LM317 Variable power supply

[audioguru2]

Hi Alun,
Your 18V transformer is rated to produce 18VAC at its rated current. It isn't a little 200mA wall-wart whose voltage doubles without a load, so its internal resistance is extremely low. So low that its voltage doesn't drop much when the rectifiers conduct their massive current.
The main filter cap should have a high enough capacity to reduce ripple voltage to about 1Vp-p at full load, or the regulator is going to have ripple in its regulated output.
I used a typical thermal resistance for the LM317 because I couldn't find spec's for an LM217 nor find a max thermal resistance rating. The LM317A has a guaranteed thermal resistance of 5 degrees C/W max. The extremely high calculated temp of your regulator's chip could be even higher.

I don't think your 45W transformer will catch fire with a load of up to 55W if the output of the LM217 is shorted. The LM217 is supposed to limit the current at about 2.2A with less than 10V across it, or about 1A with 20V across it. ;D

Hi Spagbol,
Many years ago I saw audio amps using a big light bulb for current-limiting. It worked. ;D

 

[Alun]

8)

Hi Alun,
Your 18V transformer is rated to produce 18VAC at its rated current. It isn't a little 200mA wall-wart whose voltage doubles without a load, so its internal resistance is extremely low. So low that its voltage doesn't drop much when the rectifiers conduct their massive current.

I haven't measured the secondary resistance, but the primary resistance is about 1kohms which is high enough to case a significant reducion in voltage on the secondary side, I can't see any other reason for me measuring 15V, do you think my meter's broken?

The main filter cap should have a high enough capacity to reduce ripple voltage to about 1Vp-p at full load, or the regulator is going to have ripple in its regulated output.

VRipple = Iload/(2fC)
so:
VRipple = 2.5/(2*50*10,000*10-6) = 2.5V
So perhapps 10,000uf is not big enough then.
I disagree, the ripple rejection of the LM217 is >80dB when a 10uf capacitor is added to the adj pin (I used 100uf) so as long as the DC is higher than Vout + Vdropout then the ripple on the output will be negligible.

I used a typical thermal resistance for the LM317 because I couldn't find spec's for an LM217 nor find a max thermal resistance rating. The LM317A has a guaranteed thermal resistance of 5 degrees C/W max. The extremely high calculated temp of your regulator's chip could be even higher.

I don't know, perhapps my LM217's thermal resistance was even lower than the data sheet's specification of 3oC/W and my heatsink is better than 1oC/W otherwise it would've shut down for sure.

I don't think your 45W transformer will catch fire with a load of up to 55W if the output of the LM217 is shorted. The LM217 is supposed to limit the current at about 2.2A with less than 10V across it, or about 1A with 20V across it. ;D

Those current and voltage limits you've specified have a very very wide tollerance. I hope the transformer won't catch fire but I'm also very concerned about the high case temperature causing a burn hazard.

 

[audioguru2]

the primary resistance is about 1kohms

No way, Alun. Your 18V/2.5A transformer has a 230V max primary current of 196mA. If it was 1k then it would drop 196V at full power and nearly double its output voltage without a load.

do you think my meter's broken?

Definately. It shows 1k for about a 60 ohms resistance.

I disagree, the ripple rejection of the LM217 is >80dB when a 10uf capacitor is added to the adj pin (I used 100uf)

I hope you used a diode to discharge the cap at turn-off.
The cap reduces ripple a lot. But ripple increases at high output currents and the rejection is typically only 1/300 without the cap and 1/1200 with. You will have ripple even with the bypass cap unless you have a huge filter cap.

otherwise it would've shut down for sure.

I think your LM217 got so hot it didn't know how to shutdown.

I'm also very concerned about the high case temperature causing a burn hazard.

If you are using the meatal case as the heatsink then you'll need a real heatsink that will be big enough to stay cool. ;DView attachment 37629

 

[Alun]

No way, Alun. Your 18V/2.5A transformer has a 230V max primary current of 196mA. If it was 1k then it would drop 196V at full power and nearly double its output voltage without a load, I still don't know how it was only 15V


Definately. It shows 1k for about a 60 ohms resistance.

You're right, (well almost) I measured it again it it was 100 ohms, I suppose must've got the decimal place wrong.

I hope you used a diode to discharge the cap at turn-off.

Don't worry did include a diode, and I used 100uf not 10uf but it probably won't make much differance.

The cap reduces ripple a lot. But ripple increases at high output currents and the rejection is typically only 1/300 without the cap and 1/1200 with. You will have ripple even with the bypass cap unless you have a huge filter cap.

But even if the ripple is only 40dB down which is 1/100 the 2.5V ripple will become a mere 2.5mV.

I think your LM217 got so hot it didn't know how to shutdown.

I don't know how, it can'tve got to 150 odd o that quickly, I think it's just the solder junction doing its job very well.

If you are using the metal case as the heatsink then you'll need a real heatsink that will be big enough to stay cool. ;D

That's just the problem I don't have room fo this, I think I'll add a 3A poly switch to the transformer's secondary, this will limit the continious output current to 2.21A and still allow 2.5A surges and the voltage drop will only be 0.2V

 

[audioguru2]

But even if the ripple is only 40dB down which is 1/100 the 2.5V ripple will become a mere 2.5mV.

My old fashioned slide rule says 25mV. You slipped the decimal again.

it can'tve got to 150 odd o that quickly, I think it's just the solder junction doing its job very well.

Did you solder it to a huge silver plate with liquid nitrogen for cooling? ;D

 

[Alun]

The poly fuse did the trick, the supply now allows 2.5A surges and limits the continious current to 2A or so, in fact this has worked much more well than I planned as it's temperature sensitive it limits the current more on lower voltage settings as it's hotter due to the extra heat generated, at 1.2V it will cut off the output at just over 1A but only after 5 mimiutes or so which is excellent as it stops the case becomming too hot to handle.

I've learned a few lessons from this project:

1) Don't rely on the regulator to thermally shut down to protect your transformer and rectifier unless the former is good for >4.2A and the latter is rated to >3A.

2) Always treat the heat dissipation calulations as a rough guide that predicts the worst case senario.

3) Beware when using the case as a heatsink, think about the heat being dissipated and the maximum potential temperature it could reach, note also you don't have a datasheet to give you its oC/W rating.

4) Soldering the device to a copper bus bar might be a good idea and prove worthwhile because the solder junction has a very low thermal resistance, however don't attempt this with expensive devices and they might not survive this.

5) Finally be very careful with the decimal place when performing calculations.

 

[Um...Me123]

I think I have things hooked up the right way now but as usual there are still problems.
First hooked it up and got a steady 19V out, the pot doing nothing.
Measured the ohms on the pot and it seemed pretty messed up.  I think I was getting somewhere between 1k and 5k most of the time.
I measured it the other way and got between 1 and 7k ohm so that seemed to be ok (is 7K normal?...on a 5K pot?).  This is center tapped by the way.  My output would go up just a few decimals higher than 7V and down to .1V.  Shouldn't I be getting more though?  Trans is 25V, half is 12.5V and doesnt the bridge give me an extra 3V (some poeple say it gives 3V some say it takes away 3V ???)  Seemed ok as I adjusted it except that I could not get 6V at all.  It would jump from 5V(point something) right to 7V.
I'm thinking that the pot is broken because of the burning smell and buzzing noise it made a while back. 

 

[Alun]

Are you sure you're connecting the ptentiometer up correctly?

How are you measuring its resistance?

A 5k pot should measure between 4.9k and 5.1k between termials 1 and 3.

 

[audioguru2]

He fried his pot because the LM317 was connected backwards.
A pot that gives a "burning smell and a buzzing noise" is ruined. :'(
His LM317 might also be ruined. :'(

 

[Um...Me123]

Heres how I think I had the LM317 connected  so maybe you can help me figure out if it ruined.  I had the Vin wire conected to the ADJ--the ADJ wire connected to the Vout--and the Vout wire connected to the Vin.  Do you know if this would ruin it and is there a way to test it?  I did try it not center tapped once (25-28V) and thats when the pot fried so maybe that would fry the LM317 if the center tap didn't?
Thanks
P.S. This happens to everyone doesn't it?  They fry their first project.

 

[audioguru2]

This happens to everyone doesn't it?

 

[Um...Me123]

Never even when you were really young (or whenever you started working with electronics)!
Anyway, through a tangle of wires I hooked up what I believe to be a test using a 220 resistor instead of a pot and got .59V not center tapped.  The calculator said I should get 2.5 (on a 28V supply I think) so I'm guessing your right, the LM317 is history also.  Sigh......Thanks guys.

 

[autir]

If you don't use a "star" separate-ground-wires wiring technique, its excellent voltage regulation and ripple rejection are ruined, due to the voltage drop across a common ground wire at high output current.

What exactly is this "star" topology?

 

[Staigen1]

Hi all of you

I have made thousands (or millions?)

;D Millions? Audioguru, how old are you? I thought you where 60! If you made 1 project a day, you should be somewhere between 2500 and 3000 years old! ??? ;D ;D ;D ;D ;D

circuits and haven't burnt any of them.

You didn't told us about all the components you really fry! ;D You make mistakes, and I do it! And the experts make the biggest ones, take the Hubble project as an example, it was a very costly mistake! Murphy is always, and i say always, with us! He's sitting there on our shoulder, ready to jump down on our projects! ;D ;D

What exactly is this "star" topology?

It's called so because you have only one grounding point, where all ground connections are connected to.

Hey Um...Me123, do you want me to help you with your project, and maybee your progress in electronics? In a moore palpable(maybe the wrong word, it's from my swedish-english dictionary) way!

//Staigen

 

[audioguru2]

What exactly is this "star" topology?

Hi Autir,
If you have it wired so that load current travels through a wire or pcb trace to which R2  (the pot) is connected to ground and the resistance of the wire or trace develops a voltage across it, then the regulation is much poorer.
National Semi explanes a similar problem that is caused when load current travels in a wire or pcb trace between the regulator's output and the current setting resistor R1:View attachment 37643

 

[autir]

Darn... 
So, since I cannot create PCBs by myself, the only way to build such a circuit is hard-wire?
The PSU I have built (remember that 500mA project?) is soldered on a stripboard and I have not whitnessed any problems... 

 

[audioguru2]

Hi Autir,
I also built my LM317 variable supply on stripboard. I had the load connected to the heatsink tab, with a terminal screwed on with the same bolt that attaches it to the heatsink, then connected the current-setting resistor to the output pin without any load current in it. Then the output "star" connection is inside the IC.

I connected the supply's ground, the load's ground and the pot's ground all together in one spot, called a ground "star"connection.

It works perfectly. ;D

 

[Um...Me123]

Hi all.  I was the one who put the LM317 in backwards and blew it and the pot up  ;D :. 
Anyway I got new parts and finally put it together and got it working.  I want to power a fan and a voltimeter in the unit but I don't know what to base my resistors on.  I have a pre-regulated wire that come ut of tthe circuit after the bridge and smothing cap.  I believe my trnsformer is 24V @ 2A but I can't remember cause it's been a while since I worked on this.  Thing is-I get a reading of 38.6VDC off that wire?!  I'm not swimming in parts like alot of you so I picked up any old handy resistor that happen to be a 220 1/2w and it made no difference in the reading at all?  ??? What's going on and how do I figure out what voltage to use to figure out what resistors to use for my fan and meter.

This was with no load on the regulated output wires (its a shared neg.).  I guess I never checked to see if changing the pot would do anything?
Thanks

 

[audioguru2]

Hi Um,
A 24VAC transformer might be 28V without a load. The rectifier charges the filter cap to the peak voltage minus two diode drops which calculates to be 38.2V.

The datasheet for the LM317 shows examples of extra-cost LM117 regulators but you can easily calculate resistor values for an LM317.

If you have a 120 ohm resistor from the LM317's output pin to its adjustment pin to set its current to 10mA, then a 2.8k pot (strange value) will allow it to have an output voltage from 1.2V to 30.5V. The 2.8k pot will dissipate 0.3W so will get warm.

Since the voltage across the LM317 exceeds 15V, it reduces its max output current to protect itself. With its output voltage set to 5V, it will have 33V across it and its datasheet says its max current could be only about 200mA but is typically about 600mA.
With a 5V output voltage and a 600mA load, it will dissipate 19.8W so will need a very big heatsink.

I don't know where you added a 220 ohm resistor. If it was in series with the 38.6V unregulated supply and the input to the regulator, then the current through the resistor determines how much it will reduce the voltage (Ohm's Law). Without much current, not much voltage drop. Only 100mA through the 220 ohm resistor will cause a 22V drop, and the resistor will dissipate 2.2W.

It looks like the unregulated input voltage is much too high. What regulated voltage range do you need for your meter and fan? 

 

[Um...Me123]

Wow everything you said there was very confusing.  I need 5V @ 110mA for the fan and 9V (8-12) at 1mA for the meter (but it's not working (see other topic))  The fan is a 40mm and fits nicely on top of the heatsink (to give you an idea of heatsink size) So I don't believe heat is a problem.  I just need to know if I should base the calculations for the fan on 38.5V.  Meaning I would need 300 ohm at 3.7 watts for the fan.  As for the meter, if it decides to work,  I would need 30,000 ohms at some very small wattage.  Would this be correct?