HomeНаука и техникаRelated VideosMore From: Lindsay Wilson

More From: Lindsay Wilson
722 ratings | 95633 views
Adjustable constant-current source I designed for things like laser diodes. Full details at: http://imajeenyus.com/electronics/20160530_adjustable_current_source/index.shtml More details on the grounded-wiper potentiometer feedback network: http://imajeenyus.com/electronics/20160517_potentiometer_feedback/index.shtml This was used to power a 300mW 808nm laser diode on the AxiDraw drawing machine for some rudimentary laser cutting: http://imajeenyus.com/computer/20160531_axidraw_laser/index.shtml
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sefo (4 days ago)
I have some doubts about the circuit you drew at 4:50, R2 and R3 are not in series,because there is current flowing through adj pin,so they cannot form a resistance divider.
sefo (4 days ago)
Lindsay Wilson (4 days ago)
I have explained this time and time again to other commenters. Yes, there is a TINY current flowing through the adj pin, but it is INSIGNIFICANT. This is why, to all intents, we can consider R2 and x*R3 as being in series, forming a divider, and we can forget about (1-x)*R3 being in series with the adj pin. The current flowing through the adj pin is 50uA, according to the datasheet. Suppose we were using a 1k potentiometer. Then, the maximum drop across (1-x)*R3 would be 1k*50uA=0.05V. So instead of the reference voltage being 1.25V, it would be 1.20V - only a 4% difference. This is not something to worry about, and I really wish people would understand this! There is no point in trying to calculate the adjustable range exactly, for the simple reason that it IS adjustable. If you calculate that you're going to get a range of 100-400mA, and you actually get 95-405mA instead, is that really a big deal?!? Considering that the tolerance on most potentiometers is 10-20%, a 4% error caused by ignoring the adj pin current is irrelevant. I hope this doesn't come across as harsh, but I have explained this all before ;-)
Mr Remmoz (5 days ago)
Why (1-x)R3 is neglected ? It can be greater than xR3. What is the sense?
Mr Remmoz (5 days ago)
+Lindsay Wilson OK. Thank you.
Lindsay Wilson (5 days ago)
Because the ADJ input pin of the LM317 has a very high input impedance, so any additional resistance in series with it (like (1-x)*R3) has no significant effect.
Hello! If I may ask, about the power of the resistors it's only mentioned the 3W of R1; and the others , R2 and R3, resistors? Can I use your calculations for an 1A output? It won't get on fire? :-) Joking but asking seriously! Thanks!
Thanks!
Lindsay Wilson (8 days ago)
Hi! All the other resistors can be normal 1/4W varieties since they hardly dissipate any power at all. Yes, the calculations can be used for any current level - just work out the resistor values to suit.
Dario Szlain (10 days ago)
The calculation on Ir1 is correct, but you forgot to consider that final Iout current is equal to Ir1 + Ir2... I get that you omitted Ir2 since is much smaller than Ir1. But this may really confuse people trying to learn
alephii (24 days ago)
I've got a different result here, the current is I = 1.25 (R1 + R2 + xR3) / (R1 * R2), the values in the end are not much different, but I am not sure if you calculated I1 correctly. In the end there is a R1 / R2 missing in the final equation, but since R2 >> R1 the difference is not very significant. Anyway, I think it might be confusing for people that are learning if they try to understand the calculations, that's why I am commenting it. Please Lindsay, take a look at it...
Lindsay Wilson (24 days ago)
You're absolutely correct - in theory you should include the R1/R2 term, but like you say the ratio is so small in most applications as to be insignificant. For example R1 might be 10 ohms, and R2 might be many kiloohms.
Mechatrons (2 months ago)
Nicely done. If one desired a 0amp minimum, say 0 to 10ma, how would that look?
Dice Zulu (3 months ago)
It's great
Sankalp Mandlik (3 months ago)
Hi there , Your circuit is quite easy to understand , I have a situation where my current varies from 0amos to 2.5 amps and I would like it to vary as normal but I would like it to limit till 1.9amp and not exceed more than 1.9amp , is it possible through this circuit or this circuit will give me a constant xx ampere current. Thank you
Nima Nabavi (3 months ago)
Thank you. nice.
mayank singhal (4 months ago)
Great video, quickly and clearly explained the application of lm317. Thanks
EJ-160 (4 months ago)
Is it possible to replace the potentiometer with a digital potentiometer instead?
thanks for sharing ... but it's getting to high temperature when i use it to control my 12v soldering iron voltage.. any suggestions or different solution for high loads ?
Lindsay Wilson (6 months ago)
MAKE A DEFERENCE شكل فرقاً You could try using an LM338 instead - they can handle up to 5A, exactly the same operation.
William Ashland (6 months ago)
Hi Lindsay, great video. I was curious about the equation at 5:37, and was wondering if you could go into any more detail about it. I see that R2 and xR3 combined are in parallel with R1, but since we know the voltage across R2 = the voltage across R1 + xR3 = 1.25 V, wouldn't it be more accurate to say VR1 = 1.25(1 + R1/R3)? Or is that not right. Sorry to everybody else for nerding up the comments section
Lindsay Wilson (6 months ago)
Hi, yeah, I know, it takes a bit of getting used to! The best way to think on it is looking at cause and effect, and which "direction" the effects flow in, if that makes sense. Here, the output current passes through R1, producing a voltage, which is then divided down by R2+xR3, and applied to the ADJ input. The chip then adjusts its output until the voltage at the ADJ input (well, the voltage between OUT and ADJ) is equal to 1.25V. At first glance, a lot of people (myself included, until I got the hang of it!) attack the problem from the other end, which is - The voltage across R2 is 1.25V, which is divided down by R1+xR3, so the voltage across R1 is 1.25*R1/(R1+xR3). That, unfortunately, is the wrong way to think about it - the ADJ terminal is an input, and SENSES the divided voltage produced by the resistors. Hope that makes things a bit clearer!
Pritam Saha (8 months ago)
What about high current situation like 4amos
Pritam Saha (8 months ago)
Thanks a lot ...
Lindsay Wilson (8 months ago)
Yes - the connections and operation are the same (it has the same reference voltage as the LM317)
Pritam Saha (8 months ago)
For that case , will all the connections be same ...
Lindsay Wilson (8 months ago)
Use an LM138/LM338 instead, which are rated to 5A.
Metin Ozsavran (8 months ago)
Dear Lindsay, my circuit analysis classes were 35 years ago, and this video is 2 years only. The circuit has the advantage of defaulting tp R1+R3 in case potentiometer looses contact, fine. However you loose me from min.5 onwards. Letting R4=xR3, and R5=(1-x)R3, the equivalent resistance from ADJ pin's position to OUT pin is (1/(R1+R4))+(1/R2)+R5. Therefore the bottom half of the R3 pot cannot be assumed zero, it is in fact the dominant term to determine current adjustment. Voltage across R2 is not 1,25 volts. Even if it was You can't write VR1 as you wrote. It assumes same current over R4 (ie x.R3) as R2. Since there still is a 3rd branch coming out of that junction as I5 even if you reduce R5 (ie (1-x).R3) to zero. What's really happening here is the adjustment of R5.
Lindsay Wilson (8 months ago)
According to the datasheet, the current on the ADJ pin is 50uA, so if R5 was on the order of 1k, that's only 0.05V, which is reasonably small compared with the 1.25V reference. I haven't actually got any to hand either, and unfortunately the only simulation software I know how to use (LTSpice) doesn't have an LM317 model ;-)
Metin Ozsavran (8 months ago)
Thanks for the continued exchange Lindsay. Well, we should measure both ends of R5 and see if the voltage drop is really close to zero. I have 4 old Chinese fake LM317's around, none seems good to test. Waiting for new ones. Maybe an EWB simulation would illuminate me? :)
Lindsay Wilson (8 months ago)
No problems, I don't mind ;-) Try this for visualizing how the ADJ pin functions. Suppose you have a voltmeter which draws zero current from what it's measuring - it simply measures the voltage. And suppose you use it to measure the voltage of a 1.5V battery. If you connect the voltmeter directly to the battery, it will obviously measure 1.5V. Now, suppose you inserted a resistor between the voltmeter and the battery. Because no current flows, there is no voltage drop across that resistor, so the voltmeter will STILL read 1.5V. It's as if the voltmeter was still connected directly to the battery - i.e., the resistor replaced by a short circuit. This is why it's possible to ignore the presence of R5 (which is (1-x)*R3) - since no current flows, there is no voltage drop across it, so the voltage at the ADJ pin will be equal to the voltage at the junction of R2 and R4. The LM317 will adjust its output until the voltage between the OUT and ADJ pins is 1.25V, in other words until the voltage across R2 is 1.25V. That then means the voltage across R2+R4, or across R1, is 1.25*(1+(R4/R2)), because no current flows in to or out of the junction of R2+R4. Does that make any more sense? ;-) Regarding the relative value of R2, you don't want to make it really small, because that would give you a huge adjustment range. In the current-regulator circuit shown in the video, the maximum output current is (1+(R3/R2)) times the minimum output current (which is 1.25/R1). If you made R2 very small, that would give a very large output current.
Metin Ozsavran (8 months ago)
Thanks for caring to respond Lindsay. Although I am really in deep sleep state with eyes half shut, at this wee hour here, I surely don't question the circuit does work. Just trying to clarify how it really works to brush off rust from my circuit analysis grasp, by rasping against your analysis, in friendly manner. I am guilty of not checking LM317 characteristics too. Hope you don't mind exercising with dumb old me here. :) Now, "but", if current to/from ADJ pin is close to zero, we can only consider that as an open circuit, not a short circuit. In that case, the effective resistor circle value on the output becomes equivalent to (1/R1)+ (1/(R2+R4)). Evem if we give R4 pot a huge range, the max effective resistance control range we get, that between LM317 OUT pin and overall circuit output node, vary from (1/R1) to (1/R1)+(1/R2). To increase that control range, we need a big value R1 and an R2 well below 1 ohm. Perhaps we could make R2 a 1ohm max. pot to use as a fine control over Iout? After I wake up, I should try this physically, and get back. Please feel free to explain even further... :)
Lindsay Wilson (8 months ago)
Thanks for the comment. I agree, the circuit does look a little odd at first glance. However, the important thing to remember is that no (or at least very little) current flows in to or out of the ADJ pin of the LM317. Because of that fact, it does not matter what the value of R5=(1-x)R3 actually is, so it can be replaced with a short circuit. The current through R2 will be nearly equal to R4=xR3, so the divider equations I've written in the video will apply. Believe me, I built it, it works ;-) In reality, there is a current of about 50uA on the ADJ pin, but this is negligible for most applications.
Mare Prah (10 months ago)
Hello mr. Lindsay. Can you please give the info on how to calculate the power dissipation of the R1 resistor, since I am not able to figure it out for my case. You imajeenyus site you mention that the in your case it is 1.6W? Also, sorry if you mention it in the video, but I am not able to hear any sounds (problem on my end). Thank you in advance for the answer.
Mare Prah (10 months ago)
Lindsay Wilson oh gosh, I didnt even think about that formula, shows how much of an amateur i am :). Thank you and have a nicce day. PS: the sound is my problem with the audio card, the video is fine. Watched it now on my phone and its ok.
Lindsay Wilson (10 months ago)
The power dissipation is calculated from (I^2)*R. Square the maximum output current and multiply by the resistance - in my case, the maximum output current is 0.4A, and the resistor is 10 Ohms, so the power is 0.4*0.4*10 = 1.6W. Not sure why you're not getting any sound, there's definitely sound on the video! Best regards, Lindsay.
shubham shukla (10 months ago)
sir i want to make a cc/cv using 2 lm 317 and i understood the circuit but i want a dynamic range of current i want current upto 1.25/1.5/1A so please suggest me best resistor values along with the power rating and which are commonly available
Leonard Janus (1 year ago)
Hello Could you suggest where put an optical coupler in that circuit? I mean, drive the optical coupler with a smartphone signal generator and then modulate/control the output of that circuit. Could also suggest a current controller for up to 100V? Max LM317 is 37V.
John Michael Stock (6 months ago)
Are you trying to make a Tesla coil by chance and play music through it?
Robonza (1 year ago)
Nice job on the explanation. I thought you would have needed a small cap on the output for the switching transients? Also have you tried modulating the ADJ pin instead. It will take much less current to control so cleaner switching would be easier I would guess. It would be inverted though.
Robonza (1 year ago)
Lets say you switch your laser at 100khz. That transient load is too fast for the 10khz response of the LM317, so you would need a 0.1uf cap on the output and input in theory. Grounding the ADJ pin will switch off any load (1.25v out is off for most loads) So you can switch the ADJ pin. which is easier than switching 400ma. I will probably try all this in a few weeks when I get a chance. The next TV I build needs a modulated laser. I am just researching right now.
Lindsay Wilson (1 year ago)
The LM317 is a linear regulator, not switching, so a filter cap isn't strictly needed. It would help improve transient response to changes in the load, but too large a value can cause instability. I'm not sure what you mean by "modulating the ADJ pin"?
electronics nerd (1 year ago)
https://youtu.be/QV_p4syM6UI Variable voltage and variable current using LM 317
Guedes (1 year ago)
Thanks for your explanation, good job.
Excelent!, I was looking for a adjustable current circuit based on LM317, thank you very much. Regards from México.
How to vary both voltage and current in the same circuit ?
electronics nerd (1 year ago)
https://youtu.be/QV_p4syM6UI
ANGEL BUTTER (1 year ago)
your illustrations need more explanation and diagrams... : {
Lindsay Wilson (1 year ago)
Have you looked at the webpages as well? (Links in description). My intention is not to provide a complete tutorial on how to use an LM317 regulator; rather, I wanted to show the specific application of the grounded-wiper configuration.
Darieee (1 year ago)
Thanks ! Awesome explanation
Milan Karakas (1 year ago)
Good explanations. Did you noticed that in many datasheets, Vref is 1.2 V, not as it should be 1.25 V?
v ant (1 year ago)
I do not get it, why have not you used a constant current circuit from the datasheet ?
Lindsay Wilson (1 year ago)
I am well aware of the circuits in the datasheet. There is no problem whatsoever about adding other resistances around the ADJ terminal. All the chip does is adjust its output until there is 1.25V between the ADJ terminal and the output. You can put in whatever resistive feedback networks you like. I have built many constant-voltage and constant-current regulators using the LM317.
v ant (1 year ago)
In TI datasheet there is "50-mA Constant-Current Battery-Charger Circuit" example. Basically you need only R1 to set constant current mode. R2 and R3 are ok if you need to adjust current except a resistance to ADJ terminal may (I am not sure) have some unwanted effect.
Lindsay Wilson (1 year ago)
I don't understand what you mean. The only current regulator circuit in the datasheet is Figure 14, which is a current limiter. This is totally unsuitable for my application. My design gives a linear adjustment response, plus protection against wiper contact loss.
akhil vaid (1 year ago)
can u make similar video for lm337 regulator?
Lindsay Wilson (1 year ago)
Again, same deal - circuit is the same, just flipped upside down. I went and updated my page with a circuit for the LM337. Check out http://imajeenyus.com/electronics/20160517_potentiometer_feedback/index.shtml. Also added the circuit to the current regulator page as well - http://imajeenyus.com/electronics/20160530_adjustable_current_source/index.shtml
akhil vaid (1 year ago)
Lindsay Wilson and wat about using it as voltage regulator
Lindsay Wilson (1 year ago)
If you're using it as a current regulator, then the circuit will be exactly the same - current sense resistor, potentiometer, feedback etc. Just replace the LM317 with the LM337, keeping IN, OUT, and GND connections the same.
Photonic_ Induction (1 year ago)
so if iid like to make this circut with a lm388 for a 5a adjustable current supply cloud i make that with this circut ?
Lindsay Wilson (1 year ago)
Use the equations given at http://imajeenyus.com/electronics/20160530_adjustable_current_source/index.shtml to work out the min and max current - it's most helpful if you stick them into Excel. For example, R1=2 Ohm, R2=220 Ohm, R3=2000 Ohm gives a minimum current of 625mA and a maximum current of 6.3A. Watch out: remember that R1 has to carry the full load current, so will need to be high-power. (In this example, power dissipation is nearly 80W!)
Photonic_ Induction (1 year ago)
cloud you add some practical resistances i dont seem to fully grasp the equasions
Lindsay Wilson (1 year ago)
Yes, should work exactly the same. Reference voltage is also 1.25V. (I'm assuming you mean the LM338; the LM388 is an audio amplifier)
rntesla nielsen (1 year ago)
hey can you have 6 out stable from an 12v battery whit no load on it and if so how does it take amp watt to do that in standby mode ?
John Michael Stock (6 months ago)
If your going from 12v to 6v you should use a buck converter instead. Voltage regulators work by emitting the excess energy as heat and very inefficient. Buck converters can reduce voltage with over 90% efficiency and can be bought for pennies in the UK
Lindsay Wilson (1 year ago)
No load, no current flowing, so no power drawn from the supply.
rntesla nielsen (1 year ago)
i mean if you dont put a load on it just want to have a floating 6v aut from the 12v input how much does the input vatt take to maintake the aut 6v whit no load on it
Lindsay Wilson (1 year ago)
If you are asking, first, "Can you have a stable 6V output from a 12V battery?" then, yes. You don't even need to use the LM317 - there are plenty of fixed-voltage regulators which do the job, e.g. LM7806. As for "how does it take amp watt to do that in standby mode", I have no idea what you're talking about.
peaceful1123 (1 year ago)
Yes, you're right. I got the same result as you did.
Curtis Newton (1 year ago)
I dont get it, it's acurrent source, so how do you limit the current
Lindsay Wilson (1 year ago)
Uhh....it limits the current BECAUSE it's a current source! It will put out whatever voltage is required to maintain the specified current through the load, and that current is set by the user by choosing the resistor values.
Bata Gris (1 year ago)
Why in the calculations you didn't considered the (1-x)·R3 for the other part of the potentiometer between x·R3 and the adjustment pin?? Nice video by the way!!
Lindsay Wilson (1 year ago)
Thanks! You're absolutely right - there is (1-x)R3 in series with the adjustment pin, but I'm making the simplified assumption that the pin doesn't source or sink any significant current, so that resistance won't affect anything. In practice, there is a current of about 50uA flowing out the adjustment pin, which will become significant if R3 is relatively large, but I try and keep it around 1k or so to avoid this.
Anouar Ben (1 year ago)
Hello lindsay following your calculation of VR1 I don't seem to get the same result as you maybe I'm wrong but I've got Vr1= 1.25.(R1/R1+xR3) because the resistive divider is between R1 and R3 no ??
Lindsay Wilson (1 year ago)
RIght. I see where you're getting confused, but I'm still right ;-) You can't calculate it the way you did, by thinking of R1+xR3 acting as the divider, because the same current does NOT pass through R1 and xR3. R1 carries the full load current, whereas xR3 only carries whatever small current flows through it and R2. It's R1 that's PRODUCING the voltage, which is then divided down by R2 + xR3. The IC adjusts its output until that divided-down voltage equals 1.25V (or close, when you account for the small 50uA current that flows out of the ADJ pin). I've built and used both the current regulator and voltage regulator circuits, and both work as I calculated. Not sure what you're referring to by Vout not being 1.25V when the ADJ pin is connected to ground - for a start, the ADJ pin is never intentionally connected to ground (unless you happen to short the output). It always goes to the feedback network. If you ARE using a constant-current circuit and you short the output, which DOES bring ADJ to ground, the Vout from the IC is indeed 1.25V (I just checked this). Hope that helps.
peaceful1123 (1 year ago)
You're right. He got all his calculation wrong. Vr1 = 1.25 [R1/ (R1+xR3)] and Iout = 1.25 [1/(R1+xR3) + 1/R2]. He also got his modified voltage regulator calculation wrong in another video. I did a simple experiment to verify his circuit. The problem is that when pin adj is connected to ground without any load resistor, Vout is not 1.25V as one would predict.
jp (1 year ago)
Hello Lindsay, I'd like to thank you for giving me a leg up or maybe a leg re up into an old hobby. Almost 50 years ago I was into valve amps. Google Mullard 20W - The National Valve Museum, yes I'm that old. I was looking around YouTube and came across your video above. I was particularly interested in your current control. I had not touched electronics since my valve days and the 'gubbins' you were using intrigued me. I decided to have a go. Mind you picking a 60W soldering iron up by the wrong end made for an interesting start. I had had 410V across my nose but that's another story. Thanks to people on Forums and YouTube videos like yours I am now into 2N3055's LM723 and all the bits for a bench power supply. Strangely enough though I think yours is one of the few videos that goes into a simple current control. I needed it as I wanted to play about with LED's. Things have moved on a bit since then but I just wanted to say - Thanks Lindsay.
Lindsay Wilson (1 year ago)
Hi JP, thank you so much for the kind comment! I'm really happy my videos are of some use to people ;-) I like going back to basics with simple things like regulators and discrete components - so many things nowadays just use an "all-in-one" chip, which doesn't really give you much room for understanding how it operates. I actually love "antique" technology as well - anything involving tubes, Nixies, gas-discharge stuff. I'll enjoy reading an old book from 1940 about valve technology more than a modern one on semiconductor stuff. I've never done much with tubes, apart from once building a little oscillator with an ECC83, but I have played around with Nixies and even built some gas-discharge tubes (there's some stuff on my site at http://imajeenyus.com/vacuum/index.shtml if you're interested). Best regards, Lindsay.
Stephen Roberts (1 year ago)
A Very good video, Being a novice, could you please tell me the values for R1 & R2 in an example. I realise that the circuit diagram shown is for various input voltages but an example for say a input voltage of 9v and an output voltage of 6v would help me understand the circuit better. ManyThanks
Lindsay Wilson (1 year ago)
In the laser diode driver, for example, the output voltage ranges from 0 up to a few volts, depending on the current being delivered, diode characteristics etc. In a constant-current configuration, the LM317 could in theory be used in high-voltage applications, as long as the voltage drop across it never exceeds its maximum of about 37V. The problem with an adjustable regulator running from a HV supply - if you reduce the current to such a level that the voltage drop across the LOAD is relatively small, that will then drop a LARGE voltage across the LM317, blowing it. You can use it as a higher VOLTAGE regulator, but it requires additional components - for example, have a look here: http://www.pmillett.com/HV_reg.html. That can probably be modified to suit.
Omar Azeem (1 year ago)
hello lindsay, thanks for giving such helpful video... i have one concern that in exaample of your video what are the output voltages,?? also tell me if i need to inputt higher voltages e.g 330v dc and converr it to 320vdc what we need to do to circuit......and what effect do this varying current produce on output voltages....in actual working lm317 is a voltage regulator but u r using it as a current regulator,
Lindsay Wilson (1 year ago)
There's lots of examples of LM317 voltage regulators online - for example, this one: http://www.learningaboutelectronics.com/Articles/LM317-voltage-regulator. The purpose of my circuit is a constant-current regulator, not a voltage regulator.
John Michael Stock (1 year ago)
I notice your configuration for R1,R2 and R3 is different from most CC circuits I've seen using the LM317. They usually just have a pot where R1 is and don't have R2 and R3.. Any reason why you've went with this design?
John Michael Stock (1 year ago)
Brilliant!. Good design. Thanks
Lindsay Wilson (1 year ago)
Thanks for the message. One reason is power-handling capability. Suppose I wanted an adjustment range of 100mA - 300mA. The potentiometer value is determined by the low end of the current range - in this case, 1.25V/100mA = 12.5 Ohms. But when the pot is adjusted to give the highest current, the full 12.5 Ohms has to pass a current of 300mA, which is just over 1W dissipation. I know you can get 1W variable resistors, but why risk it. The other main reason is getting a linear adjustment in response to potentiometer movement. The traditional method, with a pot in R1, gives a nonlinear response. My method gives a linear response, and avoids the need for a high-power potentiometer. In addition, it also offers protection if the potentiometer wiper happens to leave the track at any time (even briefly) - the output is forced to a low level, rather than high (with the traditional arrangement). This is especially important for things like laser diodes, which are extremely sensitive to over current. I've got much more on what I call the "grounded-wiper" potentiometer configuration on my website here: http://imajeenyus.com/electronics/20160517_potentiometer_feedback/index.shtml
Conan Krasniqi (2 years ago)
Nice video. How to increase the output current to 10 Amps with 2N3055
halo i wanna to ask u, how to regulate the intensity of the laser with regulator current LM317..?
Mathew Pendleton (2 years ago)
your the first person to actually make this clear to me. Thanks!
Daniel's Game Vault (2 years ago)
The LM338 has a max current of 5A - will your design work for that as well ? Provided of course the series resistor R1 can hold that.
Daniel's Game Vault (2 years ago)
Great, but the issue is that I doubt you'd find a resistor that can hold that amount of current through it. At 5 volts, you'd already be pulling 25w (since V times I, equals P, assuming the load is rather heavy - something like an automotive halogen lamp or medium sized motor that draws the whole 5 amps). I was thinking of a pass transistor to take away most of that current, then somehow sample the voltage using the resistor divider so you can still get the 338 to do its thing. Haven't quite mulled it over, so I don't know exactly how it should look or work - might wanna check it out :)) An op-amp might come in handy as well somewhere.
Lindsay Wilson (2 years ago)
Should do - it looks to have the same 1.25V reference voltage as the LM317. Thanks for mentioning the '338, by the way - didn't realise it came in 5A versions as well!
Harsh Bhoi (2 years ago)
nice work... but the current is not constant for variable load.... can us suggest , what should be the resistor values for current range 100mA to 1amp
Lindsay Wilson (2 years ago)
Not sure why you think the current isn't constant as the load varies, that's the whole point of the circuit. It will regulate the current, so long as the voltage drop across the LM317 doesn't become less than the dropout voltage, which is approximately 2V. Suitable values for a 100mA to 1A range (following equations given on my webpage). Minimum current is 0.1A, so R1 = 1.25/0.1 = 12.5Ohms. This carries a current of 1A, so power dissipation is 12.5W. I'd try 15W or 25W metal-clad resistor, which are available in a 12Ohm value. (If you can't get 12Ohm, put two 24Ohm in parallel). Maximum current is 1A. Min/max current ratio is 10, so (R3/R2) ratio is 9. Say you use R3 = 500Ohm (a standard potentiometer value). R2 is then 55.5Ohms - use a standard value of 56Ohms. Summary: R1 = 12.5 Ohms, 15W or 25W R2 = 56 Ohms R3 = 500 Ohms