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I am a novice with electronics, I may know what my issue is but wanted to put this out and make sure my thinking is correct.

The below circuit is one I am using to turn on an LED when charging a battery and turn off when the battery is complete for an amplifier I've designed. What is happening is the LED never turns off. 0.74V is coming from a pin on the charging module's IC. When it turns off it is about 10V. This was working fine with a 12v AC adapter as I was waiting on my 15V adapter to come from China. Foolish me, I completed my PCB design and placed the order before I was aware of this issue. Sure I could just continue to use a 12V adapter but the circuitry is more efficient with the 15V when charging and listening at the same time.

I am hoping I just need a different transistor, my current one is a 2N3906. From looking at the datasheet I am thinking my Emitter−Base Breakdown Voltage of 5 is not high enough and if I switched to something like a KSA708 with an Emitter−Base Breakdown Voltage of 8 I may get the result I am looking for with this project. Any insight would be greatly appreciated, thanks.

enter image description here

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    \$\begingroup\$ You didn't state an actual problem. Wild guess : the LED no longer turns off when charged.R2 and R3 form a potential divider. With 10V on the input this sets the base voltage positive enough to turn off the LED when V+ is 12V, but not when V+ is 15V. Work out Vbe for both cases, and find values for R2,R3 that turn off Q1 at 15V. \$\endgroup\$
    – user16324
    Commented Jul 24, 2020 at 21:05
  • \$\begingroup\$ Why the heck are you using 15V for an LED? You must use an NPN, not PNP \$\endgroup\$
    – D.A.S.
    Commented Jul 24, 2020 at 21:09
  • \$\begingroup\$ Brian you are correct, it no longer turns off. Sorry for not stating that originally. I will play with different resistor values and see what I come up with as a potential solution. \$\endgroup\$
    – user3348186
    Commented Jul 24, 2020 at 21:09
  • \$\begingroup\$ Consider using 2 or 3 green LEDs in series. \$\endgroup\$
    – JRE
    Commented Jul 24, 2020 at 21:15
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    \$\begingroup\$ emitter-base breakdown is not a the problem \$\endgroup\$
    – Jasen Слава Україні
    Commented Jul 24, 2020 at 22:32

5 Answers 5

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Your issue is that you have Q1 configured as an emitter follower. So the voltage on the emitter will be about 0.7V higher than the voltage on the base.

With your 12V adapter, when the charge ends you didn't have enough voltage across the diode to turn it on.

With the 15V adapter you'll have about 10.5V on the base when charging is done, and 11.2V on the emitter. If your diode drop is say 2V you will have 15-2-11.2 = 1.8V across R1, and 1.8mA in the diode. Probably enough that it still glows.

No easy solution by changing the transistor, you'll have the same result. It has nothing to do with breakdown.

If you don't want to change your PCB you could try a bigger resistor for R1. Increase it until the LED no longer glows and see if you get acceptable brightness during charging.

The better approach would be to use the transistor as a switch, but that's a PCB layout change or lots of cuts and jumpers. JRE's suggestion of putting some LEDs in series would also work.

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  • \$\begingroup\$ I was going for the switch approach but clearly I got that wrong as I am used to using NPNs. \$\endgroup\$
    – user3348186
    Commented Jul 25, 2020 at 2:26
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Any insight would be greatly appreciated

Try adding a 10 volt zener diode (rough and ready level shifter) like this: -

enter image description here

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  • \$\begingroup\$ Or just replace R2 with the Zener \$\endgroup\$
    – Bruce Abbott
    Commented Jul 24, 2020 at 22:42
  • \$\begingroup\$ I think R2 should stay as current limiter between input and Q1 base. \$\endgroup\$
    – user105652
    Commented Jul 25, 2020 at 0:03
  • \$\begingroup\$ I tried the 10v zener and the light is still on. Should I try a different value zener? \$\endgroup\$
    – user3348186
    Commented Jul 25, 2020 at 0:38
  • \$\begingroup\$ Did you have the zener the right way round (as shown above)? Can you measure the voltage across R3? \$\endgroup\$
    – Andy aka
    Commented Jul 25, 2020 at 7:28
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use a FET instead. a p-channel depeltion MOSFET or JFET will turn off when it sees 10V on the gate, and turn on for a low voltage there.

schematic

simulate this circuit – Schematic created using CircuitLab

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enter image description here

Even an NPN would always be only with 1K to 10K to 15V. Maybe 33K is OK or no 10k pull-up needed if using logic levels. REF

Summary:

  1. Change PNP to NPN
  2. Remove 10K for any logic level 1V to 5V ON.
  3. Invert output in software if possible.
  4. Define input signal & state when ON/OFF with Voltage and Z. Let Ic/Ib=20 roughly.

The left side works. Right side (no Good) PNP will always be ON.

enter image description here

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  • \$\begingroup\$ See REF simulation , then yank the 10K resistor off and then it works \$\endgroup\$
    – D.A.S.
    Commented Jul 24, 2020 at 21:39
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So I think I am just going to bite the bullet and amend my PCB design. I tested this design last night but welcome all feedback. Making this an appropriate high side switch with a 20K resistor on the gate and 1k as a pull-up seems to work.

enter image description here

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