Go Down

Topic: LM317 voltage regulator R2 is getting very hot? why? (Read 11034 times) previous topic - next topic

calvingloster

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

123Splat

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

KeithRB

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

JChristensen


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.

Paul__B


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!   :smiley-eek:

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.   :D

JChristensen


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.

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?

JChristensen


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.

Grumpy_Mike

Quote
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.

krupski


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......)   :)

Gentlemen may prefer Blondes, but Real Men prefer Redheads!

fungus


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).

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

fungus

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
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

polymorph

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.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8
Multitasking: forum.arduino.cc/index.php?topic=223286.0
gammon.com.au/blink - gammon.com.au/serial - gammon.com.au/interrupts

calvingloster

I find this very frustrating and hard to understand. Can I ask that someone answers the questions I am about to ask in a yes no fashion.


1) the Bulk charge of a lead acid battery is done by a constant current charge. For simplicity sake let's say we want to charge our battery with 100mA. Now in order for us to force 100mA into the battery we require a voltage of 20v. So now our power supply only puts out 16v. So we will not be able to push 100mA into the battery, BUT!!!! BUT BUT BUT!!! We will still be push some amount which is LESS THAN 100mA!!! So essentially our BULK charge will take longer than usual BUT it will still work and will NOT damage the battery. Yes or no?

2)Now an LM317 CAN!!!!! CAN CAN CAN!!!!!! Provide the top explained power supply. As a constant current source? YES or NO?


3) for the float charge we charge the battery with a constant voltage source? At 14.1v. Now the LM317 can also provide us with such a power supply!!!!!! Yes or no?


The LM317 can supply us with a constant voltage and a constant current source. I hear u guys are saying it cannot but I'm sure it can because all the first LM317 is doing is limiting voltage to the second one. So the second LM317 never see's any voltage higher than say for example 14.1v. Then the second LM317 regulates current to say 100mA. So the device can never see voltage above 17volts and can never see current above 10mA.


Even if I use one LM317 set up as a voltage regulator regulated to 14.1volts I will be able to charge the battery until it's full!!!!! This will obviously take much longer but it does not damage the battery at all and if I leave it over night then awesome!!!!!


And I know that efficiency isn't going to be great but these are the only components I have!!! I can't just buy stuff like you guys recommend. I have what I have and I want to use it to make what I need. Even if it isn't the most efficient thing on earth.

Grumpy_Mike

Quote
Can I ask that someone answers the questions I am about to ask in a yes no fashion.

No.

Go Up