LM317-based high-current adjustable power supply
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- เผยแพร่เมื่อ 4 ก.ย. 2020
- In this video, I discuss a few things about the LM317-based high-current adjustable power supply. This circuit is the exact same circuit that you can find in the datasheet of the LM317 in the 9.3.12 section. My ultimate goal is to precisely control the current of Peltier coolers, so I started to discover other power supply solutions. This is not an efficient power supply, linear power supplies have never been efficient. There is a lot of heat dissipation. However, I wanted to understand this concept, so I built this circuit. If you want to have an efficient (but more complex) circuit, you should look at the switch-mode power supplies (SMPS).
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Link to my website: curiousscientist.tech/blog/lm...
LM317T datasheet: www.ti.com/lit/ds/symlink/lm3...
2N2905 datasheet: www.mouser.com/datasheet/2/68...
TIP35C datasheet: www.st.com/resource/en/datash...
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![LM317 Current Boosting Circuits [2N3055, MJ2955]](http://i.ytimg.com/vi/FItAwcsj0OI/mqdefault.jpg)
I bought those exact same test leads. After measuring them with a milliohm meter, I can say they are probably the worst thing to use for any circuit with current. I was able to determine their resistance is high because they are using copper plated steel (checked it with a magnet) and the connection at the alligator plug is a poor crimp. I replaced the wire in the set and did a proper solder joint at the crocks. They are likely some of the reason for the changes in voltage during the test.
Yeah, they are not high-quality cables. At that time, that was what I could afford. Maybe the cables are still go to be used as heating wires, though! 😄
Cool video well explained!!
Thank you!
Nice work Dear 👍
Thank you very much!
I built a similar circuit on a breadboard but using 2N2907 (TO-92) instead of 2N2905. I also used a load resistor (unmarked on the schematic) that drew about 50mA. I tested the circuit setting the output voltage to 12V without external load (open circuit voltage). However, when I connected the power supply to a load (I used a 10 ohms 50W resistor), the output voltage dropped by a substantial value (from 12V to 7V). I expected the output voltage to remain constant (+/- a small percentage) at the preset voltage of 12V regardless of the load (of course within the limits of the components).
Hi! I don't remember noticing any significant voltage drops, but I have to admit that I haven't used this circuit for a while, so maybe I just forgot it. What kind of power supply do you use to power this circuit? Maybe the voltage drop is caused by that. Nevertheless, I will check it on my side this weekend and give you a brief update.
@@CuriousScientist : I used a laboratory grade PSU capable of delivering 20A, so I don't think the culprit is anything from upstream of the regulator circuit. Have you tried setting the output voltage without a load first then connect a significant load and take note of any change on the preset output voltage? I'll run another test once I got hand of the 2N2905. Cheers...
I haven't tried it without any load, but I will check it! I haven't forgotten your question and my promise, I just don't have time nowadays to sit down and do these tests. I will be back to you soon.
Meh, I couldn't resist, so I tested it now following your description. I get similar drop. Unfortunately, I don't know why, but I will do some research and educate myself on the possible reasons.
The lm317 needs a minimum 10ma load to regulate correctly.
0:26 What is the function of that 500R resistor in TIP base-emitor?
Please check the datasheet of the LM317 chip, I took the schematics from there. I built this very long time ago and I don't remember all the details.
I am not understand the circuit but it looks and work perfectly.
Incase i want to use this circuit for 5v or less out put. Will it work?
If I want to explain it in very simple terms, the circuit works in the following way. As you adjust the potmeter, the LM317 will draw more current. The more current draw will open the 2N2905 transistor which is connected to the input of the LM317, which opens the TIP35C power transistor. So basically you indirectly control the power transistor with the current draw of the LM317 voltage regulator.
It works also for low voltages, however you have to keep in mind that this particular circuit always has a minimal amount of output voltage.
@@CuriousScientist thanks for your full details
Hello@@CuriousScientist! This is a good practical example. It is good to see things working too.
I think i am understanding most of the principle now but there is one question i cant answer to myself yet. What keeps the output voltage at LM317 regulated level instead of power supply level which is fed to the emitter? Thank you in advance!
Hi, I have built a prototype peltier cooling box originally to cool water bottles in my room and also an experiment but now I am going to use it to transport medicines (covid vaccines) in a temperature controlled environment. Anyway I have three peltiers 12706 controlled by 3 individual temperature controllers. the box is well sealed. I am using water cooling in series with the three peltiers running to a 360mm radiator. Each peltier is mounted on a alluminium plate which is 3mm thick and three plates running on three sides of the box like a "U". On the cold side of the plates which is inside the insulated box I figured the entire plates got very cold but the the cold was not getting transferred to the rest of the enclosure so I added one central fan on top to circulate the air inside. That did give some improvement and I was able to get the temperature inside the box down to 15-16*C. then I added some pentium CPU heatsinks with fans right on the surface of the plates and now I got the temp down to 11-11.5*C. the warmest I have seen the hot water coming out of the peltiers is 28 Deg and gets cooled down to 25Deg by the radiator so I know the heat extraction is not an issue. So I am still trying to figure out where I might have gone wrong with the whole thing because I want to get the inside temperature down to 0-1Deg. Also I am using a 12v 30A common power supply because I want to keep the project cost feasible instead of going for a Regulated one. Do you feel that lack of current regulation is becoming an issue here? I measured the current flowing thru each peltier and it is around 3.5-4amps only and the actual power supply voltage is 11.7V.
Also I have 12 water bottles of 1 liter each inside to simulate the load.
Any advise would be of great help so I could acheive the temperature goal.
I forgot to mention, the room temperature is around 28Deg average. and when I switch on the unit the temperature drops from 28Deg to 14Deg in less than 50mins but after that it drops to 11.5 very slowly.
Hi! This is a video about a LM317 regulator based power supply... Next time please comment under the corresponding video.
It is simple, those three TEC12706 Peltier coolers are simply not enough to cool such a large amount of material (12 kg of water). You can calculate the heat needed to cool down the water to a certain temperature (q = c * m *dT) and then you can define a time for your desired cooling speed which you can further use to calculate the cooling power needed. Also if you are running your Peltier at nearly 12 V and 4 A it means that the DT is around 60°C. So, the system provides a very low cooling power. This suggests to me that the heat transfer between the cold side and the load is still not sufficient. Use large heat sinks together with fans on the cold side.
I have a video series on my own box, check it out, you can learn a lot: th-cam.com/video/kWiiCdc3xcw/w-d-xo.html .
@@CuriousScientist my sincere apologies for posting my questions here. The only thing I had in mind was to know if adding a current regulator to my setup while the peltiers are already just drawing 3-3.5amps would make a difference and if yes then I will add a regulator to each unit to help lower the temperature further inside the box!
As I said in my other comment, you have too low voltage. The issue is not the current. But ultimately, your Peltier coolers are the real limit. Either add like 5-10 more or replace them with a few TEC12712 or TEC12715. They provide much more cooling power at the same voltage, but obviously they'll draw significantly more current so you will need maybe more PSU and also much better cooling on their hot sides.
I have seen this type of circuit using only one npn transistor with no base resistor and directly connected to base from regulator out. So will it overheat the transistor and ic when higher load current of 5 amps is drawn ?
Sorry, but I cannot comment on the circuit you described without seeing its schematics. Why don't you use this one? It is a good circuit, I guess otherwise they would not have suggested it in the LM317's datasheet.
@@CuriousScientist if I use this circuit you described, will I need to use heavy heatsink on both the regulator and transistor or only on transistor ?
As you can see, only the power transistor is attached to the large heat sink and the voltage regulator has its own tiny heat sink. It worked fine for me so far.
:26 I can not get this design to work. Did u change it in any way to make it work
No, I used the same design in my video. Make sure you don't mess up the polarity of the positive rails (I did a few times 😄). Also, the circuit requires some minimal load in order to start up.
We used a dial that was huge compared to the potentiometer. Not sure, but I think made it easier to adjust smaller increments. I could be way off on that. It may have been just so it could be used while wearing gloves
Yes, usually a multiturn potentiometer would be more precise and sensitive. Also, this is the worst possible quality potentiometer, so I am not surprised. But also, I just put this potentiometer in the circuit without any calculations, so it also might be that the value of the potentiometer or the R1 resistor is not good to cover the range properly. I will check it.
Good night friend, I'm here again.
I decided to change my project due to the infeasibility by expensive aircoolers, and I decided to do as follows, 1 block of 160mm with 2 peltier 12715 on each side, and the cooling system for the hot part of the peltiers would be 2 blocks of 120mm, 1 in each side where I would leave approximately 2cm or a little more distance between the peltiers.
In these cooling blocks used in the hot part of the peltier, they will have 2 radiators of 240mm, reservoir of 1.5L of acrylic preferably metal, a fuel pump that had left here, in the range of 150 ~ 200l / h, the temperature of mine room is in the range of 20 ~ 25 Celsius with air conditioning, here it is very hot, do you think that changing this way would make better use of the peltiers? or would need even better cooling
Hi! I haven't forgotten about you, but I need some time to carry out these experiments, and I am still waiting for a part. As I said, 1 TEC12715 Peltier already has a very high thermal output at the specifications that you mentioned. I was using a 12x24 cm CPU cooler radiator for a single TEC12715, and it was on the borderline. If you are planning to add one more, you won't be able to cool your Peltiers properly. Use smaller Peltiers (TEC12708 for example) where the Joule heating is smaller and you can benefit from the cooling power more. Keep checking back to my channel, I will upload more videos, but as I said, I am waiting for a part and it takes time to produce these videos.
Curious Scientist even using TEC12715 with a lower voltage and amperage would it be better to use Tec 12708 close to the limit?
Hmm. I did the math, and running both devices at 6 A, expecting 40°C DT, TEC12715 produces ~44 W Joule heat, while TEC12708 produces ~70 W. BUT! 12708 provides 36 W cooling power, while TEC12715 provides 26 W. So, yes, you produce significantly higher amount of heat with the TEC12708 at low current, but it gives you more cooling power.
Now you have to decide. Do you want to sacrifice or not? If you use 4 units of TEC12708, you will have 144 W of cooling power and a 12 V, 30 A PSU would be a perfect fit for this task. If you use 4 units of TEC12715, you will "only" have 104 W of cooling power. Do you need that extra 40 W cooling power in exchange for ~120 W extra heat that you have to get rid of? If you can cool properly, you should go for the TEC12708. In an upcoming video, I will test a very cheap but potentially very powerful radiator. If that works as I expect it, then the waste heat of Peltiers will not be an issue anymore.
If you want to replicate the cooling performance of the TEC12708 at 6 A, you have to bump the current to ~7.5 A on the TEC12715. This will need roughly 8.6 V. You will have roughly 39 W of cooling power and 64 W rejected heat. The problem is that you will need a more powerful power supply. This is 30 A if you use let's say 4 units (4*7.5 A) and I personally would not run it with a 30 A, but rather a 35-40 A power supply to have some safety margin. It is up to you. And if you don't mind, I will not write such a detailed answers anymore, because I am not a free support service. Watch my videos, I show every details in them. For example, how to calculate the above numbers using the performance chart. Or wait 1-2 weeks, I will publish videos on tools for performance charts (and the tools too).
Dear, my advice is not to touch any electric or electronic components with fingers , always use proper tools i.e. heat sensors temperature. This is hazardous even with low voltage and current
Thank you! I already knew what will happen, so I was not worried about being burned or shocked. I tested the circuit prior to this video. Despite all of these, your advice is still useful of course, one should be careful when touching wires under (high) current.