# LM317 not getting results I expected

Thread Starter

#### mopar

Joined Oct 21, 2015
16
I have a device that is usually powered by soldered on 1.2v 80maH NiMH coin cell. The device is difficult to take apart and replace the battery, so I want to modify it to receive power from an external source. The source is 6v which I've been told the device will operate on, but nobody has done long term testing to see if the device will burn out eventually, so I want to drop the incoming voltage to about 1.5v to be safe. On an 80maH battery the device can stay running for over 12hrs so its current draw is quite small. I've done some testing with an LM317 with an R1 of 220ohms, and an R2 of 47ohms (which should get me 1.5v roughly) and with my multimeter with or without the device connected I'm seeing 4.5v. My understanding is that the LM317 requires a 10maH load to work correctly, so apparently my device isn't creating enough of a load.

My question is, how do I create a large enough load to get the correct voltage?

I read somewhere that if I divide my resistor values by 10 and use those new values it will drop the voltage to what I need without a load. Tried this and the voltage does come close at 1.8v, but it doesn't power my device.

This device is less than 3/4" square and quite lite and I'd like to keep the package small because it will be held in place by velcro. The LM317 and the 2 1/4 watt resistors weight much less than the battery itself and current housing (that I wont be using) gives me some weight to work with (1 or 2 ounces maybe) so I can add some weight I just don't want to add anything to bulky.

Not great at this kind of stuff so any help is appreciated.

#### atferrari

Joined Jan 6, 2004
4,279
I presume that you could (should) use the small version of the LM317 able to cope with 100 mA load, maximum. Its case is TO92.

Can you implement it in a perfboard (or a protoboard) until you see it is providing the minimum, 1,25 V?

Once you have it working you could start thinking of reducing the size of the whole thing. You will be surprised of how small a circuit can be built with careful design. As I use to say in Spanish "más paciencia que ciencia".

Edit to add:

Have you got the datasheet? In other words, are you sure about the pins distribution?

Can you show the actual circuit you are tying to implement?

/Edit

#### GopherT

Joined Nov 23, 2012
8,012
Double-check the pinout of the LM317. The flow is not logical.

Use a 120 ohm from out to adj
and 22 ohm from adj to ground.

#### Dodgydave

Joined Jun 22, 2012
9,938
You need 10mA minimum in the adjust pin, so as said before use a 120 ohms for R1, and 24 ohms for R2,(nearest value is 22ohms) that should give you 1.5V,

Thread Starter

#### mopar

Joined Oct 21, 2015
16
Just found out some more info on this device. Apparently if the device goes into what I guess you would call sleep mode (ie; when the battery gets below minimum voltage or the device becomes disconnected from the battery) the device needs to see 3v before it will "wake up" and function correctly. So that may explain why I couldn't get the device to work when I divided my resistor values by 10 to get the voltage to 1.8 with no load.

With that in mind, is there a way to get a momentary burst of at least 3v to the device to make it up before regulating the power back down to 1.5v?

A little more info on these devices. When they aren't being used they are placed in a charging rack to charge the internal battery so the battery won't go below cutoff voltage and the device won't go into sleep mode. The external power source I'm using to power these gets turned on and off repeatedly so the device will always be going into sleep/shutdown mode hence the reason to find a way to send a short 3v signal to the device

#### mcgyvr

Joined Oct 15, 2009
5,394
So with the existing battery the devices NEVER goes into "sleep mode"?
What happens if the battery runs out of charge? What "wakes" it up again..

What about just replacing the soldered on battery with one that can be easily replaced when its reached its end of life?
Then nothing changes other than the initial problem is now solved?

or what about just a momentary push button tied right to the 6v source to where ever you need this "pulse" of min 3V

personally I think replacing the soldered battery with a wired pigtail/quick connect is the best solution though..

#### Dodgydave

Joined Jun 22, 2012
9,938
To get 3V out, you need to replace R2 for a 180 ohms, you could use a changeover push switch, or a transistor to switch in the resistor.

Thread Starter

#### mopar

Joined Oct 21, 2015
16
The 220 and 47 ohm resistors I was using were the results I got from several different online calculators to get me approx. to 1.5v. I will try the values you both mentioned for R1 and R2 tonight.

I assume that when cutting the resistor values roughly in half is what will create the load needed to drop the voltage? I'm not great with electronics, but I do understand that resistors create a load and that their resistance value determines how much of a load they create, I'm just not clear on how resistors with half the value will create more load. Guess I'm in for more reading.

Thanks for the help guys.

Thread Starter

#### mopar

Joined Oct 21, 2015
16
So with the existing battery the devices NEVER goes into "sleep mode"?
What happens if the battery runs out of charge? What "wakes" it up again..

What about just replacing the soldered on battery with one that can be easily replaced when its reached its end of life?
Then nothing changes other than the initial problem is now solved?

or what about just a momentary push button tied right to the 6v source to where ever you need this "pulse" of min 3V

personally I think replacing the soldered battery with a wired pigtail/quick connect is the best solution though..
Correct, as long as the battery doesn't die before the device is placed back on the charging rack it never goes into sleep/shutdown mode.

When I inspected the charging rack a while ago I noticed the voltage was at 3v, so I assumed that when the device is placed back in the rack the charging circuit senses the device and drops the voltage back to 1.5v to charge the battery correctly (probably uses an LM317, but I've never taken it apart). With this new info of needing 3v to wake up it explains why the rack is a 3v and not just the 1.5v needed to charge the battery. 1st it wakes the device up if needed, then it starts charging it, so I would say that 3v continuous probably wont hurt the device.

It's the disassembling of the device to replace the battery that is difficult, and the fact that the correct batteries are sometimes hard to get and the extras I keep are hassle to cycle so the battery stays healthy. Another reason for doing this is that the packaging is bulky and is held in place by a pin that breaks off. I've had to replace the pins on all of the housings and even using a strong epoxy to hold the replacement pin on the pins still break off so I want to get away from the housings altogether.

If I can find a small momentary switch and I can find a way to attach it to the device I think that may work.

I've tried to upload a pic, but for some reason the file browser doesn't show my file. I'll keep trying though as it may help show why I'm trying to do what I'm doing.

Thanks for the help.

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
I'm just not clear on how resistors with half the value will create more load. Guess I'm in for more reading.
This all relates back to Ohm's Law, V = I * R. The load you are talking about is a current load. So, using Ohm's Law again, I = V / R, a smaller resistor gives you a larger current. The larger current at the specified voltage will give you a larger power, P = V * I.

Thread Starter

#### mopar

Joined Oct 21, 2015
16
Yeah, I should have known that. Some of this is starting to come back to me.

How would I use a transistor to switch in the second resistor? My only experience with transistors is a photo transistors that I used to measure the rpm of a shaft with an Arduino a year ago. I'm aware that you can use them as a switch to complete a circuit if you have enough voltage (or lack of if on the low side), but my understanding is that as soon as I connect the external power source to the LM317 I will only have the 1.5v to work with so I don't know how that will give me a 3-6v burst to wake the device up.

Thanks guys.

#### Dodgydave

Joined Jun 22, 2012
9,938
If you put the 22 ohms and the 180 ohms in series, with the 180 at ground and the 22 on the adj pin, the transistor can go across the 180 C/E and biased on so the Lm317 will give out 1.5V, then a momentary push switch N.O across the base emitter will switch off the transistor and put the 180 in circuit to raise the voltage to 3V,

OR you can use a normally closed switch across the 180 ohms and this will do the same, when pressed.

#### ian field

Joined Oct 27, 2012
6,539
I have a device that is usually powered by soldered on 1.2v 80maH NiMH coin cell. The device is difficult to take apart and replace the battery, so I want to modify it to receive power from an external source. The source is 6v which I've been told the device will operate on, but nobody has done long term testing to see if the device will burn out eventually, so I want to drop the incoming voltage to about 1.5v to be safe. On an 80maH battery the device can stay running for over 12hrs so its current draw is quite small. I've done some testing with an LM317 with an R1 of 220ohms, and an R2 of 47ohms (which should get me 1.5v roughly) and with my multimeter with or without the device connected I'm seeing 4.5v. My understanding is that the LM317 requires a 10maH load to work correctly, so apparently my device isn't creating enough of a load.

My question is, how do I create a large enough load to get the correct voltage?

I read somewhere that if I divide my resistor values by 10 and use those new values it will drop the voltage to what I need without a load. Tried this and the voltage does come close at 1.8v, but it doesn't power my device.

This device is less than 3/4" square and quite lite and I'd like to keep the package small because it will be held in place by velcro. The LM317 and the 2 1/4 watt resistors weight much less than the battery itself and current housing (that I wont be using) gives me some weight to work with (1 or 2 ounces maybe) so I can add some weight I just don't want to add anything to bulky.

Not great at this kind of stuff so any help is appreciated.
The LM317 is basically just a 1.25V 3-terminal regulator, if you ground the Vset pin it will output 1.25V.

You can actually use the fixed voltage 3-terminal regulators in the same way as the normal application of the 317. Of course a 7805 will have a minimum adjustment range of down to 5V, but some manufacturers do occasionally use the fixed regulators with the GND pin returned to a voltage divider across the output.
An example springs to mind of a 2M candlepower halogen hand lamp, in the charger a 7805 with GND returned to a sample of the output voltage was made just right for charging the 6V SLA battery.

Thread Starter

#### mopar

Joined Oct 21, 2015
16
Ahhh, that makes some sense dd. I have several of these to do so for right now unless I can find a way to burst 3v when external supply is powered without user intervention I'm going to set it up for 3v's and see what happens. Thanks for the information.

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
You can actually use the fixed voltage 3-terminal regulators in the same way as the normal application of the 317.
Have never thought about it but it makes sense. Pretty interesting. I love these kinds of tips.

Thread Starter

#### mopar

Joined Oct 21, 2015
16
Thanks to everybodies suggestions, using a breadboard I was able to get the 3v I was looking for and the device worked correctly. But now that I soldered the LM317 and 2 resistors to the device it's not working. I double checked to make sure I have the right resistors on the right legs and I show 6v across the Vin post and my negative wire, but I get nothing on the Vout leg.

Maybe it's how I attached everything to the device so please correct me if something is wrong.

I soldered the Vout leg of the regulator directly to the Vin side of the device (same pin the + side of battery was soldered to).
I soldered a piece of wire from the positive side of my 6v supply to the Vin leg of the regulator.
I have the R1 resistor soldered to the Vout and Adj leg o the regulator.
I have the R2 resistor soldered to the Adj leg of regulator and to the negative pin on the device.
I have the negative wire from my 6v supply soldered the the negative side of the device.

Even if resistors are flipped I should still see some kind of voltage. The iron I used was just hot enough to melt the solder so I doubt I did any thermal damage to the regulator (but it is possible I suppose).
I did shorten the legs on regulator and I cut the tab off the top flush to the housing to make it more compact, but that shouldn't affect anything should it?

Any ideas would be appreciated.

#### Dodgydave

Joined Jun 22, 2012
9,938
The tab is also the output, make sure you have it isolated from the case, can you post pictures of your project.

#### ian field

Joined Oct 27, 2012
6,539
Thanks to everybodies suggestions, using a breadboard I was able to get the 3v I was looking for and the device worked correctly. But now that I soldered the LM317 and 2 resistors to the device it's not working. I double checked to make sure I have the right resistors on the right legs and I show 6v across the Vin post and my negative wire, but I get nothing on the Vout leg.

Maybe it's how I attached everything to the device so please correct me if something is wrong.

I soldered the Vout leg of the regulator directly to the Vin side of the device (same pin the + side of battery was soldered to).
I soldered a piece of wire from the positive side of my 6v supply to the Vin leg of the regulator.
I have the R1 resistor soldered to the Vout and Adj leg o the regulator.
I have the R2 resistor soldered to the Adj leg of regulator and to the negative pin on the device.
I have the negative wire from my 6v supply soldered the the negative side of the device.

Even if resistors are flipped I should still see some kind of voltage. The iron I used was just hot enough to melt the solder so I doubt I did any thermal damage to the regulator (but it is possible I suppose).
I did shorten the legs on regulator and I cut the tab off the top flush to the housing to make it more compact, but that shouldn't affect anything should it?

Any ideas would be appreciated.
You're probably OK on voltage headroom, but the 317 needs about 2V to spare for it to work. Someone told me the negative 337 has the same series pass structure as a LDO regulator and only needs about 1/2V headroom - but I haven't checked it out for myself.

A soldering iron that only just does the job means you take longer to make the joint, that opens the possibility of cooking components. A good temperature controlled iron lets you complete the joints quickly and cleanly without dwelling on it long enough to fry components.

Thread Starter

#### mopar

Joined Oct 21, 2015
16
Dave, I cut the tab because I knew it was the output and didn't want any chance of hitting anything on the board. As a temporary spacer I had a piece of foam to support the regulator while I soldered it.

My wording of my iron settings isn't quite correct. What I meant was that my iron was hot enough to melt the solder as soon as it made contact with the solder. Sorry about that.

I tried to get a pic with my camera, but it was to grainy to see anything. I'll try again with my tablet which has a better camera.

On the breadboard I was showing about 3.15v when everything was connected temporary including the device. All I did was transfer the regulator and resistors to the device. Gonna take it apart and start over tonight unless I figure out the problem.

#### MikeML

Joined Oct 2, 2009
5,444
As AudioGuru would tell you if he were still here, a lightly-loaded LM317 will have a higher output voltage than spec'ed unless R1 (in my schematic) is less than 125Ω to meet the min. 10mA load current through the LM317... RTFDS!

Here is a plot of V(out) and I(R1) vs the value of R2.

Last edited:
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