I was playing around in LTSpice with this "current booster" circuit I found on this paper but it doesn't seem to work as I expected.

I set up R2 and R3 in order to get a Vout of about 14.82V. If we neglect the voltage drop on D1 (just for this example's sake) and fix R4 to 14.82 Ohm then I would expect around 1A flowing through R4. But that's not exactly what happens, instead Vout drops to about 12V, through R4 there's about 0.8A which is fine, and Q1 is basically useless since almost no current flows through it. If I try to get more current through R4 the result is that Vout goes down even further and Q1 doesn't turn on.

I expected Q1 to turn on and Vout to stay close to the expected 14.82V. Why is this not happening?

I simulated the circuit both here and LTSpice so I think I might be the one missing something.


simulate this circuit – Schematic created using CircuitLab

EDIT: I looked a bit on the datasheet and at page 17 there's a circuit that does the same thing as the one that I proposed above and it works fine in LTSpice. If you don't mind the fact that Q2 is dissipating about 33W (ie probably frying) the circuit works as expected: the voltage right before D1 is about 14.80 V as expected and about 2.8 A flow through R4. Here's a picture of the new circuit: enter image description here

  • \$\begingroup\$ Your output voltage is about \$13.8-13.9\:\textrm{V}\$. (Depends some on ADJ pin current.) With the drop across \$D_1\$ (through a small signal diode, cripes), this means perhaps as much as \$900\:\textrm{mA}\$. That's about \$630\:\textrm{mV}\$ via \$R_1\$ and that's not really enough to turn on \$Q_1\$ (much.) Of course, \$Q_1\$ is a small signal BJT. But I guess you are just trying to get it to turn on. So lower \$R_4\$'s value a bit? \$\endgroup\$
    – jonk
    Commented Jul 10, 2017 at 22:40
  • \$\begingroup\$ @jonk If I lower R4 however the voltage on Vout drops too much. Anyway I think I found a circuit i can start to work with (see edits). \$\endgroup\$
    – mickkk
    Commented Jul 10, 2017 at 23:03
  • \$\begingroup\$ If you'd just increased \$R_1\$ in the first schematic to twice (or larger) the value, it probably would be simulated. Or used a higher current BJT that kicks in at a lower \$V_{BE}\$. (Not that I'm arguing it was a good cricuit, regardless.) In your latest circuit, of course \$Q_2\$ is bearing the brunt. Most of the current is through it and it has to drop \$10\:\textrm{V}\$! But that's linear regulation for you. \$\endgroup\$
    – jonk
    Commented Jul 10, 2017 at 23:30

2 Answers 2

  1. Loose the diodes on the output. I can't even guess what you think they do for you.
  2. Increase R1. You want the transistor to kick in well below 1 A. It probably won't do much useful until you get over 1 A thru R1 as it is. I'd use at least 1 Ω.
  3. Use a real power transistor.

I know it's a simulation but what do you expect with puny BJTs (200 mA peak current for the 2N3906) and 1N4148 (300 mA) signal diodes. May I also suggest you go to google and type in: -

LM317 current boost circuit

Then select images and pick one that doesn't use diodes in the outputs AND is supported by good information from the web site it comes from

  • \$\begingroup\$ Ok I'll look for it, however, I don't understand why the BJT does not turn on. \$\endgroup\$
    – mickkk
    Commented Jul 10, 2017 at 22:18
  • \$\begingroup\$ And if considering such a circuit to construct, first consider a switchmode supply. SimpleSwitcher parts and a dozen more are easy to use and are about 95% efficient, whereas any BJT and 317 will waste lots of power as heat. \$\endgroup\$
    – rdtsc
    Commented Jul 10, 2017 at 22:18
  • \$\begingroup\$ @mickkk your new circuit - why it gets so hot - your input is 24 volts and your output is 5 volts - something has to drop 19 volts for the math to add up and that gets dropped at a current of several amps. If current is 1 amp then power lost is 19 watts. Please consider using a switch mode buck converter as suggested by rdtsc. If you need help choosing just ask. \$\endgroup\$
    – Andy aka
    Commented Jul 11, 2017 at 7:08
  • \$\begingroup\$ @Andyaka Ok, I understand. Thanks for the suggestion I'll surely consider using that. I'm experimenting a bit on how to build a very simple and basic charge controller. My idea is to make some sketches, then a few simulations, then if everything looks good I'm going to buy the parts and make some real tests. \$\endgroup\$
    – mickkk
    Commented Jul 11, 2017 at 20:32
  • \$\begingroup\$ You have been here long enough to know what the most appropriate form of thanks is. \$\endgroup\$
    – Andy aka
    Commented Jul 11, 2017 at 20:37

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