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For the LM317 there is a precision current regulator circuit. This circuit uses an LM317 and only one resistor between the out and adjust pins. The operation of this circuit completely satisfies me, since I do not need to regulate the voltage, but I need more current than 1.5 A.

How do you recommend doing this? In all the suggested circuits there is also a part for voltage regulation, which I do not need; only current is needed.

I have seen proposals for parallel connection of LM317s, but not everyone approves of this, and there is also a part for voltage regulation.

My idea is to use an LM2596-ADJ as a low-loss voltage regulator and after that use an LM317 for current limiting. At the same time, I will place the connection point of the feedback output already behind the LM317 output. But the whole problem is in the maximum current of 1.5 A for the LM317. I need about 2.7 A to charge a battery.

schematic

UPD: I have seen circuits using PNP and NPN transistors, but there is a voltage adjustment using LM317. I want LM317 to correct current only, and 2596 to control voltage.

UPD2:

Now my diagram looks like this. new schematic I did not buy 2n3906, but I have bc327 and bc557. Perhaps there is a problem with them, or this circuit is not designed for a battery. According to the recommendations here, I abandoned the lm317.

Now the problem is that I set the voltage at idle, and when I connect the battery, it goes down and the current also becomes less. If I increase the voltage with the trimmer, the current also increases. But as the battery charges, the voltage increases and the current, respectively, also increases. They probably want to reach idle values.

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  • \$\begingroup\$ Why does it need to be a linear LM317 regulator, and what batteries are you charging? \$\endgroup\$
    – Justme
    Jan 18 at 0:22
  • \$\begingroup\$ Is the LM317 already integrated into the circuit so you can't change it? I really don't understand this. If you can change the voltage regulator, why can't you change the LM317? \$\endgroup\$
    – pipe
    Jan 18 at 1:41
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    \$\begingroup\$ Please do not clarify your question in comments, add that information by editing your question. Down here it will be overseen. You might want to take the tour to learn how this site works, and read some of the help center. -- Did you consider or even try the common current boosting with a PNP? I think it should work. \$\endgroup\$
    – the busybee
    Jan 18 at 7:05
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    \$\begingroup\$ "but I need more current than 1.5 A" Forget about LM317 then, it will melt through the floor. As for battery chargers, they typically work with PWM controlling a MOSFET. No need to reinvent the wheel. \$\endgroup\$
    – Lundin
    Jan 18 at 7:57
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    \$\begingroup\$ @EdinFifić As much as I sympathise as well, this comment field is not really the place for these kind of discussions. Please keep it on topic. As for helping out interactively, the main chat is probably a good place for that. \$\endgroup\$
    – Lundin
    Jan 18 at 12:21

3 Answers 3

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Add a PNP transistor or other voltage detector into the output to feed current to the FB node when the current hits the limit. In this way the switched-mode regulator chip will throttle back when the current limit activates.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Didn't you misplace the output freewheeling diode? It goes right next to the OUT pin. Also, let him know he needs at least a 2W resistor or a combination of resistors, due to the power dissipation across it (the R1). Plus large input and output capacitors (min. 220uF in, 470uF out, low ESR). \$\endgroup\$
    – Edin Fifić
    Jan 18 at 10:32
  • \$\begingroup\$ I sure did! thank you. \$\endgroup\$
    – Jasen Слава Україні
    Jan 18 at 10:38
  • \$\begingroup\$ @Jasen, in general, this circuit resembles a typical circuit for xl4015 or xl2596, except for the transistor and resistor that are added. How to calculate the current in this circuit? \$\endgroup\$
    – Deltinos
    Jan 18 at 10:54
  • \$\begingroup\$ @Deltinos The current is calculated by taking into account that 0.6V is needed across the base-emitter junction for the transistor to conduct. \$\endgroup\$
    – Edin Fifić
    Jan 18 at 10:55
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    \$\begingroup\$ @EdinFifić, I deleted one of my comments when the thought came to just try to apply Ohm's law, because regulation is secondary, as I understand it. First of all, the current simply flows through the resistor. And I came to this meaning. Thank you! \$\endgroup\$
    – Deltinos
    Jan 19 at 6:44
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The LM317 is more than you need, yet not enough. It has a high overhead voltage so will have a lot of loss, and it’s limited to 1.5A.

A simple current limiter can be constructed from two NPN transistors, which will have much less overhead (only a half a volt or so) and can be scaled up to high current by choosing an appropriate pass NPN.

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  • \$\begingroup\$ Thanks for the advice! I'll think about it, look for an example circuit, and try if necessary! For now, I'll try the above diagram. \$\endgroup\$
    – Deltinos
    Jan 19 at 6:47
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National Semiconductor has other 3-terminal regulators with higher current ratings.

LM350 - 3 A

LN338 - 5 A

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  • \$\begingroup\$ Thanks, I've been thinking about them. It is even possible to order them. This is the easy way. I'll see what works best. Logically, PWM is indeed more correct. \$\endgroup\$
    – Deltinos
    Jan 18 at 19:43

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