Power Supply, new design.

You may have seen my previous post regarding the Power supply I'm working on. I've thought about it and have decided it wasn't a lost cause... So now I've been trying to come up with a design that will work with the parts I have already bought. My transformer has a center tap on it which will give me 12.5 V or so at a maximum of 3A (the fuse will blow with any more... I think)

So here is my new design:
**broken link removed**

I'm don't know much more about AC other than it alternates in a sine wave pattern... So, will this work. My idea is, is that a DPDT switch will select the 25V or the 12V transformer output and the corresponding resistor needed to get the correct voltage range.

First, will my idea of using the center tap(12V)/normal(25V) selector work to get 2 different input voltages to the rectifier. And how much will my capacitor up the voltage. Something I read about the center tap being 180 degrees out of phase... I don't really know what that means or how that would effect my design. Even if it would not work, can I buy some type of transformer with two different output voltages? Something someone suggested in my previous thread.

Second, will my idea of using the DPDT switch to select the voltage and the corresponding resistor work?

Third, output amperage? This was the problem in my first design. There was too much voltage dropped at 1.5A when the regulator was set to 1.5 volts that the heat coming off the regulator was not practical to try and dissipate with any heat sink. So, I'm thinking that 1A or 0.75A is all I would need. So, how would I create such limitation? Where would I need resistors, what resistance, what wattage rating. Also, how do I calculate those for future reference?

Just as a side question, how would I make the output amperage variable?

With your 25VAC transformer the input of the LM317 will be 33.4VDC.
With the center-tap 12.5VAC the input of the LM317 will be 15.7VDC.
The input of the LM317 must be 2V or more higher than its output voltage.

It limits the current to 1.5A to 3.4A if its input to output voltage is 15V or less.
It limits the current to only 0.3A to maybe 0.5A if the input to output voltage is 40V.

If you want to adjust the current then you will need many more parts.

Hank102938 said:

Just as a side question, how would I make the output amperage variable?

Click to expand...

Do you mean an adjustable current limit?

Lab supply

I think you need a bigger filter cap to be happy. Below shows the ripple at 1.5 amp load with 1000ufd. and 6800 ufd.. You need to allow 2.5 volts for the regulator so 6800 ufd. would let you get to 12 volts on the low tap at 1.5 amps. Now for the second part: Don't worry to much about the current limit, the IC has built in thermal protection, so if it gets to hot it will shut down. Look up the data sheet there will be some info there. So how much current can you expect? Well it depends on if you use a heatsink. The part in free air can dissipate about 2.5 watts. Now mind you this puts the junction at 150C. (50 degrees C per watt) So with no heatsink you have about 15 volts into the regulator and 12 volts out so it can supply .8 amp maximum. .8 amp X 3 volts = 2.4 watts. Using the same math at 6 volts out you can supply .2 amps at 25C ambient. Not enough I bet. So add a heatsink, lets ay 6 degree C per watt. Now you have to look at thermal resistance junction to case. For the TO-220 part this is 5C per watt. Add this to the 6C for the heatsink and you have 11C per watt.
Now your cooking! You can dissipate 11 watts. Now you can get full capacity at 12 volts and about 1.2 amps at 6 volts. A better heatsink or a fan and you can get more at the lower voltages. The same ratios will apply with the high side switch. Full output at about 24 volts and less as you go down to 12.

https://www.electro-tech-online.com/custompdfs/2010/08/LM117-1.pdf

Hope this helps.

ronv said:

I think you need a bigger filter cap to be happy. Below shows the ripple at 1.5 amp load with 1000ufd. and 6800 ufd.. You need to allow 2.5 volts for the regulator so 6800 ufd. would let you get to 12 volts on the low tap at 1.5 amps. Now for the second part: Don't worry to much about the current limit, the IC has built in thermal protection, so if it gets to hot it will shut down. Look up the data sheet there will be some info there. So how much current can you expect? Well it depends on if you use a heatsink. The part in free air can dissipate about 2.5 watts. Now mind you this puts the junction at 150C. (50 degrees C per watt) So with no heatsink you have about 15 volts into the regulator and 12 volts out so it can supply .8 amp maximum. .8 amp X 3 volts = 2.4 watts. Using the same math at 6 volts out you can supply .2 amps at 25C ambient. Not enough I bet. So add a heatsink, lets ay 6 degree C per watt. Now you have to look at thermal resistance junction to case. For the TO-220 part this is 5C per watt. Add this to the 6C for the heatsink and you have 11C per watt.
Now your cooking! You can dissipate 11 watts. Now you can get full capacity at 12 volts and about 1.2 amps at 6 volts. A better heatsink or a fan and you can get more at the lower voltages. The same ratios will apply with the high side switch. Full output at about 24 volts and less as you go down to 12.

https://www.electro-tech-online.com/custompdfs/2010/08/LM117-2.pdf

Hope this helps.

Click to expand...

OK, so a bigger cap you say? Alright, you convinced me with the simulation.
On to the current and heat dissipation. The problem I had with my previous design was If it wanted to run something at 12V pulling all of 1.5A, then the regulator would have to drop the 30 some or more voltage creating too much heat. So now with the selector that's great, I can get 12V at full capacity. But what if I want 1.5V or 15V at full capacity? Would that be possible? And If so, with what kind of heat sink?
And what about the 12V from the center tap? If the fuse I have before the transformer is 0.5A, then wouldn't that mean the max current on the 12V output would be 5A? (120/12=10, 10*0.5=5) Or something... idk. I just don't want to fry something when I hook this up. And the center tap being 180 degrees out of phase, what's that mean and does it have an effect on the design?

Oh, and for resistors 2 and 3, ideally 2 would be 550 ohm for a 12.5V max and 3 would be 270 ohm for a 24.5V max. What wattage rating would these resistors need to have?

The LM317 has a max output current of 3.4A when the input to output voltage is 15V or less, but some limit the current to 1.5A. It is stated in the datasheet.

The resistor between the output and the ADJ pin of an LM317 is supposed to be 120 ohms or less. If it is a higher resistance then the output voltage will rise without a load. The datasheet shows the more expensive LM117 that can use a resistor of 240 ohms or less.

An input fuse of 0.5A will blow if the power to the circuit exceeds (0.5A x 120V)= 60W. For a 12V output, 60W will occur when the current is 5A which will never happen.

The power rating of a resistor is simply calculated with "I squared R". The I is only 10mA or less so the power rating of the resistors is small.

Alright.... sure, so, what resistance values then do I need for R2 and R3 to correspond to the raw input voltage of 12V for R2 and 25V for R3 in order to get a range of min voltage to 12V for R2 and 12V to 24V for R3? I'm, I guess what you would call an experienced novice at this stuff, so I don't really understand much, but I get the basics.

So.... what do I need to do to get this thing to work? What changes need to be made (Capacitors, wtv else)? What resistance values do I need (and what Wattage rating. I fried a 220 ohm 1/4W when I ran it at max voltage at 1.5A with my original design so...yeah)? What head dissipation rating do I need on a heat sink for the regulator in order to get 1.5A out no matter what the output voltage (if that is possible)?

Please help me out so I can get this thing working. Then, I can use it to learn something, and maybe one day I'll fully understand how the heck it works...

An LM317 cannot supply 1.5A when it has an input to output voltage of 13V or more. It gets hotter than a huge heatsink can cool it. A high velocity fan or liquid nitrogen will help.

The calculation of the resistor values and their power dissipation requires grade 4 arithmatic.

audioguru said:

An LM317 cannot supply 1.5A when it has an input to output voltage of 13V or more. It gets hotter than a huge heatsink can cool it. A high velocity fan or liquid nitrogen will help.

The calculation of the resistor values and their power dissipation requires grade 4 arithmatic.

Click to expand...

I know it requires simple arithmetic but you said something about the voltage increasing over time with resistance over 120 ohms. So, say I use a 120ohm resistor then that could give me a max voltage of about 53V. But if my input is only 30V then that leaves it adjustable from ~1.5V to 28V, would that not leave a lot of headroom on my 5K pot? As in after it's turned half way it's already at full voltage and any further would result in no effect??? Thus, making it a more coarse adjustment, or no? I don't even know what I'm talking about...

So, if I can't get the 1.5A at all my voltages what current can i get at all voltages (using a medium sized heat sink)? Can I get 1A, 750mA, 500mA what? I don't know the equations to figure it out or I would do it myself. So, with such amperage known, how can I add a resistor or something so more current than said amperage is never pulled?

There is an easier way to get half DC voltage from the C.T. transformer and a bridge rectifier.

Leave the bridge rectifier hooked up without the switches you have. The full DC voltage will be present on the positive output of bridge rectifier. Half the DC voltage will be present at the center tap of the transformer. You can put filter caps on both points for two outputs.
In the C.T. DC output configuration, the two rectifiers to ground in the bridge on transformer outside terminal connections make the full wave rectifier for the half voltage output.

When output of regulator is set for lower voltage you can SPDT switch to the transformer center tap to reduce the voltage drop across the regulator and therefore its power dissipation.

As mentioned, you need a lot larger filter cap then 1000 uF for amperage you are trying for.

RCinFLA said:

There is an easier way to get half DC voltage from the C.T. transformer and a bridge rectifier.

Leave the bridge rectifier hooked up without the switches you have. The full DC voltage will be present on the positive output of bridge rectifier. Half the DC voltage will be present at the center tap of the transformer. You can put filter caps on both points for two outputs.
In the C.T. DC output configuration, the two rectifiers to ground in the bridge on transformer outside terminal connections make the full wave rectifier for the half voltage output.

When output of regulator is set for lower voltage you can SPDT switch to the transformer center tap to reduce the voltage drop across the regulator and therefore its power dissipation.

As mentioned, you need a lot larger filter cap then 1000 uF for amperage you are trying for.

Click to expand...

Ok, I kinda see what you mean. But how is there DC voltage at the center tap? Circuit diagram/further explaination for clarification?

And just a heads up I don't need 1.5A, really at the most 1A, I could probably even be ok with 0.5A... I just don't know how to limit the current... would I put an appropriate resistor on the + output?

Hank102938 said:

... I just don't know how to limit the current... would I put an appropriate resistor on the + output?

Click to expand...

What limits the current is what load you connect to the supply. Suppose you have a 10V supply. If you connect a 5Ω resistor across it, 2A will flow into the resistor If you connect a 10Ω resistor, 1A will flow. If you connect a 20Ω resistor, 1/2A will flow. (Ohms Law: I = E/R)

Now, it is entirely different question to ask what happens if I have a 10V supply which is capable of only supplying 1A due to heating limitations, or loss of regulation, and I attempt to connect a 5Ω resistor? Something has to give...

You can start your learning experience right now with this project. Start with ohms law.

This is how I learned it:

E Where E = Voltage, I = Current and R = Resistance. If you remember the little picture you can remember
----- E = IxR, I =E/R and R=E/I
I R

Same thing with P=IE with P=Power in watts.
Armed with this information you should be able to find out why you smoked your 240 ohm 1/4 watt resistor when you placed it across 30 volts. Hint: To much power.

To get the most out of your supply get a big heatsink - maybe from ebay 2X3 inches with fins or add an ac fan to a smaller heatsink.
Nothing would have kept you from burning up your 240 ohm resistor except ohms law, (don't put a 240 ohm 1/4 watt resistor across 30 volts) but as for the regulator see the data sheet.

In addition to higher performance than fixed regulators, the LM117 series offers full overload protection available only in IC's. Included on the chip are current limit, thermal overload protection and safe area protection. All overload protection circuitry remains fully functional even if the adjustment terminal is disconnected.

Ohms law right, so at max 30v/220=0.136... so P=0.136*30=4.08, so I would need a 5W resistor then?

On to the heat and output current. How do I calculate the amount of heat in Watts that is generated by the regulator at E voltage and I current? I don't quite understand the ratings of the heat sinks (like 6 degrees C per Watt). I want to figure out what currents I can pull when set to common output voltages (like 1.5V, 3V, 6V, 12V... etc) with a decent sized, but not too big (larger than 2" x 2"), heat sink... That way I'll know the limits when I'm testing.
Ronv you kinda touched on it here but I'm still confused.

Resistors are made from stuff that can get very hot. A 5W resistor can withstand 4.08W but your fingers and many plastic capacitors and insulation on wires cannot. I would use a 10W resistor because it will not be so hot.

A heatsink with a thermal resistance of 6 degrees C per Watt increases its temperature 6 degrees C for every Watt dissipated in the transistor attached to it. If the transistor dissipates 10W then its case will be 60 degrees C higher than the surrounding air.
The transistor also has a spec'd thermal resistance from its chip to its case that might be 3 degrees C per Watt. Then the total thermal resistance is 6 + 3= 9 degrees C per Watt. If the surrounding air is 30 degrees C then the chip that is dissipating 10W has a temperature that is 30 + (10 x 9)= 120 degrees C which might be the absolute max allowed temperature.

Ok so, let me get this strait. A lower thermal resistance is better? I got the stats and chip to case is 4°C/Watt and 125°C is the max temperature. So say I had a heat sink with 6°C/W. So I have 10°C/W total thermal resistance. So say for example, with an input of 30V, I set the output to 20V and I pull 1.5A. 30-20=10 @ 1.5A so I have 15W of heat to dissipate.... the heat at the junction would be 180°C in 30°C ambient air which would trigger the thermal shutdown...? Ok so If I go to 20V at 1A then that would still create just a little too much heat....

How big would a heat sink with a thermal resistance of 6°C/W be?

Hank102938 said:

How big would a heat sink with a thermal resistance of 6°C/W be?

Click to expand...

Look at the many heatsinks at a good electronic parts distributor like Digikey or Newark. They have links to manufacturers' sites who show sizes.

I ended up answering my own question about the heat sink size... alright so now that I kind of have a grasp on the heat calculations, might I be better off finding a LM117K? It has a higher heat tolerance (150°C max operating temp) and only 2°C/W Junction to Case rating. With a good heat sink, I might be able to achieve all voltages at 1A and most at 1.5A. Thoughts?

Hank, I think you got it!
Yes the K package (TO-3) would be better. If you look at the difference you can see why. It is in an all metal larger base case so the heat can get away from the circuit to the heatsink better. It is harder to work with and more expensive than the TO-220 case however. Good job!

If you want more current, look at **broken link removed**. You will need a bigger filter cap and heatsink, but will have a more useful lab power supply...