or so you'd think but then you build it, and it doesn't work at all

Thanks! Glad you liked it. Not to split hairs, but technically a rheostat is a 2-terminal device (variable resistor) whereas a potentiometer has 3 terminals (potential divider) ;-)

@@imajeenyus42 yeah, but for some reason the stabilisation part of your scheme didn't work for me until i connected it like a pot, with R2 on the wiper

It will work, but it's less efficient because of the larger feedback voltage.

I practised them for a long time 😂😂

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 imajeenyus.com/vacuum/index.shtml if you're interested). Best regards, Lindsay.

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!

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.

Thanks. This circuit is acting as a current regulator, not a limiter, so it's not really possible to have an LED somewhere to indicate when current is being limited because, well, it's not being limited!

@@imajeenyus42 Yes, I see what you mean now. Thanks.

@@imajeenyus42 sorry to bug you again, just wanted to re-phrase my question. I understand it is regulated but in a way, it is a limited current range by the adjustable trimmer R3 (125mA to 400ma). If I was to lets say short the output with a 10ohm resistor, then no more than 400ma would flow through it. Where could I add a LED to indicate that the current had reached the maximum regulated current set by R3 ? Does that make more sense ? Thanks for your time, I really enjoy your videos, they are always clear and explained very well.

Are you trying to make a Tesla coil by chance and play music through it?

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.

@@imajeenyus42 Interesting I'm glad you kept it simple seriously thanks for going deeper and explaining it so well great vid thanks

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.

Thanks!

starting by reading and comprehending the Ohm's law.

Thanks for the comment! Unfortunately, the design procedure I've shown in the video relies on R1 being much, much smaller than R2 & R3. So if you did want to use say a 10kOhm for R1, you would need to make R2 & R3 much larger, in the region of say a few hundred kOhm. An additional complication is the leakage current out of the ADJ pin - although it's small (50uA), it would create significant voltage drops when using larger resistor values. It's really difficult to say exactly what the equation would be! If you want to play around with some values, I'd try 10kOhm for R1, 100kOhm for R2, and 1MOhm for R3 (the variable).

@@imajeenyus42 Hello, unfortunately it didn't work for me, but I have found a different solution for low current at this link: electronics.stackexchange.com/questions/211249/lm317-%C2%B5a-constant-current-source-possibility Thank you again for your help!

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

There's lots of examples of LM317 voltage regulators online - for example, this one: www.learningaboutelectronics.com/Articles/LM317-voltage-regulator. The purpose of my circuit is a constant-current regulator, not a voltage regulator.

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: www.pmillett.com/HV_reg.html. That can probably be modified to suit.

Might be possible, I've seen a few people use a thermistor + LM317 for a fan speed controller (thing heats up, fan runs faster, cools it down). If you put a thermistor (NTC) in place of R2, then as it heated up, its resistance would drop, which would make the LM317 decrease its output voltage, thus reducing the heater power. You'd need to play around a bit with values though to get something that worked at just the right heater level.

Yes, correct. Search for "digital potentiometer", there's lots of options. Usually I2C or SPI interface.

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 ;-)

​@@imajeenyus42 my fault,i had just written without reading the comments.

@@imajeenyus42 Quite right, there's always one!

MAKE A DEFERENCE شكل فرقاً You could try using an LM338 instead - they can handle up to 5A, exactly the same operation.

Linear regulators like this are NOT the best solution for high current or when large voltage differences. That is because the excess voltage is ALL converted to heat by the regulator. For soldering iron it is much better to use a switch mode buck regulator with constant current - these cost a few dollars from eBay etc.

You could, but that would give some weird response because it's then a sort of blend between a current regulator and a voltage regulator. I.e. it would regulate neither. If it's a current regulator you're after, one end of the pot needs to go to the adjust pin as shown.

@@imajeenyus42 thanks so much for responding! Looking forward to building and testing this circuit!!

@@imajeenyus42 this may be a basic question, but it's been my experience that laser diodes are best driven with current, not voltage. That being said, if the resistor values are chosen for a current range, should there be a concern for voltage limits of the laser diode being exceeded? Or is it safe to assume that if the current ratings are within the datasheet spec, there should be no cause for concern? Was the circuit you built for the laser cutter simulated in LTSpice, Multisim, etc before being built or were the hand calculations sufficient enough to get the gist of the circuit behavior? Thanks again for an excellent tutorial and quick support!

@@libanhussein7433 Laser diodes, like LEDs, have essentially a constant voltage drop regardless of the current passing through them, so it's not like there's really a voltage limit that can be exceeded. The current is the important thing, and there is a current limit above which damage will occur (usually from overheating). So if the current is kept in the safe range, nothing really can go wrong. I didn't bother simulating anything here because it's a relatively simple circuit and the behaviour can be predicted just from the equation.

@@imajeenyus42 really appreciate you taking the time to reply to these comments! Thanks again!🙏🏿

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"?

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.

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.

Confimed - I reached the same result. The mistake Lindsey made was when he talked about "a resistor divider" when he computed V_R1. That is not correct; what happens instead, is this: (a) the voltage across R2 is 1.25V (b) that means the current across R2 is I_r2 = 1.25/R2 (c) the *same* current goes through xR3, since there's no current going in the Adj pin (very high impedance) (d) that gives us the total voltage drop V_total across both R2 and xR3 - since the same current goes through them: V_total = I_r2 * (R2 + x*R3). Now *that* voltage is the one driving the "secondary" current through R1: i.e. I_r1 = V_total/R1 = (1.25/R2)*(R2+x*R3)/R1. The two currents add to form the total I_out, which ends up as you computed. Sadly, the way KZfaq works, Lindsay can't fix the video - many people will be confused and learn the wrong thing about circuit analysis here.

It's because of the small current that flows out of the ADJ pin on the LM317. This is approximately 50-100uA. I haven't included its effects because it makes the equation more complicated. If you do include it, the effect is to increase the output current, and in your case it would increase the minimum from 12.5mA to around 20.

@@imajeenyus42 this is what I figured. It just seemed like a big increase, but I guess not. Thanks for the reply

The LM350 and LM317 have exactly the same specifications for the adjustment pin leakage current (50-100uA), so that's not the problem. It sounds like the input voltage isn't high enough. If you have a R1=10Ohm, that will drop 25V, and the LM350 needs at least 2V to operate, so that's 27V so far, then you need to add on whatever voltage your load requires.

@@imajeenyus42 you are perfectly right, the input voltage is about 20V, I was stupid 🙃. Thanks a lot for the fast answer!

@@valichesu No problem, glad to help!

Don't know how you arrived at those figures, that's not what the equation gives. Try R1 1k, R2 100, and R3 10k. Should give 1-100mA range, but note that the adjust pin current might become significant. Easiest is just to try it.

@@imajeenyus42 thank you for your quick response. I'll give it a try. I will share the result. Thank you

Yes - me. I used it to drive a 300mW laser diode on my AxiDraw plotter. From what I remember, the range was pretty close to what was calculated, maybe 150-380mA or so. It'll be a little off with component values, tolerances etc. imajeenyus.com/computer/20160531_axidraw_laser/index.shtml

There are a few examples shown in the datasheet of either connecting multiple LM317 in parallel or using an external pass transistor to increase the current capacity, but they are configured as voltage regulators, and I'm not sure how you'd do it for a current source. However, at these sort of current levels, you'd probably be far better using a switching regulator instead. There are lots of constant-current LED driver chips around, e.g. www.ti.com/lit/ds/symlink/lm3409.pdf

@@imajeenyus42 Thanks, I ordered a 8amp buck converter which has adjustable current and voltage at the output. It was only $4.99.

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)

cloud you add some practical resistances i dont seem to fully grasp the equasions

Use the equations given at 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!)

The calculations shown assume that R1 is small compared with R2 & R3, which is not the case in your situation. Try 10x the value of both R2 and R3, i.e. 625 and 2500 ohms.

Hmm, it shouldn't do. Something to check is that you have a sufficiently high input voltage - the LM317 has a dropout voltage of 2-3V, then there's the voltage drop across the current sense resistor, then finally the voltage drop across the diode load. If the supply voltage is insufficient, then you could still get an adjustable current at the output, but it wouldn't be as high and would behave strangely.

@@imajeenyus42 Thanks for the reply. I'm using a 9V input. For the blue diode laser, using my calculated values I just got 1.2 V measured across the diode max. So then I just kept slowly reducing the resistors until I got the diode lasing ... but according to the calculations using the new R's, it should be sourcing 125 mA. I'm pretty sure it is not because the diode is rated for 50 mA max and I barely made it emit light, so I'm definitely missing something. Right now my circuit is working, but I don't know why :(

@@Danyel615 What's the value you're using for the main sense resistor (R1 in the video)?

@@imajeenyus42 I'm using R1 = 500 Ohms, so I min = 1.25/500 = 2.5 mA. R2 = 100 Ohm, R3 = 1.5 kOhms, so R3/R2 = 15 and Imax = 2.5 mA*16 = 40 mA, at least that's what I had at the start.

@@Danyel615Here's the reason it's not working with a 9V supply. Say you drop 3V across the LM317, and 1V across the load, 4V total. That leaves 5V maximum that can be dropped across R1, so the maximum current that can possibly flow is 5/500 = only 10mA. You won't be able to reach the 40mA maximum current, because that would require 20V to be dropped across R1. This is actually a point which I should have mentioned. The main resistor R1 is sized based on the lowest current output, but it still has to carry the maximum current as well. For large min/max ratios, this means that R1 is quite large, drops a large voltage at maximum output, and hence you'll need a higher input voltage. A better circuit for what you're trying to do might be this: imajeenyus.com/temp/current_source.jpg. That's a single-transistor current source. The 10k pot applies a 0-9V voltage to the base of the transistor. It ensures that its emitter is always about 0.6V below the base voltage, so a 0-8.4V voltage appears across the 200R resistor. This will give a current range of 0-40mA or so. Neat thing is it's adjustable right down to zero (note if the base voltage is below 0.6V then the transistor will be off totally). I've shown both NPN and PNP versions, in case the diode needs to be connected to the 0V rail. Any common small amplifier transistors should work fine. The linearity and stability probably won't be as great as with an LM317, but it's maybe enough.

I don't understand your question. Different people will of course use different values of resistors, since they will have different voltage/current requirements. If you are asking about R1, it is the current-sense resistor so obviously it will be small.

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.

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.

With those values, R1 will be dropping 16V at maximum current, so you'll need an input supply capable of delivering at least that above the level and voltage.

@@imajeenyus42 now that explains a lot as I'm testing 12v supplies. Guess I'll have to recalculate with maybe 4 or 8 ohms.

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!

This helped me out understanding the formula, thanks for sharing! @@imajeenyus42

@@imajeenyus42 I fell there too, so VR2=1,25V=VR1 * R2/(XR3+R2) from here VR1=1,25*(R2+XR3)/R2. It is right? thanks

@@angelodemichele Yes, that's correct - that's what I show in the video.

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: imajeenyus.com/electronics/20160517_potentiometer_feedback/index.shtml

Brilliant!. Good design. Thanks

It would work, but unfortunately isn't very efficient. If I_min is 125mA, then R1 (the series current sense resistor) needs to be 10R. I_max/I_min is 1400/125=11.2, so R3/R2 has to be 10.2. Your choices of 500k + 47k are the right ratio, but I would choose 5k + 470R instead - with high value resistors, the leakage current out/in of the LM317 adjust pin can throw things off. At the maximum 1.4A, R1 will be dissipating 20W, so it'll need to be able to handle that power. Again at maximum 1.4A current, the LEDs will drop 10V, R1 14V and the LM317 needs 3V of headroom, so that's a minimum supply voltage of 27V - a 28V supply would probably be the closest.

@@imajeenyus42 Thanks Lindsay. I'm very new to all of this, and just happen to have 500k & 200k pots. I will try and obtain some other analog pots in the lower range, closer to the above, and possibly start with the LEDs in 1x3 to see if that improves the situation. The desire was to reduce the foot print somewhat. I do have some XL4015 / LM2596 CC drivers. If I cannot control this thru a 12v supply then I may just go the module route.

@@imajeenyus42 I found some 1K pots to use with 100 ohm as R2. Would this be ok? Also is the loss of wiper contact a concern and so should I look closer into your other topic regarding that? Thanks a lot.

@@PanteraPersa They should do the trick. With the circuit in the video, any loss of contact will cause the current to drop to the minimum value, so that won't be a problem.

You can't control both voltage and current at the same time, no matter what circuit you use. You can control one parameter, but the other will then be determined by the load resistance. Depending on the configuration used, the LM317 can act either as a voltage or current regulator.

@@imajeenyus42 hi sir, if we want to control voltage and current same time how?

@@nagarajub4011 You can't. Did you even read my comment?

@@imajeenyus42 ok thank you

kzfaq.info/get/bejne/h7yPo5ek3rOahX0.html

I have explained the equations as best as I can in the video. Can you please explain which particular equation you want to know about?

@@imajeenyus42 I understand your explanation but I would like to know there is the equation coming from. How do you arrive to that solution from the original equations. I have been able now to calculate the values taking into account all resistors. But I don't know how to get to that formula you provided. Sorry I am the kind of person that needs to understand how to get into the results. I would thank you if could explain it to me. Thanks.

Never mind Lindsay. I just found out where is your equation coming from. Thanks :)

I'm not sure what you mean - each regulator is only capable of handling its own polarity (positive or negative), I don't think there's a way of combining them so the output is continuously adjustable from negative to positive.

@@imajeenyus42 I meant that I want a circuit which can perform current limiting for both negative and positive supplies. Since LM317 can only work for positive supplies we need LM337 to work for negative supplies. So just by using a switch we can make them work right?

@@vishal_on_yt Possibly, I'm not sure.

You could try this, it looks suitable: www.microchip.com/wwwproducts/en/MIC29152

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.

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

No load, no current flowing, so no power drawn from the supply.

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

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.

Lindsay Wilson and wat about using it as voltage regulator

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 imajeenyus.com/electronics/20160517_potentiometer_feedback/index.shtml. Also added the circuit to the current regulator page as well - imajeenyus.com/electronics/20160530_adjustable_current_source/index.shtml

Power supply smoothing/decoupling

@@imajeenyus42 Thanks

Hey, one thing more. Is there any limit to the value of capacitor which we can use? I mean on what basis should we decide?

@@vishal_on_yt A lot depends what sort of smoothing you already have on the power supply feeding the regulator. If it's already well smoothed, you don't need much. Few uF low-Z film would probably do.

@@imajeenyus42 Thanks again. :)

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.

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.

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.

What voltage of power supply are you using, and what is the wattage of the bulb?

@@imajeenyus42 Hello Wilson, thank you for answering. The power supply is an ATX 600W and the light bulb is 21W 12V. What i am trying to achieve is a battery charger with a constant output current of 1A and a voltage of 14V, but the voltage always drops...

@@hichamtassi1753 If you're regulating the current, then the output voltage will depend on the load resistance. You can't have a fixed output current AND a fixed output voltage, that's impossible. Your 1.5 Ohm resistor will give a current of 1.25/1.5 = 0.83A. If the bulb is rated 21W at 12V, that means its resistance is around 6.8 Ohm. With a current of 0.83A, that's a voltage drop of around 5.5V. However all that is extremely approximate since bulb resistance is highly nonlinear. A voltage of 4V across the bulb would seem quite reasonable.

@@imajeenyus42 very clear, why impossible to have Constant Current and Constant Voltage at the same time ? because if i consider the lamp as a battery to charge, with 4V it will never charge. I am more of a business background but dream of understanding electronic science. Thank you for taking your time to explain to me

@@hichamtassi1753 For a resistive load, which you can assume a lightbulb is, voltage and current must be related by Ohm's Law, i.e. the voltage across the load is the current flowing through it multiplied by its resistance. If you vary the current, the voltage must vary, and vice versa. A battery is more complex, because it's not a resistive load. Very approximately, the voltage remains constant until the battery is fully charged, then it starts to rise (overcharging). A power supply will normally put out a fixed voltage (e.g. 12V) and the current drawn will depend on the load applied. If instead you want to produce a constant current, that's when you'd use this regulator circuit. It will attempt to keep the current constant, within the limits set by the power supply. For example, if you have say a 12V power supply, you set the LM317 to regulate at 1A, but connect a 20 Ohm resistor, you are never going to get 1A flowing, because that would require 20V across the resistor, and the power supply is only 12V. I could go into more detail, but that should help with the basics.

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.

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... :)

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.

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? :)

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 ;-)

Try R1=2R, R2=1k and R3=1k. Should give an adjustment range from about 0.6-1.2A.

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.

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.

Yes - the LM317 doesn't care what input voltage it sees (within its safe limits)

i have no clue how to measure if what i am doing is correct. if i wire it up like the schematic. and i test the resistance over the output and the input. it does not change (635 ohm or 1153 depending on the polarity of my probes) no matter what i do to the potentiometer. if i test the resistance between out Iout an adjust on the lm317. i get a range of 161 ohm. to 533 ohm. is i test the resistance between lm317 out and lm317 adjust. it ranges from 157ohm to 543ohm. if i connect the 12v plus of a led strip. to Iout. 2 9volt battery's + to lm317 in. and the ledstrip negative to the negative terminal of the battery. then the leds light up bright. and changing the potentiometer does nothing. hoever if i connect the 12v+ of the leds to lm317 adjust. i can actually manage to change the leds brightness. do you have a video of you actually building and using this thing so i can double check that i am doing it right instead of some written down schematic that might or might not be correct as it is now i am not hooking this up to a laser diode. i am 99.9999% sure it will just say poof.

@@darkracer1252 I do not have a video of my "actually building and using this thing" - that's the whole point of a schematic diagram. Suffice it to say, I have built and used this exact circuit for driving laser diodes. I have no idea why you are trying to measure the resistance between the OUT and ADJ terminals of the LM317 - that is going to be completely meaningless, because you have absolutely no way of knowing how the internal circuitry of the LM317 will respond to the test signal from the meter. Trying to measure the resistance of components in-circuit is always going to give strange results when they're connected to "active" components like semiconductors. Do NOT connect the load to the ADJ terminal - you are likely to damage the LM317 since it isn't meant to source a large current through the ADJ pin. All I can suggest is you double-check your circuit. The laser diode driver shown at 7:00 works perfectly well.

@@imajeenyus42 yeah it's not needed either. i'm just a freaking dumbass. i guess the load i was testing with had some sort of corrective circuit aswel. (you can hook it up to a straigh dc power source all day without issue) after i grabbed a single led diode. your schematic worked. when put on Iout i can adjust the brightness of the led. now i need to use something other then a 9V battery so it won't get drained and give me increasingly dimmer results. using 2 9v battery's seems to burn out the led though. wich kindof has me worried about hooking up an actual laser diode.

@@imajeenyus42 if i connected up the previous load. (the led strip with resistors and what not on it) it seemed the lm317 did nothing at all. it did not heat up in any way i wired it up. and i tried everything. but with the straight up single led diode. everything did what it's expected to do. turning the pot changed the brightness (although in reverse, clockwise was dimmer guess that's just a matter of reversing the outer wires of the potentiometer.) and the lm317 actually started heating up when i dimmed it. so i guess there is some black magic happening in the led strips i tested this circuit with before.

This circuit is a linear regulator, not a switchmode regulator, completely different.

Sir Plz advice me for completing this project

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.

@@imajeenyus42 OK. Thank you.

Use an LM138/LM338 instead, which are rated to 5A.

For that case , will all the connections be same ...

Yes - the connections and operation are the same (it has the same reference voltage as the LM317)

Thanks a lot ...

Really? How so?

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.

Oh excuse me for taking the time to explain the topic in a methodical manner. Next time I'll do it in under a minute with a thumping electrobass soundtrack, that keep you happy?

@@imajeenyus42 lol. I like that reply lollol

@@imajeenyus42 OH NOoo I have seen a couple of those. please whatever you do no electrobass or music during the video unless you want to torture me. I will take it slow and gentle as you have it. Lololo

Yes I did. What's your problem?

Show me where in the LM317 datasheet they describe using a "grounded-wiper" potentiometer arrangement to achieve a linear output adjustment. THAT is my idea. Not the concept of a constant current regulator itself.