Proper use of LM338 in "Relatively High" current

[zdelarosa00]

Salutes, I'm making an assignment for college, and the instructions were "make a 55w power supply" no more questions asked, it can be whatever amperage and voltage we want, no further specification, but the overall expectation is a linear supply, its hard to get less-than-common parts, so together with the class contents, a switched supply is out of the question.

Fast forward, I have a salvage 12v 10amp common 1 phase step-down transformer, (127v line in my country), and with a little luck we managed to get an MB156 (15amp diode bridge),

The question is, how exactly must I handle my voltage regulator so the supply its as stable as possible?, from the datasheet I dont understand as concrete as I need, calculating resistor values or handling the top ratings well with protection circuitry, I see a couple example diagrams but all they tell me is overall how the regulator resistors connect and that sometimes certain capacitors or diodes for discharge are needed, would you help me understand?

From the datasheet I see that the metal can package can dissipate top 50w (starting when given a voltage difference between adj and Vin to 15v and higher or always top power?); so I connected a couple in parallel, side to side parallel, same traces, same direction, and both outputs to the same metal, to "expand the stress of dissipation" in words of my teacher, I know it doesnt work like that, I know theyre not ideal nor the load, but thats my best bet with the request and those parts.

So my question is, do I need the diodes and capactitors in my case wich values and why?, apparently it can handle somewhat good without them as i've tried in the past with less power supplies, but Im almost sure thats not my case.

That was the biggest transformer I had, and the other ones were 24v 1-Amp top, so in my basics I say I can get as high as 24w in a linear top?, and thats the reason i have to stick with my 12v 10-Amp.

The traces are made that way to decrease the chances of having issues about the current with them, and the little ceramic capacitor on the bottom is a past try to understand how it works.

I know isolation is wrong, I know the parallel is not a too coherent idea, I know trying to get power like that is wrong, I know we're handling current wrong, I know the load test is wrong, it's an assignment, and sadly my teacher is not for negotiating, so all I'm trying to do is making things works the less ghetto as possible and as right as possible given the sources.

Some photos were sticked to complement the situation.

TL;DR: Built a +55w power supply, need concrete info on handling LM338 voltage regulator

[Audioguru]

Your LM338 ICs will get too hot without heatsinks. Your filter capacitor is too small for much output current and the resistor and pot to set the output voltage are missing.

[zdelarosa00]

I think I'm not making myself clear, first, I havent said it but the load will be 16 (4x4) ac bulbs as charge, like 20w or 30w each, last time I connected it (with both regulation resistors, wich I still dont know exaclty how to calculate at wich powers, 5k pot and 330 reference did the job for me), it puled at max voltage (like 13-14v) a current of 2.4Amps, then one went off, no heat, no smoke, what I'm missing?, didn't surpace current or voltage ratings, no shorts were made, thats why I made those questions first, how should I operate the regulator on theory?

[DaJMasta]

I don't know about the LM338 specifically, but many chips have an auto shut off for very high temperatures, and that could certainly mean that after briefly functioning as intended, they got too hot and one (or both) shut off.


I think doing a basic thermal analysis will show you that you'll need some real heatsinking on there (as mentioned above).  If you have 2.4A going through with your test load, take your regulated output voltage and subtract it from the rectified voltage going in (is that 15V in and 12 out?  I'm not 100% from the post) and multiply by your 2.4 amp current, that is the power dissipated by your regulator.  Then since you're not using the heatsink, you use the junction to ambient thermal resistance figure on the datasheet, which for a TI LM338 in a TO-3 package is 35 deg C per Watt.  So if it was actually a 3V drop in the regulator, you're looking at about 7W dissapated, which means that without a heatsink, the die will raise 7W * 35C per Watt... and running you get a temperature of 245C above ambient - well over 2x what it takes to cook a lot of ICs.

If you get a nice TO-3 heatsink or a big plate to bolt it to, you can use the thermal resistance from the die to the case (only 1C/W!), then from the case to the sink, then from the sink to ambient and you can dissipate much more heat without damaging the chip.  Even if the chip is capable of dissipating 50W, it can only do that when cooled properly, it can't just eat up all the heat that entails.


I'm also no power supply expert, but the problem with regulators in parallel like that is that they will fight each other - the tolerances means one will be regulating slightly higher than the other, which means a vast majority of the loading will be one one of the parts - instead of sharing equally.  That would mean a different topology is a better choice - there should be many schematics available online which can be a good reference for other techniques.  I think the recommendations for larger caps is quite appropriate too, since 10A is a lot of current.

[mariush]

Ok, let's take it step by step.

You have a power transformer which outputs  Vac = 12v and Iac = 10A.

AC voltage means the voltage of the transformer goes goes from a positive voltage to a negative voltage and back about 60 times a second... 60 because that's the mains frequency in your country.

A bridge rectifier converts this AC voltage into DC voltage by making the voltage always be above 0v, instead of voltage going below 0v, it now goes from 0 to a positive voltage. This is called full wave rectification :



As you can see, the number of "hills" is twice the mains frequency, or 120 in your case. The peak DC voltage is equal to  1.414 x Vac  - 2 x Vdiode , where Vdiode is the foward voltage of an individual diode in your bridge rectifier.
Common values for this forward voltage are 0.7v to 1.1v, but you can look in the datasheet and get a more accurate value. Here's the datasheet : http://www.futurlec.com/Datasheet/Diodes/MB1510.pdf
On the second page, you can see a graph called "Instantaneous forward voltage" so you can just follow with your eyes that curve and you can see that at 10A, the curve intersects the 0.9v point, so that means you should expect a forward voltage of 0.9v (but let's make our lives easier and just work with 1v)

So anyway, the peak DC voltage will be 1.414 x 12v - 2 x 1v = ~ 15v
The maximum current available won't be the same as the AC current (10 A), the DC current can be estimated using the formula  Idc = 0.62 x Iac = 0.62 x 10 = 6.2A  - but just to be safe let's round it down to 6A

So, after the bridge rectifier, you have a DC output with a peak voltage of about 15v and capable of up to 6 A , so a total of 15v x 6A = 90 watts.

But as you can see in the picture, lots of times a second the voltage goes down all the way to 0v, so you have to add a capacitor (or several) to get charged with energy when you're at the peak voltage, and then when the output from the transformer would be close to 0v, the capacitors can provide the energy instead.
There's a simple formula you can use to figure out about how much capacitance you would need so that the output voltage would always be above a certain threshold :

Capacitance (in Farads) =  Maximum Current /  [ 2 x AC Frequency x ( Vdc peak - Vdc min) ]  where  Vdc min is the minimum you want to see in your circuit.

So for example, we determined that the transformer can output up to 6A , we know the mains frequency is 60 Hz, and let's say you want at least 12v DC all the time.  We put the numbers in the formula :

C = 6A / [2 x 60 x (15v - 12v) ]  = 6 / 3 x 120 = 0.016666 Farads or 16'666 uF 

The capacitors would have to be rated for at least 25v, because the peak DC voltage of 15v is very close to 16v (and if the mains voltage in your country goes up at some point, the peak dc voltage may also go up over 16v for brief periods)

You can use several capacitors in parallel, for example a bunch of 3300 uF or 4700 uF 25v capacitors, in order to reach at least that amount of capacitance.  With capacitors in parallel you simply add the capacitance.

Now you have a DC voltage between 12v and 15v, and a maximum current of 6A.
Since your power supply has to work at up to 55 watts, you must design something that can output a combination of voltage and current up to that amount, for example let's say maximum 10v at 5.5A

Linear regulators produce an output voltage by dissipating the difference between the input voltage and the desired voltage as heat. Linear regulators also require that the input voltage must be a bit higher than the output voltage.. that voltage difference is called dropout voltage.

For the LM338, you have the dropout voltage specified in the datasheet, and it's a bit over 2v at more than 5A at normal operating temperatures (50-80c)  - see Figure at the bottom of page 4 here: http://www.ti.com/lit/ds/symlink/lm338.pdf

So basically if you want the power supply to absolutely be able to output 10v, you would have to raise the capacitance after the bridge rectifier a bit, to have at least 12.5v or more at the input.

Keeping in mind that the difference between input voltage and output voltage is dissipated as heat, you HAVE to put the linear regulator on the heatsink.
In the best case scenario where you have 12.5v at the input and 10v at the output and you output 5.5A , you're dissipating  (12.5-10) x 5.5A = 2.5*5.5 = ~ 14 watts of power.
If you want to make it adjustable, the regulator would dissipate even more power as heat as high currents. It would be safe for example to adjust the output to 5v and connect something to it that uses up to 2A  (because then the regulator would dissipate (12.5v - 5v ) x 2 =  15w which is still reasonable) but you couldn't make it output 5.5A at 3v for example, there would be too much power dissipated as heat.

[Audioguru]

The datasheet for the LM338 says that it has "thermal overload protection" where it shuts down when it gets too hot.
I do not know what is "16 (4x4) ac bulbs as charge, like 20w or 30w each". If they are incandescent light bulbs then they draw about 10 times their rated current before they warm up.

[zdelarosa00]

Ok, let's take it step by step.

You have a power transformer which outputs  Vac = 12v and Iac = 10A.

AC voltage means the voltage of the transformer goes goes from a positive voltage to a negative voltage and back about 60 times a second... 60 because that's the mains frequency in your country.

A bridge rectifier converts this AC voltage into DC voltage by making the voltage always be above 0v, instead of voltage going below 0v, it now goes from 0 to a positive voltage. This is called full wave rectification :



As you can see, the number of "hills" is twice the mains frequency, or 120 in your case. The peak DC voltage is equal to  1.414 x Vac  - 2 x Vdiode , where Vdiode is the foward voltage of an individual diode in your bridge rectifier.
Common values for this forward voltage are 0.7v to 1.1v, but you can look in the datasheet and get a more accurate value. Here's the datasheet : http://www.futurlec.com/Datasheet/Diodes/MB1510.pdf
On the second page, you can see a graph called "Instantaneous forward voltage" so you can just follow with your eyes that curve and you can see that at 10A, the curve intersects the 0.9v point, so that means you should expect a forward voltage of 0.9v (but let's make our lives easier and just work with 1v)

So anyway, the peak DC voltage will be 1.414 x 12v - 2 x 1v = ~ 15v
The maximum current available won't be the same as the AC current (10 A), the DC current can be estimated using the formula  Idc = 0.62 x Iac = 0.62 x 10 = 6.2A  - but just to be safe let's round it down to 6A

So, after the bridge rectifier, you have a DC output with a peak voltage of about 15v and capable of up to 6 A , so a total of 15v x 6A = 90 watts.

But as you can see in the picture, lots of times a second the voltage goes down all the way to 0v, so you have to add a capacitor (or several) to get charged with energy when you're at the peak voltage, and then when the output from the transformer would be close to 0v, the capacitors can provide the energy instead.
There's a simple formula you can use to figure out about how much capacitance you would need so that the output voltage would always be above a certain threshold :

Capacitance (in Farads) =  Maximum Current /  [ 2 x AC Frequency x ( Vdc peak - Vdc min) ]  where  Vdc min is the minimum you want to see in your circuit.

So for example, we determined that the transformer can output up to 6A , we know the mains frequency is 60 Hz, and let's say you want at least 12v DC all the time.  We put the numbers in the formula :

C = 6A / [2 x 60 x (15v - 12v) ]  = 6 / 3 x 120 = 0.016666 Farads or 16'666 uF 

The capacitors would have to be rated for at least 25v, because the peak DC voltage of 15v is very close to 16v (and if the mains voltage in your country goes up at some point, the peak dc voltage may also go up over 16v for brief periods)

You can use several capacitors in parallel, for example a bunch of 3300 uF or 4700 uF 25v capacitors, in order to reach at least that amount of capacitance.  With capacitors in parallel you simply add the capacitance.

Now you have a DC voltage between 12v and 15v, and a maximum current of 6A.
Since your power supply has to work at up to 55 watts, you must design something that can output a combination of voltage and current up to that amount, for example let's say maximum 10v at 5.5A

Linear regulators produce an output voltage by dissipating the difference between the input voltage and the desired voltage as heat. Linear regulators also require that the input voltage must be a bit higher than the output voltage.. that voltage difference is called dropout voltage.

For the LM338, you have the dropout voltage specified in the datasheet, and it's a bit over 2v at more than 5A at normal operating temperatures (50-80c)  - see Figure at the bottom of page 4 here: http://www.ti.com/lit/ds/symlink/lm338.pdf

So basically if you want the power supply to absolutely be able to output 10v, you would have to raise the capacitance after the bridge rectifier a bit, to have at least 12.5v or more at the input.

Keeping in mind that the difference between input voltage and output voltage is dissipated as heat, you HAVE to put the linear regulator on the heatsink.
In the best case scenario where you have 12.5v at the input and 10v at the output and you output 5.5A , you're dissipating  (12.5-10) x 5.5A = 2.5*5.5 = ~ 14 watts of power.
If you want to make it adjustable, the regulator would dissipate even more power as heat as high currents. It would be safe for example to adjust the output to 5v and connect something to it that uses up to 2A  (because then the regulator would dissipate (12.5v - 5v ) x 2 =  15w which is still reasonable) but you couldn't make it output 5.5A at 3v for example, there would be too much power dissipated as heat.

Excellent, i didnt know all those considerations as such, but you seem to have alk that figured out, i did add the heatsink, and as i seem keeping the regulated voltage at max seems to be a little the safe point, im gonna search another cap to add, but why the 10A of the ac cannot be withdrawn??

[zdelarosa00]

The datasheet for the LM338 says that it has "thermal overload protection" where it shuts down when it gets too hot.
I do not know what is "16 (4x4) ac bulbs as charge, like 20w or 30w each". If they are incandescent light bulbs then they draw about 10 times their rated current before they warm up.

well, i actually dont know, im kinda lost with that, ive drawn 2.6~Amp in dc with all of them in parallel and they have written 20w...

[mariush]

Excellent, i didnt know all those considerations as such, but you seem to have alk that figured out, i did add the heatsink, and as i seem keeping the regulated voltage at max seems to be a little the safe point, im gonna search another cap to add, but why the 10A of the ac cannot be withdrawn??

Well think of it like this... you have 12v ac and 10 A , for a total of 120 watts.
Let's say the bridge rectifier has diodes with 1v forward voltage each. At any point in time, two diodes in the bridge rectifier are conducting, so there's 2v drop inside the bridge rectifier.
In an ideal world, the voltage would be converted from 12v AC to 17v DC without losses but even then we still have only 120 watts in total, so obviously can't have higher voltage AND same current as we can't pull energy from thin air, we will have have 120 watts / 17v = about 7A
We also live in a not so ideal world where we have the voltage drops on the diodes inside the bridge rectifier, so we're down to 15v peak dc voltage (and when 7A go through the bridge rectifier, the bridge rectifier itself wastes about 2v x 7A = 14 watts of power).
There are better solutions to convert AC voltage to DC , for example combining a ideal diode controller LT4320 with four mosfets to create a bridge rectifier that can have less than 0.1v voltage drop, but it costs much more (probably about 6-8$ at least) 

So in the end, you're down to 15v peak dc voltage and a theoretic 7A of current. There are other losses as well but it's more complicated.. I used the "magic number" 0.62 as it's a reasonable compromise and works well with small transformers.   Remember, it's "peak" voltage. The minimum voltage you'll get for a particular amount of current will depend on how much capacitance you have after the bridge rectifier.
Also it's important to know that the capacitors when empty appear like a short circuit to the bridge rectifier and the transformer so for a brief moment (a few milliseconds) while they charge, both the bridge rectifier and the transformer may push a lot of current through them, more than the normal expected. This matters when you decide what fuse type and size to use before your transformer ... for example, you know you have a 120w transformer and your mains in 120v , so you may think you can do with a fuse a bit higher than 1a ( 120v x 1a = 120w), but if you have too much capacitance, your transformer may pull way more than 1A from the mains for a few milliseconds.
With a huge amount of capacitance, the current may be even too much for the diodes in the bridge rectifier.. see datasheet, your particular rectifier has a normal current rating (continuous) like 15A for your rectifier, but can handle up to 300A for about 8ms (see first graph on page 2)

You can get more current but less voltage by adding a choke (inductor) at the output of the bridge rectifier, for example. See image below.

[Dave]

There's a lot of misinformation in this thread. Let's try to clear it up a little.

If the transformer secondary is rated 12V at 10A, it's a 120VA transformer, not a 120W transformer. The specified power is apparent power, not real power. You could load it with 120W, if you kept the phase angle of the load at exactly 0, so the power factor would be 1 and you would have zero reactive power. You could only do this with a purely resistive load, like a resistive heater or a light bulb.

A bridge rectifier followed by a capacitor bank does NOT behave like this.

Here's what a current waveform looks like from a transformer with this kind of load.


It's far from a perfect sine wave, it has lots of odd harmonics in there. In terms of power transfer, these harmonics provide nothing on the output, but they do increase the RMS current of the transformer secondary quite significantly, which is responsible for the heating of the winding.

The 0.62 factor from the guide that mariush pasted is purely a figure of thumb.
___________________________________

tl;dr You're not going to get 120W from a 120VA transformer in a linear power supply. Not even with an "ideal diode controller".

[mariush]

The 0.62 factor from the guide that mariush pasted is purely a figure of thumb.
___________________________________

tl;dr You're not going to get 120W from a 120VA transformer in a linear power supply. Not even with an "ideal diode controller".

I never claimed 0.62 to be an exact value, it's just a number that helps you approximate, let's you know quickly about how much to expect from a transformer. Works well for transformers in the 50-200VA range. It's also the number Hammond (see picture)  - a transformer manufacturer - uses in their application notes.

I never claimed he would get 120 watts from a 120va transformer, not even with ideal diode controller. But an ideal diode controller has the benefits of reducing heat as you no longer have almost 2v x current dissipated on the bridge rectifier and potentially gives you the choice of using smaller capacitors increasing the ways you could design a circuit.

I was also simplifying things a bit, as it's obvious he's not that documented. The power factor thing probably went over his head anyway or may be too hard to comprehend.