How to Configure the LM317 Voltage Regulator
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- In this video we give you an overview of the LM317 Voltage Regulator and how to configure it. The LM317 is popular voltage regulator among makers because of its low cost and its adjustable output voltage. You can purchase the LM317 at forcetronics.com
Thank you Sir😊
LM317 is a linear voltage regulator. It has the advantages of low output noise and easy configuration.
Yes, you are right
Hi, thanks for the tutorial. One thing you want to show is how the regulator actually regulates (or not) the output voltage regardless of the load connected to it. It supposed to maintain (regulate) the voltage with or without load and any current in between. I have tested a LM317 and was disapointed with its performance, perhaps I used a damaged component.
Hey thanks for the input and insight into your experience with the LM317
ForceTronics haircut at
Your explanation is awesome, good job.
Great video man, I've been looking for something like this. Most videos are just "lets put together this quick Lm317 circuit!" I like the use of the data sheet. I think it's very important if you are going in to electronics to be able to decipher data sheets. Some of the are literally like reading Chinese! (that was a 1/2 joke :] )
th-cam.com/video/3ZMfVpUieFc/w-d-xo.html
thank you. may the Force be with you
4.35 : the input voltage should never be much higher then the output voltage. This will cause the LM317 to generate excessive amounts of heat
Thank you sir. Quite useful!
Thanks for this great video! I just wanted to add that these get hotter as you either draw more current or drop more voltage. Functionally, they regulate voltage by dissipating extra power as heat, EXACTLY like a series resistor would. As such, they are inefficient and are not a good choice for battery powered projects. Also, since the heat sink tab is connected to Vout, DO NOT bolt it to anything grounded (metal case or PCB pad) or you'll get a short circuit. They do shut down to try to save themselves from overheating, but don't count on this. :)
Great video very well explained
One thing to remember it if you want to use this to build a battery charger at 12 volts or above the variable pot will be destroyed while you are testing or using it.
Joseph Nicholas true. my pots kept burning while testing over 12 volts regardless of the power rating. but why? and how to overcome this problem?
You need to use a higher value resistor on the pot.
Question- I have a heavier 24v ac transformer and a high voltage UV lamp that requires a constant AC regulated 24v 1A on its input side. Would this lm317 chip work?
The LM317 is a DC voltage regulator. It does not work with AC voltage
Great job.
Well explained thanks
In equation Why you have taken 1.25V as 1.25? Does this V value always remain 1?
V stands for Voltage. V is not an unknown variable
Very nice video and teacher ! does the LM317 adjusts current or voltage ? if i want to draw more than 1.5Amp, how to do ? thanks.
My video configures the LM317 as a constant voltage power regulator. It's output current is set by load. Lower resistance load more current. I would suggest reviewing Ohm's Law
@@ForceTronics Ok, did you make a video of Ohm's Law ?
You can put a transistor in parallel with the lm317 using a resistor to turn on the current at a certain level from the transistor
what is the 100 ohm resistor on the output for? You talked about it but it is not in the circuit diagram...
Thanks
The 100 ohm resistor is just there for a dummy load so that current is flowing out of the LM317 and I show that it is maintaining the voltage we set it for. It basically represents the thing you are trying to power. Probably could have used a light bulb or a fan for something a little more exciting.
Thank you. How would you limit the current to say, 50 mA with this set-up?
Well it depends what you mean by "limit." If you mean maintain a steady current for a static load like an LED then you can use a series resistor and ohms law (you know the voltage and the current so calculate the resistor value). The data sheet also provides some constant current suggestions www.ti.com/lit/ds/symlink/lm317.pdf, refer to 8.3.3 or 8.3.7. If you mean prevent the current from going over 50mA you would want to either use a fuse or see if you can find a regulator that allows you to set a current limit.
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Can you give the name of any books from which I can study about IC's......???
Well what type of ICs do you want to learn about? If it is power ICs like linear regulators and DC to DC converters then you probably want to look for a power electronics book. I don't have any to recommend because the books I used are pretty old by now, but I am sure Google can help you find a book
SchemecuLm312
you should give the link for that data sheet and do you have a site you prefer for data sheets? search engines i use typically does n't have the old school components. i found the LT117A. works even better output voltage tolerance is a maximum of ±1%, allowing an overall power supply tolerance to be better than 3% using inexpensive 1% resistors. and same price
For data sheets I either go on manufacturer website or to a distributor site like Mouser and search for the exact part to access the datasheet from there
Great video. I am however math illiterate and would love it if you could do a run through of how to do the R2 calculation. Sorry if it is basic but I've forgotten how one does sums with brackets in them.
I'm not great with math either so sometimes I use Excel to help. The original formula is Vout = 1.25V x (1 + R2/R1) (we drop the Iadj * R2 because its value is negligible). Since we already know what Vout we want (in this case, 3.3V) and we know what R1 is (in this case 330ohm) we have to rearrange the formula to solve for the only thing we don't know, which is R2.
We do this by first stripping off the 1.25V on the right hand side of the equation by dividing both sides by 1.25V to give us
Vout/1.25V = 1+ (R2/R1)
Next, we strip off the +1 on the right hand side by subtracting 1 from both sides giving us
(Vout/1.25V) -1 = R2/R1
Finally, we break off the R1 divisor on the right hand side by multiplying both sides by R1 to give us
R1 * ((Vout/1.25V) - 1) = R2
Now we can solve for the missing value R2 in Excel. In all of the above except the last form, the parenthesis aren't strictly necessary because the rules of math say we always do what's in parenthesis first, then multiplication then division before addition then subtraction. I've included them so you know you always operate on what's inside the parenthesis first. So in that final equation you first divide Vout by 1.25, then you subtract 1, THEN you multiply by R1. Now on to your spreadsheet.
1. Create a spreadsheet with 3 columns labeled Vout, R1, and R2 in cells A1, B1, and C1
2. In cell A2, put in your desired output Vout (3.30 in this case)
3. In cell B2, put in your known R1 value from the demo (330 in this case)
4. In cell C2, put in your formula which calculates R2. In this case, the formula is =B2*((A2/1.25) - 1)
4a. That means: first, divide your desired Vout by 1.25, then subtract 1 from that answer. Then, multiply it by your known R1 value. The result is R2.
Now you can play around with the different values of desired Vout or R1s to come up with different R2s needed for that output. Hope that helps.
Thanks for this video - I needed to see how you laid it out on the breadboard to make sense of the Vin and Vout -- it's confusing when the Vin is on the left pin 3 --
Very Very useful video and clear explanation for the equasion.
This is the first Circuit I want to make however...I'm new to circuits and electricity in general. please somebody for the love of god explain the path of the current here that makes these resistors and reference voltage actually change the consistent output. I don't understand how a reference voltage is being created and how it's actually changing the output. For example. 12v comes in Vin to LM317. Does the path go out Vout or Adjust first or at the same time. Is it that r1 and R2 are drawing current so it comes at it both ways then once it goes through what I presume is the potentiometer in R2 it goes up the adjust terminal and out the Vout? Can current flow against itself? I'm beyond lost here.
The lm317 voltage regulator regulates its output to maintain a 1.25 v at the adjust pin. You want to set resistors r1 and r2 so that you get the desired voltage at the output and a 1.25 voltage at the adjust pin. The regulator is not monitoring it's output voltage it's monitoring the 1.25 voltage that you set with the resistor divider Network. The 1.25 volts is created by the resistor divider Network and the current flowing from the output to ground through the resistor divider Network. The adjustment current can essentially be ignored when you're using the lm317 as a voltage regulator and not a current regulator.
Excellent details!
Good video and very helpful. Thanks much.
👍hi
THANK YOU
awesome tutorial
How high of an input voltage will it accept? I have 120v dc will that work?
No. The datasheet will give you it's input voltage. I don't have exact range memorized
Sir I have query on the circuit.i had similar design I used tantalum capacitors ,the problem is when the capi goes bad it is shorting the supply how to overcome this ,by using same type capi
If this issue is happening over and over than you probably are not using a cap that is right for your design. Is the voltage rating too low? Are you using it out of its temperature range?
7161
why the input voltage must higher than the output voltage at 1:96
+Elaine Yeoh that is true for all linear regulator s like the LM317, input voltage must be higher than output. If you want your output voltage to higher than input voltage than go with a boost DC to DC converter
All Voltage Regulators, (and most/if not all IC's in general), require a small voltage just for the IC to operate. In the case of the 317 it's around 1.25 Volts for the IC to do it's thing. I believe it's called the "Reference Voltage". Check out the datasheet of the 317, there's a schematic of what the inside of the IC looks like. There's a Lot of Tiny components making up the internal circuit of the 317, (short circuit, thermal, overload protection...), so it takes a little over a volt for all that to work, and it's all built inside of a pretty small, easy to handle, easy to use IC.
Hope that helps some....
it looks to me you are using a book to show circuits,?? which one is it??
The circuit diagram comes straight from the LM317 datasheet