First off, 13.75V is very close to the voltage you will get by using a 220 ohm in place of the 240 ohm resistor and 2.2K in place of the 2K resistor so there is no problem there.

Second, the use of a full-wave bridge and a capacitor will not provide you with the level of voltage regulation that you will need for your fans.

Can you provide the part number and manufacturer of the fan you are using?

The LM317 does have a thermal overload protection so the shutdown you are experiencing if probably this feature kicking in.

It sounds like your heatsink may not have sufficient surface area to dissipate the heat built up in the part due to the load. I tried to look up the specs on the heatsink you indicated you are using but I can find anything about the heatsink at www.radioshack.com.

A picture is worth a thousand words so if you could take a jpeg of the circuit and post it here that would be very helpful.

hgmjr

 

hgmjr,

I will take a pic of the circuit and upload it. Here is the link to the heatsink I am using.

http://www.radioshack.com/product/index.jsp?productId=2102857&cp

Here is the link for the fans.

http://www.radioshack.com/product/i...ng+Fan&kw=micro+cooling+fan&parentPage=search

Radio Shack Catalog # 273-240


Thanks
ScarEye

 

Here is a pic of the circuit. It's not pretty. But I am just testing. The final product will be much nicer.

 

Your circuit looks good from what I can see. Your heat sink appears to be ample at least for power one fan.

If the LM317 is getting hot there are two things that could be causing the heat up.

A. The current being consumed by the load is too great for the heatsink to dissipate the heat built up in the LM317.

or

B. The DC voltage at the input of the regulator is much larger than it needs to be.

Recall that I mentioned early on that the power being dissipated by the LM317 can be computed with the formula (Vin minus Vout) times the current being drawn by the load. If the input voltage is 2 times the output voltage then the amount of power being dissipated in the LM317 will equal the power consumed by the fan. To make the LM317 operate at a cooler temperature will mean that the DC input voltage to the LM317 will need to be reduced since there is nothing you can do about the power drawn by the fan.

Can you take a voltmeter and measure the DC voltage at the Vin terminal on the LM317?

hgmjr (1000)

 

Okay, I am going to check to see what is the input voltage at the LM317T and we will take it from there.

I am currently able to run 2 fans at the same time but the heat sink is freak'in cook'in I can literally throw and egg on it and cook it. lol


Thanks
ScarEye

 

What would happen if I hooked up TWO LM317T together ?

I am a n00b just asking a n00b question. =)

Thanks again
ScarEye

 

VIN @ The LM317T is 36V.


Thanks
ScarEye

 

VIN @ The LM317T is 36V.

Thanks
ScarEye

Aaaahhhh !! That explains the heat.

Since your regulator is set to around 12V, the difference between 36 and 12 volts is being dropped across the LM317. Assuming that the fan takes about 1.5W (taken from the datasheet you sent me a link to earlier), that would mean that the poor regulator is dissipating around 3W.

Have you considered placing the regulator and its heatsink in the airstream coming from the fan? That would set up a forced air situation which would in turn allow the heatsink to more efficiently pull heat away from the regulator.

That would be a short term solution. You could use a bigger heatsink but I think that as long as you are looking at a linear regulated supply then your best bet is to find a transformer that has a lower output voltage than the one you are using at the moment.

Can you get your hands on a transform that outputs 40VAC peak-to-peak? The one you are using must be able to output close to 100 VAC peak-to-peak.

hgmjr

 

hgmjr,

Wouldn't be easier for me to put together the circut below and then drop the voltage ?

I purchase a bridge rectifer a while ago from Radio Shack # 276-1185

Or Click Below To Get Specs.

http://www.radioshack.com/product/index.jsp?productId=2062584&cp


Without using a cap I get about 30 something volts DC. Should I just drop
this voltage ? Or would I be at square 1 because I would need a voltage regulator which would still procduce a crap load of heat ?


Thanks
ScarEye

 

I don't think the use of a circuit made of a simply a full-wave bridge and a capacitor is going to provide you with the quality of regulated DC voltage for use to power you fan.

Here is a 25.2VAC transformer available from RadioShack that should be capable of providing a reasonable AC voltage and current rating for your circuit.

hgmjr

 

hgmjr,

Right now I have a 24VAC power supply that is coming in. If I add this transformer would it be the same thing ?

This might be a dumb question but I am still learning. =)

Thanks for your help brotha.

ScarEye

 

Right now I have a 24VAC power supply that is coming in.

Greetings ScarEye,

When you say 24VAC power supply, I am going to assume you mean 24 VAC peak. If my assumption is correct then you will not need a transformer.

Since you are purchasing a 24VAC power supply, is there any reason why you did not go ahead and purchase a 12VDC supply? That would solve your problem in one blow.

However, if your goal is to learn more about designing a DC power supply that converts an AC voltage to a regulated DC voltage then this will give you that experience.

hgmjr

 

hgmjr,

This circuit is going to go into 100 of our stores. We already have a power supply that is 24VAC power supply. This power supply supplies power to CCTV security cameras. Instead of purcashing a transformer I would like to utilize the power that we have coming from the 24VAC power supply already installed in all the stores.


Thanks
ScarEye

 

I once built something with a 24 V power supply that also needed 10 V for CMOS logic devices. I used a 7810 voltage regulator for them but it got destroyed because the input voltage was too high. The way I solved the problem was to add a 180 ohm, 5 W resistor between the +24 V line & Vin on the regulator. I also put a 470 uF, 25 WV filter cap. between Vin & the circuit common [ground]. I don't recall what the 7810 input voltage was after this change but it's been working 24/7 ever since!

 

hgmjr,

This circuit is going to go into 100 of our stores. We already have a power supply that is 24VAC power supply. This power supply supplies power to CCTV security cameras. Instead of purcashing a transformer I would like to utilize the power that we have coming from the 24VAC power supply already installed in all the stores.


Thanks
ScarEye

Sounds like a reasonable goal.

Maybe there is another solution that might just buy you a bit of relief.

Instead of using the full-wave bridge, it may be worth considering taking a step backwards and using half-wave rectification. That would mean that you would only need a single diode instead of four diodes. The average voltage would drop by half leaving less voltage out. You would need to double up on the filter capacitance at the input of the regulator to keep the ripple voltage small. Even if you have a bit more ripple voltage at the input of the regulator, the regulator would take most of what remained out just through its regulator action. Even if a small amount of ripple managed to escape through the regulator, the fan would tolerate a little bit of fluctuation in its supply voltage without too much ill effect.

If this works, the heat should be significantly reduced. Maybe by as much a 50%.

It would be a simple thing to try since you have all the ingredients.

hgmjr

 

In my experience, a transformer marked 24VAC, outputs 24VRMS. With a full wave rectifier and a decent filter cap, the DC output voltage approaches 35V. This is right at the upper limits for a LM317. If you only need 12 volts output, why not drop 10 - 15 volts across a input resistor and reduce the power dissipated in the regulator ? (Vin-Vout)*It =Pd. (125C-70C/Pd for a LM317T should be >> than 4 C/W when a heat sink is used.)

Spoggles

If you are very sure of you power requirements, calculate the input resistor so as to have a 3.5 Volt difference between the input and output voltages.

 

Guys,

Sorry for the late response, I was sent off to Chicago for another project but I am back now. Thanks for your replys,


Spoggles,

Can you draw up a quick example of how I would accomplish what you are saying. How would I go about dropping 10-15 volts on the input resistor ?


hgmjr,

Can you also draw up a quick exmaple of what you are suggesting.


I apprecaite your responses and I would like to try everything so I can learn from experience.

Again, I value your input, and appreciate your suggestions.


Thanks
ScarEye

 

Hello:

The idea would be to drop the 'excess' power across an input resistor, rather than across the voltage regulator.
You have to be somewhat careful not to drop too much though. If you know what your expected load current is, and you know what your input voltage is, it would be a 'ohms law' calculation to figure out what the value of this resistor should be and still keep the regulator's input voltage at least 3.5 volts greater than its output voltage.. Just be sure to calculate how much power you need to dissipate in the resistor. (P=IE). Select a resistor equal to at least twice this calculation, and you should be in business. I like to use the 'too warm to touch' rule (140F) as a max temp rise for the device.

Also review the spec sheet at national semi, with regards to temp rise. (Thermal resistance)

Hope this helps.
Spoggles

 

Don't know if anyone will ready this. But I'm doing the same thing.

Only difference is I need an 12V DC output.

The circuits shown are what I came up with already in MS Paint LOL
So I guess I'm not doing too bad
But I missed the resistors.

The device I'm powering will only be using about 500ma, but I'm building the circuit to handle 1Amp. My transformer is rated 1.2Amps


I have two questions:
1) How do I re-calculate the resistors I need?

2) I bought a 12V DC transformer, but someone on another sight said that would be insufficient. Due to the gains/losses in voltage through the bridge.

3) Just for my own info if someone feels like explaining, what are the resistors doing in the circuit?

I'm attaching what I have so far.

Thanks,

Kenny

 

In this case you do not need external resistors. Your 12VAC transformer rectifies to about 17VDC on the first cap. This gives a volt or two more than the minimum you need to make the linear regulator work. You couldn't ask for much better.

If you had a 24V transformer, the voltage to the regulator would be closer to 35VDC. With a 12V output, this would drop 35v-12v=23V across the regulator. At one amp, this would be 23 watts. ->Hot! Better to dissipate some of this heat outside the regulator, thus the external resistors. In your case, 17V-12V=5V. At one amp, this is 5 watts. This is much more manageable. Just use a version in a large package (not a TO-92, more like a TO-220 or similar) and a heat-sink. You should be fine.

If you did use resistors they would be somewhere around 1Ω. Any more than this would drop too much voltage for the regulator to perform at the max current end. This would remove, at most, one watt from the regulator, as opposed to the 24V version using a 20Ω resistor to remove around 20 watts. In your case it is probably not worth the trouble.