This circuit is very popular on the Internet. I understand it but I have a question about the PNP transistor. The base of the transistor is connected directly to the regulated output of the transformer, so when the transformer produces 9 volts the base voltage will be greater than or equal to the emitter voltage and the transistor will open the circuit.

That's OK. But when there is no power in the transformer, the output of the transformer is floating and the transistor's base as well. I tried to find a path that connects any negative voltage to the base so that it closes the circuit and the LEDs turn on but I didn't find one!

Would you tell me from where the transistor's base gets its negative voltage? Or at least a voltage that is less than the emitter voltage by 0.7 volts?

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1 Answer 1


The quiscent current through the LM317 (from its input to its ADJ pin) is what pulls the base of T2 toward ground.

Note that the datasheet gives this current as 50 µA typical, 100 µA max — and doesn't specify a minimum value at all.

This means that there's no guarantee that there will be enough base current to turn T2 fully on. For a typical \$h_{FE}\$ value of 100 (BD140 datasheet), this means that LEDs can only draw between 5 and 10 mA.

It would be a good idea to add a resistor from the regulator input to ground in order to supply more current to the transistor when the mains power is off. If those are 20 mA LEDs, you'll want to be able deliver a total of about 240 mA, so you'll need a base current of at least 2.4 mA. With a 6-volt battery, you'll need a total resistance of about 2200 Ω. Since R15 is already 1000 Ω, your additional resistor should be 1200 Ω.

When the line power is on, this resistor will have 9 V × 1.414 = 12 V across it, so it will need to dissipate \$\frac{(12 V)^2}{1200 \Omega} = 120 mW\$, so make sure you use a 1/4 W resistor.

  • \$\begingroup\$ Does that mean a voltage regulator must be used in automatic emergency led circuits ? I can not use just a capacitor as a regulator, can I ? \$\endgroup\$ Commented Sep 22, 2015 at 14:31
  • 1
    \$\begingroup\$ A capacitor is not a regulator. You need a DC path for the base current. See the additional analysis above. \$\endgroup\$
    – Dave Tweed
    Commented Sep 22, 2015 at 14:38
  • \$\begingroup\$ What's the purpose of the T1 plus Zener circuit? Over voltage protection for the battery if someone adjusts VR1 too high? Looks like it would pull the adjust to 0V if I understand it right. \$\endgroup\$
    – confused
    Commented Sep 22, 2015 at 15:24
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    \$\begingroup\$ @confused: Yes, it seems to be a redundant voltage limiter that will kick in at about 7.5V, which is approximately the desired "float charge" voltage for a 6V lead-acid battery. Ordinarily, I'd expect to see the LM317 wired as a current limiter, but in this circuit, is wired as an adjustable voltage regulator. \$\endgroup\$
    – Dave Tweed
    Commented Sep 22, 2015 at 15:41
  • \$\begingroup\$ In actuality, to attempt at maintaining regulation, the LM317 will try to keep the original 1.25V across R1, which would round to 7mA if it could, but of course it can't. But in the end, from experience, it'll become semi-transparent as presenting 60% to 150% of its output load resistance to the base resistance in this setup (and more likely around 80%), so with VR1 at maximum the base is loaded with about 5k worst case. Still not enough, but far from as bad as you estimated without helping resistor. Otherwise, since adding guarantees is good in all cases: +1 \$\endgroup\$
    – Asmyldof
    Commented Sep 22, 2015 at 15:56

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