What the heck does this circuit do?

[poodyp]

So I pulled apart the flashlight that came with my set of tools and found this board inside. I can't tell what it does other than possibly a super complicated 12v regulator. It's an 18v Makita li-ion battery with a 12v incandescent bulb. I'd like to replace it with a high power LED and I don't want to accidentally pull out some protection circuitry.

Attached is my attempt at reverse engineering it, and here are pictures of the back and front. It's super late so it's highly possible I failed at it, but I can't wrap my head around what it's supposed to do. I couldn't get the right models for all the parts so I tried it with other parts. Q1 is a C9015, Q2 is a B773, and U1 is a TL431A. All it seemed to do is drop 6 volts, and when I lowered the input voltage the output dropped proportionally. Also somehow V1 was sinking 5KA...

Can anyone please help?

[Psi]

You've got some errors in there, but i think it's intended as a regulator and also a protection system to stop the lightbulb from discharging the lithium battery below it's min volts and causing battery damage.

[BravoV]

Although I have not try to analyze that circuit, I recognized that you're using LTSpice for drawing the circuit.

Actually if you could substitute some of the transistors with similar one provided in the LTSpice library, and try run the simulation to have an idea how it works, provided that you've captured the schematic correctly.

Btw, that TL431 can not be replaced by LT1009.

I included as attachment in this post the TL431 model for that LTSpice, just save the TL431.ASY inside the \SYM folder or your own custom made folder and the TL431.SUB in the \SUB folder.

Hope this helps.

[amspire]

Lots of errors in the circuit.

Q1 will be NPN I think - a 2SC9015 transistor.

Q2 will be PNP - a 2SB778 6A power transistor that will control power to the light bulb.

I suspect that the emitter of Q2 probably connects to the battery +ve and the collector connects to the light with the other end of the light going to the battery -ve if it is a switching regulator.
That probably means R5 is an emitter to base resistor.

U1 probably has the reference and cathode leads swapped.

I think Psi will be right, the torch has to turn hard off when the lithium cells drops below 3V each.  It is also possible that a switching circuit is used so that instead of regulated 12 V, the battery voltage is pulse width regulated to make the average volts 12 V. If this is right, then the capacitor must form the time constant of the switching regulator to set the switching frequency.

A few too many errors to take a stab at exactly what the circuit is meant to look like.

You better check the transistor and TL431 lead pinout again and have another stab at it.

Richard.

[poodyp]

I included as attachment in this post the TL431 model for that LTSpice.

Thank you for the model, I've added it to my circuit.

Lots of errors in the circuit.

Q1 will be NPN I think - a 2SC9015 transistor.

Q2 will be PNP - a 2SB778 6A power transistor that will control power to the light bulb.

I suspect that the emitter of Q2 probably connects to the battery +ve and the collector connects to the light with the other end of the light going to the battery -ve if it is a switching regulator.
That probably means R5 is an emitter to base resistor.

U1 probably has the reference and cathode leads swapped.

I think Psi will be right, the torch has to turn hard off when the lithium cells drops below 3V each.  It is also possible that a switching circuit is used so that instead of regulated 12 V, the battery voltage is pulse width regulated to make the average volts 12 V. If this is right, then the capacitor must form the time constant of the switching regulator to set the switching frequency.

A few too many errors to take a stab at exactly what the circuit is meant to look like.

You better check the transistor and TL431 lead pinout again and have another stab at it.

Richard.

You're absolutely correct about Q2 and U1, I had those mixed up. But are you certain about Q1? Searching for C9015 comes up with this, a pnp transistor, but I can't find anything for 2SC9015.

I took the pictures of the board and mirrored the bottom pic and used it to draw the traces on the top one you can take a look here if anyone wants to try their hand at simulating it. I made another attempt at simulating the circuit and it still seemed to not do much. Tried with Q1 as both a pnp and npn. This time the output pretty much matched the input, and never dropped out. I found a multimeter (mine's in a box somewhere) and measured the output and currently the battery is at 26v (Isn't that really high for 5 li-ions? Maybe this multimeter is a little janky), and the output is about .5V less. Maybe I can set up a battery logger and see exactly what happens.

[amspire]

If the Makita battery is a 6 cell lithium battery, it would be between 25.2 and 25.8 when charged, and 18v when fully discharged.

Now if you replace the 700K resistors with 100K, then the circuit starts to make sense.  700K always sounded suspicious as it is not a normal value.

U1 and Q2 the form a 12v linear regulator and it looks like C1 is to have a slow start up which will be easier on the switch, the bulb and Q2.

If it is a pre linear regulator, I am a bit surprised, as then are throwing away 50% of the battery power in heating up Q2.  That is why I was think it must somehow work as a switching regulator.

And you are right Q1 has to be a PNP transistor. It is off when the battery is applied. when the battery is switched off, it turns on getting its base current through R1 to discharge C1. Change Q1 to a pnp and change the 700K resistors to 100K and you have a working circuit.

All makes sense, but very inefficient.

Richard.

[amspire]

It looks like there is nothing to shut the circuit down when the battery cells reach 3V, (about 18V on the battery)  so hopefully, that is built into the batteries themselves.

It is dangerous to let the lithium cells go below their minimum rated voltage.

Richard

[poodyp]

Well I changed the batteries in the multimeter and got 22.3v in, 22v out. I know the pack is 10 cells, arranged 2x5. There's also a chip inside, it monitors temperature, limits current, and stores the number of charges so they can check how old the battery is. I also know if a cell goes bad it's impossible to repair and have it charge from the original charger. If this circuit doesn't do protection then it must be the battery because the light will just shut off, so I guess it's safe to just eliminate it.

[amspire]

It looks like the Makita batteries have 5 pairs of Sony 18650V Li-ion cells. The pairs are in parallel to make 3.2AHrs.

These are 4.2V charged, 2.5V discharged.

So charged, the battery should have 5x x4.2 = 21V absolute maximum.  It is dangerous to exceed this, so you shouldn't have 22V on the batteries.

It could be that your multimeter is reading a bit high.

Discharged, you will get 12.5V out - so your regulator can keep 12V to the bulb all the way from fully charged to battery cutout.

Still pretty inefficient. Using a switching regulator circuit, you could probably get 30% - 40% more energy to the bulb from the battery.

But I guess it works, and you get a nice even light out.

Richard.

[Zero999]

700K always sounded suspicious as it is not a normal value.

Yes, it's not a standard value. It's possible he misread the colour code or his meter is slightly off.. 680k and 698k are the nearest 5% and 1% values respectively.

[amspire]

In this case, the 700K resistors have to be really 100K resistors.

With 100K, you get 12V output. With 700K, you get 3.8V output.

Either a brown band somehow looked purple in the light, or "100K" was printed on the board with a "1" that looked a bit like a "7 I guess.

Richard.

[poodyp]

I think I just read it backwards. Violet Orange Black Black Brown is a 1% 700k, but reversed it's a 100k .1% resistor. I haven't tested anything as I only just found this multimeter and it's got a half dead battery. I'll do some testing once I have a chance to get my multimeter from home.

At any rate I'm just going to replace it and find a suitable LED/driver combo, or make my own driver.