LM317 circuit doesn't supply enough current (only 20mA)???

[mac72]

Greetings! My first post...
Sorry I'm such a newb, there may be a really simple answer to this:

I made a voltage regulator using an LM317 and the circuit design at the top of this page: http://www.uoguelph.ca/~antoon/circ/ps-lm317.html.

It works perfectly, except that it only passes about 20mA, and I need at least 80, preferably 90-95.

I've tried to read up on what I'm doing, as this is a newly found hobby, but I'm still not clear on lots of things.

Any ideas? Thanks in advance,
Mike

 

[Dazza]

Hi mac72 and welcome ,
you shouldn't have any problem supplying 0.95A with the LM317 variable voltage regulator. what are you using as a power source to supply the LM317.

 

[audioguru2]

Hi Mike,
What happens to your LM317's output voltage when you try to get more than 20mA from it?
Does it just drop?
Or do you get lots of hum and ripple?

 

[mac72]

Dazza:

For testing, I'm using a variable voltage wall wart that puts out a couple of amps. I've got a resistor on it to hold the current at 250mA for this project. Ultimately, I'd like to use a 12V camcorder battery.

audioguru:

I can't get more than 20mA from it. I've hooked up various light bulbs to it, and the current draw is never more than 20mA. [edit] Sorry, the voltage is just as I want it to be, around 9v, and it doesn't change. It seems like I got the voltage part right, but the current draw is killing me![/edit]



I'm hoping I can get to the bottom of this. I'm going to recheck my circuit and make sure I didn't screw anything up.

 

[surajbarkale]

Use ohm's law to check if you can get 100mA current you will need to connect a 0.9Ohm resistor accross output. I think by connecting one 1Ohm 1/2 Watt resistor at o/p you will be able to measure it.

 

[mac72]

Suraj:
Thanks, but I may need a more in-depth explanation of what you said. I'm unclear as to where I'd be connecting and what I'd be measuring.

Thanks,
Mike

 

[audioguru2]

Hi Mike,
Get rid of your series current-limiting resistor. It is messing things up since it can supply 250mA only to a short, but not enough current at 12V for the regulator to operate properly. The regulator needs at least a 12V input to produce a good regulated 9V output.
The resistor isn't needed because the regulator has current limiting and thermal shutdown to protect itself.

Hi Suraj,
Please check your math. 9V into a 1 ohm resistor will draw 9A! That's 81W for the poor little resistor.
The LM317 would current-limit long before that much current and go into thermal shutdown.

 

[mac72]

audioguru:
I removed the current-limiting resistor, but I'm still getting the same current readings out of the voltage regulator. It's putting out 9v, but it doesn't seem to be supplying very much current.

I've checked all my paths and connections, and it seems right. The potentiometer works, and I can vary the voltage between 6v and 9v, but no matter what the setting the amperage allowed through is the same: not enough.

Thanks,
Mike

 

[audioguru2]

Hi Mike,
The 5K ohms value of the pot in that circuit is much too high for a 9V output. The total of the pot plus the 560 ohm resistor should be 1364 ohms, so the pot should be only 804 ohms for a 9V output. Maybe a 1K pot would be much better and be able to use more of its range to adjust the output voltage.

If you can adjust the output voltage down to 6V instead of a calculated 4.4V, something else is wrong. What is its lowest output voltage with a load? Maybe its output is "floating" up to 6V without a load.

You should also be able to adjust its output voltage higher than only 9V, if your input voltage is 12V or more.

If the output voltage remains at 9V when you connect a load that draws more than 20mA, the current through the load must be more than 20mA. How are you measuring the load current?

 

[mac72]

audioguru:
I used the 5k pot because that's what was listed on the circuit design I borrowed. That's what I get for not doing my own math, right?

Anyway, my measurements are taken as follows:

[V+] === |------------------------| === [ammeter] === | (+) |
|regulator circuit board | | test |
| | | lamp |
[V-] === |________________________| ================= | (-) |



Thanks,
Mike

 

[audioguru2]

Hi Mike,
You shouldn't have to do the math, the author of the circuit should have done it, right? ;D
You are measuring the load current correctly, as long as the resistance of your ammeter is much less than 470 ohms.

1) What is your minimum output voltage with and without a load?
2) What is your maximum output voltage with and without a load?
3) What is the voltage of the LM317's input when its output is fully loaded?

 

[CRE]

Hi everyone,

I was just curious, mac72, what type of LM317 have you got (I'm referring to the package)? If you don't mind the brand and suffix would help as well.

Thanks.

 

[audioguru2]

Hi CRE,
Good point! ;D
But even those tiny surface-mount LM317's are good for 500mA, because they have a heatsink tab!
I have used little 78xx regulators in a small TO-92 package and they are good for about 100mA.

It doesn't make sense to measure a current of only 20mA through a load that should draw much more and the regulator's output voltage stays the same without dropping.

 

[CRE]

Well, that's kinda where my thoughts were heading... As I've said before I'm a newbie with only enough knowledge to be dangerous. ;D But I recalled that they weren't all rated for 1.5A. looking at National's datasheet packages TO-39, LCC and TO-252 are only rated for .5A. But I was just curious to see what make and look if the pinout might vary, or if it has thermal protection. I don't expect this to help you guys, I just wanted to look at it from another angle for a sec.

 

[audioguru2]

Hi CRE,
It doesn't matter if you are a newbie or an oldtimer, all ideas are welcome here, and you've had some good ones.

You can't get dangerous with an LM317 because they are "newbie-proof"!
1) Current-limiting. If the output gets shorted or a load goes wrong and tries to draw too much current, an LM317 will limit the current to at least 0.5A for the small ones and 1.5A for the big ones. Current limiting is for protection so is not very accurate. The big one is guaranteed to provide 1.5A or more and some could give 3.4A. The current-limiting point changes a little with temperature.

The LM317 also protects itself by reducing its output current when the voltage across it exceeds 15V. A big one that is guaranteed to provide at least 1.5A with up to 15V across it will limit the current to as low as 150mA with 40V across it.

2) Thermal shutdown. If an LM317 gets too hot (between 125 and 150 degrees C) it will turn-off. When it cools, it will turn-on. This feature is for protection. I wouldn't use it to flash a light because the continuous thermal stress would probably shorten its life.

 

[MP1]

Perhaps I missed this in the posts, but I do not see where the input current was measured.

The regulator will not give you current that is not available ahead of it. Measure the current ahead of the regulator. If you only have 20 mA available ahead of the regulator, you are not going to get more current at the output of the regulator.

MP

 

[audioguru2]

Hi MP,
Measuring the current ahead of the regulator won't test anything. Of course it will equal the load current, plus a small amount to operate the regulator. A 2A (available) power source will provide only 20mA of current into a 20mA load.

Dazza: (asked,"what are you using as a power source to supply the LM317?")

For testing, I'm using a variable voltage wall wart that puts out a couple of amps. I've got a resistor on it to hold the current at 250mA for this project. Ultimately, I'd like to use a 12V camcorder battery.

audioguru:

I can't get more than 20mA from it. I've hooked up various light bulbs to it, and the current draw is never more than 20mA. [edit] Sorry, the voltage is just as I want it to be, around 9v, and it doesn't change. It seems like I got the voltage part right, but the current draw is killing me![/edit]

I asked him to remove the current-limiting resistor. He sketched a proper connection of a current meter.
It just doesn't make sense. A current limiter, whether it is the power source or the LM317, would limit the current by reducing the voltage, but the voltage remains "around 9v, and it doesn't change".

Maybe he is not measuring the voltage while it is loaded.
He is probably using his only meter to measure the 20mA current, when the power source or LM317's voltage is dropping but he isn't measuring the voltage at this time. Then he moves the meter to measure the voltage without re-connecting the load. That explains it.
Now what is limiting the current by reducing the voltage under load?

 

[MP1]

Audioguru,
If he only has 20 mA available before the regulator, he is not going to get 90 mA at the output of the regulator. I do not understand your comment that measuring the current ahead of the regulator will not tell him anything. It will tell him a lot. For example, when he is not able to measure 2 amps@ 12V like he thought he had.

The point being: If he has only a few milliamps available to the input pin of this regulator for a specific voltage, then he will never get more than that out of it.

It is my guess that his wall wart is not capable of what he wants. It might even have some current limiting capabilities in it. But this is only a guess since I cannot make any measurements or see his schematic. Personally, I would cut open the wall wart and see what is inside of it. "A variable voltage wall wart, did he say"? He does not need any of the parts inside this unit except the transformer if he is going to put it into the LM317. All other parts are just in the way of achieving what he wants.

MP

 

[audioguru2]

MP,
If the wall-wart is capable of supplying only 20mA (plus some more to operate the regulator) at 12V, then its voltage will drop if you try to get more current won't it? You don't need to test the wall-wart separately with its own current-test resistor, the regulator and its load are its current-test resistor. If it can't provide the current, its output voltage will drop.
But Mike says that his regulator's output voltage remains at 9V with changing loads which doesn't make sense. If it can't provide the current, its output voltage will also drop.
I think that either the wall-wart's output voltage is dropping when the regulator is loaded, or the regulator's output voltage is dropping when loaded, but Mike is measuring those voltages in error only while it is unloaded.

Hello Mike,
Did you measure the voltage from the wall-wart and from the regulator while the regulator was loaded with something that should draw more than 20mA?
Where is Mike anyway? He hasn't replied for nearly 2 weeks.