In order to test a Li-Ion battery charger on our board (the BQ24074) I built a battery simulator circuit which was initially designed by TI. The circuit "pretends" to be a battery and can provide different voltages to study the behaviour of a battery charger. I created a PCB with the following schematic:

Battery simulator circuit designed by TI

I'm providing a variable voltage to the VIN input using an LM317 regulator. I connect the BAT output to the battery charger like suggested in the application note of TI. I've tried to measure the battery charging cycle of the BQ24074 PMIC, but while testing I came across a few problems regarding the power supply for the battery simulator.

Since I don't have a dual voltage supply to provide +/-12 V I tried to build a voltage rail splitter. I use a 24 V supply and the following circuit I found online and in other posts:


simulate this circuit – Schematic created using CircuitLab

I first tested the battery simulator without connecting the battery charger. I measured the expected +12 V and -12V on the voltage rail splitter. I also measured the expected VBAT output voltage of the battery simulator. Until now everything is working fine.

When I connect the battery charger and actually start the real test I notice that Q1 (BD239) of the rail splitter gets very hot! It ramps up to 100 degrees celsius and I'm forced to turn off the 24 V supply. The battery charger has been setup with a charge current of 450 mA. From the datasheet of the BD239 I see that the collector current (Ic) is 2 A, so it should be able to handle this current?

I was actually using a BC548 for Q1 of the rail splitter but I've already burned 2 of them while testing. What is the reason that this NPN transistor is getting so hot and burns? Does it have to sink that much current from the battery charger? I have tried adding some schottky diodes to the +/-12 V lines to prevent potential back powering but it didn't help.

I haven't been able to solve this problem and was thinking of just using an old PSU to provide the circuit with +/-12 V. Any suggestions?

  • 2
    \$\begingroup\$ Do your power calculations. If you've got 450 mA (let's say 0.5 A) through and 12 V across the transistor then P = VI = 6 W. That will require a decent heatsink. \$\endgroup\$
    – Transistor
    Jun 20, 2022 at 17:27
  • \$\begingroup\$ Neki, a rail splitter will want NFB and look a little more complicated. Like this, perhaps? And it will dissipate power. \$\endgroup\$
    – jonk
    Jun 20, 2022 at 18:00
  • \$\begingroup\$ I guess it would indeed require a decent heatsink now when I look at it @Transistor, thanks. \$\endgroup\$ Jun 21, 2022 at 10:20
  • \$\begingroup\$ Regarding the rail splitter with negative feedback @jonk, I didn't have an OpAmp at hand to build a rail splitter with a voltage follower. But in your example, what is all the extra circuitry for (Q5, Q13, Q14, the potmeter)? Wouldn't something like this be sufficient? \$\endgroup\$ Jun 21, 2022 at 10:20
  • \$\begingroup\$ @NekiDzemaili No, I don't think so. And especially not incorporating Transistor's 500 mA figure in the comments. \$\endgroup\$
    – jonk
    Jun 21, 2022 at 17:13

1 Answer 1


Here is a very simple idea:


simulate this circuit – Schematic created using CircuitLab

Rail splitter

It might be possible to make it more efficient by using a 24 V to 12 V DC-DC converter.

(edit) I ran a parameter sweep for R1 and R2, The simulation shows that the 12V regulator can sink current as well as source, although that might not actually be the case. Anyway, it should be possible to adjust loads on the positive and negative rails so that the regulator will not have to sink current.

Parameter sweep R2: Parameter sweep R2

Parameter sweep R1: Parameter sweep R1

  • \$\begingroup\$ Please explain the reason for the downvote. Is there a reason why this would not work? \$\endgroup\$
    – PStechPaul
    May 29 at 23:21
  • \$\begingroup\$ Maybe because the regulator output cannot sink current from Vdd? (it wasn't me!) \$\endgroup\$ May 30 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.