Homebrew Power Supply

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Hey Folks,

I am building my own power supply and had a few questions.

First, I am going to use a transformer and then some diodes and a capacitor to make it into DC. Then, I will use the LM317 voltage regulator. This can supply up to a maximum of 1.5A so I was wondering if I could connect a couple in parallel to double the current handeling. Is this possible and does anyone know of a circuit that can accomplish this?
 

russ_hensel

Joined Jan 11, 2009
825
Parallel connections may not share current well. Pretty sure there are app notes on the web that show how to add a transistor as a booster. I have seen these many times for different regulators. Of course there other chips etc. available.
 

beenthere

Joined Apr 20, 2004
15,819
Page 15 of the LM117/317 data sheet shows how to do that. The LM195 is no longer available, but the LM395 is an available substitute.
 

kkazem

Joined Jul 23, 2009
160
I think he meant that that page has an example of how to connect an external bipolar transistor to the LM317 to increase it's current output and not an example of how to parallel 2 LM317's. It's a bad idea to try to parallel them. Use a bipolar that has a current rating at least 20% higher than what you need and use the LM317 or LM117 datasheet from National Semiconductor for the application circuit. THe LM117 is simply the military temperature range version or the prime version of that series. The LM317's are the commercial temperature range version. They are exactly the same inside, except that the LM117's are tested to ensure they can work at mil temp ranges and the ones that can't become LM217's (industrial temp range) or LM317's (commercial temp range). The application notes and example circuits apply to all 3 of them.

Good luck,
Kamran Kazem
kkazem
 

hgmjr

Joined Jan 28, 2005
9,027
Make sure that you provide a substantial heatsink for the LM317 regulator or its equivalent.

You will notice that the difference between the input voltage to the LM317 and the output voltage from the LM317 is dropped across the regulator. That voltage times the maximum current delivered to the load (in your case you are looking at 1.5 Amps) calculates the power that the LM317 will have to dissipate in Watts. For example, if you set the output of the LM317 for 5 Volts from an input voltage of 12 Volts, you will have a difference of 7 Volts. That 7 Volts times the maximum 1.5 Amps yields 10.5 Watts.

hgmjr
 

Ratch

Joined Mar 20, 2007
1,070
ELECTRONERD,

I dunno, I might have to use the transistor approach.
Were you not given the answer to that last month? See post #7 http://forum.allaboutcircuits.com/showthread.php?t=25976&highlight=hewlett
.

If you want to go the LM395 route, you can parallel them all you want. See Linear Brief #28 http://www.nalanda.nitc.ac.in/industry/appnotes/Natsemi/LB-28.pdf . It might be more expensive to use LM395's and you have to heat sink them. The first method with transistors uses less heat sinking because most of the heat is dissipated by a big power resistor.

Ratch
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
ELECTRONERD,



Were you not given the answer to that last month? See post #7 http://forum.allaboutcircuits.com/showthread.php?t=25976&highlight=hewlett
.

If you want to go the LM395 route, you can parallel them all you want. See Linear Brief #28 http://www.nalanda.nitc.ac.in/industry/appnotes/Natsemi/LB-28.pdf . It might be more expensive to use LM395's and you have to heat sink them. The first method with transistors uses less heat sinking because most of the heat is dissipated by a big power resistor.

Ratch
Nope, that was for my radio. You see, this one is going to be a variable power supply so I can power the stuff I'm building and testing. I've gotten to the point where batteries just don't cut it :D, they can't supply more than 24mA which is very low. I have a power supply that will supply 7A of current for my radio which is enough for 5W of power. It can do 50W but I hardly use it.

Thanks for the link. 10A is great! I can simply put that LCD voltmeter I was talking about on the output to find out what the voltage is and wallah!
 

Heavydoody

Joined Jul 31, 2009
140
In the datasheet ( http://www.national.com/ds/LM/LM117.pdf )
on page 17 you will find a schematic labelled High Current Adjustable Regulator using three LM195's (I believe this is the design mentioned by beenthere). I am not certain what constitutes "High Current". On page 22 you will find an Adjutable 4A Regulator with three LM317's in parallel. I havn't used these yet, but am looking into it. I don't have enough knowledge to understand how these designs function and I don't like throwing stuff together if I don't have at least a basic understanding of what is going on. But, when I get around to building something I will share my results.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Ok, I've decided to use the LM338 for the main supply and the LM317 to power my voltmeter (5V required). I have a transformer that takes 120V AC in the primary and changes it to 24V on the secondary. I need 30V; how can I boost the voltage 6V? Could I use an op amp?

Thanks!
 

SgtWookie

Joined Jul 17, 2007
22,230
Ok, I've decided to use the LM338 for the main supply and the LM317 to power my voltmeter (5V required).
Not sure about your particular voltmeter, but many such meters need an independent supply. In other words, you might need a completely separate power supply for your voltmeter.
I have a transformer that takes 120V AC in the primary and changes it to 24V on the secondary. I need 30V; how can I boost the voltage 6V? Could I use an op amp?
If you really need 30v out, then you're going to need a higher voltage secondary on your transformer. No, an opamp won't do it for you.

You're going to lose about 2v going through a bridge rectifier, leaving you with around 22VAC. Then comes your filter capacitor(s); the peak voltage they'll see is around 31v (22v x 1.414 = 31.1). However, the voltage won't stay that high very long if you have a load on the supply, even with a large filter capacitor.

The LM338 has a minimum dropout of around 1.7v; but at full current it'll be more like a 2.6v dropout. Under very light loading, you might get 29.5v out.

Most hobbyists don't typically need a lot of voltage & current; the majority of projects can be completed at 12v and under.

A problem that you will run into is regulator power dissipation when you are trying to output low voltage at higher currents.
If you are powering a 1A load that requires 15v, the load is dissipating 15W. If your filter capacitors have an average of 30v charge, your regulator will also be dissipating 15W.
However, if you are powering a 1A load that requires 5V, the load dissipates 5W, where the regulator is dissipating (30v-5v)*1A = 25 Watts. You will need a very large and efficient heat sink.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Thanks Sgt Wookie,

I appreciate your help. I suppose I'll just keep the transformer. For the digital voltmeter, I will be using a seperate voltage regulator (LM317) and I'll be using the LM338 as the main supply. Your saying I might have to have a seperate tap on the transformer with another rectification and filteration circuit for the 5V? Couldn't I just use the seperate regulators in parallel if the transformer can supply the current?
 

SgtWookie

Joined Jul 17, 2007
22,230
For the digital voltmeter, I will be using a seperate voltage regulator (LM317) and I'll be using the LM338 as the main supply. Your saying I might have to have a seperate tap on the transformer with another rectification and filtration circuit for the 5V? Couldn't I just use the seperate regulators in parallel if the transformer can supply the current?
Often, electronic voltmeters simply aren't able to measure from their own power source. Yes, I know, it seems like your separate LM317 should be a separate power source - but it isn't.

Just be prepared for it to not work.
 

SIcam

Joined Aug 9, 2008
61
Make sure that you provide a substantial heatsink for the LM317 regulator or its equivalent.

You will notice that the difference between the input voltage to the LM317 and the output voltage from the LM317 is dropped across the regulator. That voltage times the maximum current delivered to the load (in your case you are looking at 1.5 Amps) calculates the power that the LM317 will have to dissipate in Watts. For example, if you set the output of the LM317 for 5 Volts from an input voltage of 12 Volts, you will have a difference of 7 Volts. That 7 Volts times the maximum 1.5 Amps yields 10.5 Watts.

hgmjr
Your absolutely correct on the wattage dissipated as heat across the voltage regulator. Another solution is to use a transformer with multiple taps on it and a rotary switch to allow to use the different taps to lower the voltage out of the transformer. This will allow the regulator to not have to drop the massive wattage across it when the voltage is lowered and high current is drawn.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
I see...well then I might just have to get a transformer with a 30V pin and a 5V pin. Sounds like the best way to go. Since I would have the 5V pin and since It will go through rectification and filteration; do I need a voltage regulator? I could put a zener diode at the end to maintain a steady voltage couldn't I? How would you guys go about this?
 

SgtWookie

Joined Jul 17, 2007
22,230
I see...well then I might just have to get a transformer with a 30V pin and a 5V pin. Sounds like the best way to go. Since I would have the 5V pin and since It will go through rectification and filteration; do I need a voltage regulator? I could put a zener diode at the end to maintain a steady voltage couldn't I? How would you guys go about this?
That isn't the way it works.

You'd need a completely separate winding for the supply to the voltmeter, or a separate transformer altogether. You'll still need to rectify, filter, and regulate both secondary windings.
 

jpanhalt

Joined Jan 18, 2008
11,087
You'd need a completely separate winding for the supply to the voltmeter, or a separate transformer altogether.
Are you saying the problem is getting both a 5V and 28V'ish supply from the same transformer winding, or is that in reference to the common ground issue one faces with voltmeters?

There was a whole other thread last week dealing with the panel meter (cart before the horse,
you say ;)). In that thread several specific suggestions were made for meters that will work off a common ground. For those that don't, one can often get around that by adding a simple negative supply, but we did not go into that aspect, because it was not necessary to do.

John
 

Wendy

Joined Mar 24, 2008
23,415
I'll pick this up in my own thead (when I start one), but I've bought a couple of these for under $15 ea to do something similar with.



PWR1278
24V/6.5A Cosel
Power Supply

I'm thinking of a simple switching power supply that can handle 13A.

It makes a great foundation for a variable power supply, you will have to filter the switching noise out though.
 
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