LM317 voltage regulator R2 is getting very hot? why?

Ok so let me ask you this, if I use a 240 ohm resister as R1 then I can supply a voltage whatever depending on R2 but my maximum current I can provide is 1.25/240 is 0.005 Amps. Now what if I need 250mA?

calvingloster:

123Splat:
E/R= I
IIR=P, not IRR

Can you please talk in words and not letters? I don't know what you trying to say?

A person needs to know a little algebra and common variable names to be successful in this hobby. Do the substitution yourself, E = voltage (or you can use V), I = current, R = resistance, P = power.

Heat dissipation is Amps multiplied by resistance.

No, power (heat) is volts times amps. Or: P = V * I

With a little algebra (you passed algebra?), we also have P = I2 * R and P = E2 / R

Very useful and essential formulas, should be memorized. Used properly, they are guaranteed to prevent that burning smell :wink:

calvingloster:
Ok so let me ask you this, if I use a 240 ohm resister as R1 then I can supply a voltage whatever depending on R2 but my maximum current I can provide is 1.25/240 is 0.005 Amps. Now what if I need 250mA?

Incorrect. Where in the world did you get such an idea?

Sorry I am abit slow I really do wish I was as clever as many of you people, please forgive me.

I see now that I made the mistake of multiplying amps and current to get power dissipation instead of squaring amps and then multiplying it by current.

I got that idea from if you put a 240ohm resistor between output and adjust pin then u get a constant current source of 0.005A. So there is no way you can get more amps from using a 240ohm resistor. I am however confused why the LM317 does not act like a current limiter if you wire it up as a voltage regulator.

AMPs am CURRENT,,, It's Voltage , Resistance, and Current (in AMPs).
The 1.25 V and 50uV (not 5mV) are what is passed by the Adjust pin of the regulator. You can read the datasheet to find out how it works. But that current path is totally different from the reference loop setup by R1 and R2. That sets the Highend voltage the regulator will pass. with proper heat sinking, the 317 will pass a max of something like 1 AMP, as long as you source it with Vin=Vout+1.25V, or more and at least 1 AMP.

calvingloster:
Sorry I am abit slow I really do wish I was as clever as many of you people, please forgive me.

No worries, just slow down and try to understand the fundamentals like Ohm's Law. Not difficult at all and as ever, GIYF. It's also very important to read the datasheets of the devices in the circuit.

I got that idea from if you put a 240ohm resistor between output and adjust pin then u get a constant current source of 0.005A. So there is no way you can get more amps from using a 240ohm resistor. I am however confused why the LM317 does not act like a current limiter if you wire it up as a voltage regulator.

It doesn't act as a current limiter (well, except to protect itself when current exceeds its limits, or it gets too hot) because it's a voltage regulator. Which do you want? If you want both, then something other than a simple LM317 will be needed.

The main current to the load flows through the LM317 from its input terminal to its output terminal. The Adjust terminal works in conjunction with the resistors to set the desired output voltage. The currents involved in doing this are relatively small and have nothing really to do with the output current to the load. Does that help?

The 317 can also be configured as a current regulator (for relatively Constant Currents BELOW 1A). And,, you canconfigure one 317 as a voltage regulator (for desired Vout+1.25V) and follow it with a second 317 configured as a current limiter. It works. It kills batteries. It wastes parts. I've done it many times and with regret, will probablly continue to do so...

Yes that does help thanx. What I am trying to make is a voltage regulated constant current source.

I thought I would use 2 LM317's wired up after eachother. The first one would be a voltage regulator feeding into a constant current source. But I thought if the voltage regulator used a 240ohm for R1 then the second LM317 would never be able to provide more than 0.005A because the first LM317 was limiting the current as well. But I am wrong and I do not understand how this LM317 works

Good luck. we learn more by out mistakes. Think about having another go at the data sheet. it is pretty explicit.

Work out the math. You can't have a voltage regulated (constant voltage) constant current source. What is your load?

calvingloster:
What I am trying to make is a voltage regulated constant current source.

Like KeithRB says, there is no such animal. Understand what a constant current source is: By varying the voltage, it tries to always keep the current through the load constant, even as the load varies.

Say that I had a one-amp constant current source. If I just hook up a simple 10? resistor as a load, it will push one amp through the resistor, by increasing the voltage to 10V. Ohm's law tells us V = IR = 1 amp * 10 ? = 10V.

Now remove the 10? resistor and connect a 50? resistor. The constant current source will still push 1A through it, now increasing the voltage to V = IR = 1 amp * 50? = 50V. A 2? resistor will only require 2V to push 1A through it.

Constant current sources are good for certain applications, like high-power LEDs, that like their current to be, well, constant.

So yes please clue us in on what this power supply will be powering.

A SLA battery charger. See his other thread.

Unfortunately, someone has distressingly suggested using another LM317 in series as a current control, and in his proposed design, he put this after the voltage regulator, rendering it completely unusable.

I had a half-hearted attempt at contributing to that thread, but gave up because it was simply impossible on so many counts! :astonished:

His original proposal was to charge a 12V SLA from 16V which is perfectly, easy, but requires a completely different component set and design.

Perhaps someone might point him to a properly designed version of same (as against a random "instructable" or "makey") and he can learn the principles over time. :smiley:

Paul__B:
A SLA battery charger. See his other thread.

Aha, thanks for that. I do strive to know what I don't know, and I can say that I am pretty much ignorant of battery charging algorithms. In general maybe there is a constant current phase for most of the charge but then this tapers off and/or switches to a trickle mode to keep things topped off, but details matter as does battery chemistry. Interesting stuff actually, I should educate myself a bit sometime. So I might look for an IC built for the purpose, although I always understand the attraction of rolling your own.

That said, I'll probably sit the rest of this one out, other than reiterating that before actually attempting to charge any batteries, the OP should Google up some tutorials and ensure they have a very solid understanding of Ohm's Law and also power relationships. Burning up a resistor may smell bad but failure of a lead acid battery could be quite a bit less enjoyable.

Ok so hear me out now. I do believe it is possible to create a constant voltage constant current source. The first LM317 limits volts to 14.1v and the second one limits current to let's say 100mA. When the "device" requires 50v to draw 100mA then the constant voltage will only allow it to draw as much current as 14.1v would allow according to ohms law. So as the "device" changes it's properties like resistance, the 2 LM317's will NEVER allow it to draw more than 100mA or be subjected to more than 14.1v. And I do believe a lead acid battery could be charged this way. It might take long but will still work. Does that not make sense?

calvingloster:
Ok so hear me out now. I do believe it is possible to create a constant voltage constant current source. The first LM317 limits volts to 14.1v and the second one limits current to let's say 100mA. When the "device" requires 50v to draw 100mA then the constant voltage will only allow it to draw as much current as 14.1v would allow according to ohms law. So as the "device" changes it's properties like resistance, the 2 LM317's will NEVER allow it to draw more than 100mA or be subjected to more than 14.1v. And I do believe a lead acid battery could be charged this way. It might take long but will still work. Does that not make sense?

Not much, I'm afraid. Sounds more like a current-limiting situation. Do the maths. V=IR. If V and I are constants, so too must R be constant. So we can only have voltage and current constant with one specific resistance, i.e. the load resistance cannot change.

But don't let me discourage you. I am perfectly willing to be convinced. Design the circuit, build it, test it, then bring us the schematic and the V/I curves so that we can understand it and duplicate the results.

I am perfectly willing to be convinced.

Well I am not.
This is a classic beginners mistake thinking you can have a constant current and constant voltage at the same time.

What the OP is describing is a current limiting supply, the sort of thing you get in bench power supplies. The voltage is constant up to a certain current, then as that current is reached any further reduction of the load to try and increase the current results in a reduction in output voltage.
You can do all that with one regulator and a bit of feedback from a current sensor.

calvingloster:
I have set up a LM317 to output 13.7v. From the output to adjust I have used a 5ohm resistor and then my R2 which is from the R1 to ground I have used a 50ohm resistor. When I put a 16v power supply on the 50ohm resistor gets smoking hot!!! Like I can even smell it! Why is this?

Read this post: http://forum.arduino.cc/index.php?topic=240371.msg1725782#msg1725782

(the LM-317 data sheet may be a help too......) :slight_smile:

calvingloster:
Does that not make sense?

Yes, but from the device's point of view it won't be a constant voltage supply. The voltage the device sees will vary (up to the maximum allowed).

PS: You can get pre-built, adjustable devices for doing this:

eg. http://www.ebay.com/itm/111323886108

You can even get them with three adjusters so they turn on a LED and shut down when they reach a certain voltage:

eg. http://www.ebay.com/itm/371059092051

This can be done with two LM317, but the current limiter must come first. Essentially what you are doing is building a voltage regulator, then powering it with a current regulator. The current regulator drops the voltage to the voltage regulator if the current reaches the setpoint, and so the voltage regulator's output drops because it just isn't getting enough voltage.

It can be done, but each LM317 requires a relatively large headroom. You need about 5V headroom for each one, so for 14.1V maximum output you'll need nearly 25V. Rather inefficient.

For the current regulator, all the output current flows through the resistor between Output and Adjust, so it must be a higher wattage. At 100mA output, 1.25Vx100mA = 125mW. You always at least double the rating for resistors, so a 1/2W resistor will work. But for the voltage regulator,none of the output current flows through the voltage divider that sets the output voltage, so they can be low wattage resistors.

The LM317 each require a heat sink.

However, as has been pointed out:

  1. There are better ways to do this
  2. The price for failure can be high (dead or exploding batteries)
  3. The OP really needs to read up on the basics: Ohm's law, Kirchoff's current and voltage laws, the voltage/resistance/current/power formulas, etc.