°C/W - Case to radiator thermal resistance (I just assumed this one)
6°C/W - Thermal resistance of the radiator

(125-30)/(0,7+6+2)= 11W
Oh hell no, something is wrong here - I decided to look up some radiators that will be small enough to fit my case
but be big enough to dissipate more heat. The best one I found was a 1.6°C/W one.
(125-30)/(0,7+1,6+2) = 22W - still too small
Decided to stick to the 24v-30v:
(33-24)*6=54W - still too much power.

Lm338 wasnt an option anymore so I took a look at the lm317t pdf and calculated this:

5°C/W - Junction to case thermal resistance (wasnt sure if i used the right one)
2°C/W - Case to radiator thermal resistance (I just assumed this one)
1.6°C/W - Thermal resistance of the radiator
1.5A - Maximum output current

(33-1.25)*1.5 = 47,6W (already assumed its too big)
(33-24)*1.5 = 13.5W (seems promising)
(125-30)/(5+2+1.6) = 11,04W :c
A sudden thought went through my mind:
-What if I'm calculating this the wrong way? (english isnt my first language but I tried learning about thermal resistances from english articles, so maybe I interpreted something wrong.)

So that's the second question, am I calculating this right?

And the last question is, do you recommend any other voltage regulators?
Maybe there's a better alternative that I haven't heard about, one with 5A output would be great (but probably
unachievable with my current radiator, right?) - it can also be 3A/2A/1.5A - but I feel like I'm wasting the potential of my transformer using a 1.5A one.
I hope that some of you are willing to help and everything I wrote is understandable to some extent

Thanks!

 

So that's the second question, am I calculating this right?

Basically, yes. I haven't checked your numbers all in detail but the basic idea is wattage to dissipate = amps x (input DC voltage - output DC voltage).

And the last question is, do you recommend any other voltage regulators?

Have you read the data sheet for the LM317? I believe it shows an arrangement to accomplish what you want. You use a power transistor (or several) to dissipate the heat. The LM317 supplies the smarts to control it.

 

Basically, yes. I haven't checked your numbers all in detail but the basic idea is wattage to dissipate = amps x (input DC voltage - output DC voltage).

Have you read the data sheet for the LM317? I believe it shows an arrangement to accomplish what you want. You use a power transistor (or several) to dissipate the heat. The LM317 supplies the smarts to control it.

Sorry, I was looking at the datasheet by ST

Is this the circuit you were talking about? (source : http://www.ti.com/lit/ds/symlink/lm317.pdf)

I have 2 questions, should the 22ohm resistor be rated for any specific wattage?
And will using other transistors with similar specs but different hFE impact the operation of this circuit?

 

I'm not familiar with the circuit, but if the LM317 were running at 1A, that resistor would need to be well over 22W!

But by design I don't think that would ever happen. I'm too lazy to work it out but the first transistor turns on when the voltage drop across that resistor exceeds some value that makes the base 0.7V lower than the input voltage. Pretty sure that happens at a current far less than 1A. Do the match and you'll have your answer.

 

Hello, I'm fairly new to building psu's and I encountered a problem - thermal resistance/power dissipation to be exact.
I was trying to build a lm338 psu, that would output from 1.25v to 30v dc (I would mainly use 24v as I want to use this psu for an audio amplifier) and I thought a small 35x35x10mm fin plate radiator would be enough (oh boy was I wrong). Luckily, I decided to read up on the thermal problems before building it. So, I have few questions, but first I'll give you some specs:

I have a 230v-24v toroidal transformer that can output around 6 amps, some p1000 diodes,
1x 50v6800uF, 2x50v 4700uF, lot of 100nF caps, lm338 (I also have some lm317t) and other basic components like resistors.

I tried to calculate power loss on lm338 and picked the worst case scenario values, and here is my first question:
When calculating power loss what value should I put in for Vin?
Should it be the rms value - 24v, 33v - peak amplitude voltage 24*√2, or somewhere in between? (my guess is the 3rd option as the capacitors wont charge and discharge at exactly 33v but im not 100% sure)
I hope it makes sense

I used 33v as the worst case scenario, just in case.
Here are my calculations: (33-1.25)*6A = 190,5W
I thought to myself, wow, there is something wrong here, but I decided to calculate maximum power dissipation of lm338 anyway.

125°C - Maximum operating temperature
30°C - Ambient temperature
0,7°C/W - Junction to case thermal resistance (wasnt sure if i used the right one)
2°C/W - Case to radiator thermal resistance (I just assumed this one)
6°C/W - Thermal resistance of the radiator

(125-30)/(0,7+6+2)= 11W
Oh hell no, something is wrong here - I decided to look up some radiators that will be small enough to fit my case
but be big enough to dissipate more heat. The best one I found was a 1.6°C/W one.
(125-30)/(0,7+1,6+2) = 22W - still too small
Decided to stick to the 24v-30v:
(33-24)*6=54W - still too much power.

Lm338 wasnt an option anymore so I took a look at the lm317t pdf and calculated this:

5°C/W - Junction to case thermal resistance (wasnt sure if i used the right one)
2°C/W - Case to radiator thermal resistance (I just assumed this one)
1.6°C/W - Thermal resistance of the radiator
1.5A - Maximum output current

(33-1.25)*1.5 = 47,6W (already assumed its too big)
(33-24)*1.5 = 13.5W (seems promising)
(125-30)/(5+2+1.6) = 11,04W :c
A sudden thought went through my mind:
-What if I'm calculating this the wrong way? (english isnt my first language but I tried learning about thermal resistances from english articles, so maybe I interpreted something wrong.)

So that's the second question, am I calculating this right?

And the last question is, do you recommend any other voltage regulators?
Maybe there's a better alternative that I haven't heard about, one with 5A output would be great (but probably
unachievable with my current radiator, right?) - it can also be 3A/2A/1.5A - but I feel like I'm wasting the potential of my transformer using a 1.5A one.
I hope that some of you are willing to help and everything I wrote is understandable to some extent

Thanks!

I don't know where you are located, but a switching regulator similar to this would be better than a linear regulator for the wide voltage range.
http://www.ebay.com/itm/DROK-XL4016...732045?hash=item1a26c7194d:g:RtIAAOSwol5YzMb0
These can be built, but it is usually cheaper to buy one.

 

Would putting a step down converter before the voltage regulator be a viable option?
And thanks for all the quick answers.

 

Would putting a step down converter before the voltage regulator be a viable option?
And thanks for all the quick answers.

You could, yes. They are called pre - regulators. But you probably don't need to do that. You can just use the switching regulator.
The problem with the linear regulator is that you need some voltage above the desired voltage - lets say 3 volts. So no matter what you do at 6 amps you will need 18 watts of heatsink.