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I'm trying to alternate between two 3W LEDs (forward current of 750mA, forward voltage of 3.6V). I've got a 555 set up in astable mode, and am driving the LEDs via npn and pnp transistors. In practice, the LED that's run via the pnp transistor gets more current when it's switched on compared to the LED connected to the npn, and I'm not sure why. I'd just do away with the transistors all together and let the 555 sink/source the current for me, but I believe it can only handle up to 200mA, and the LEDs need more. FWIW, I'm running this circuit off of 2 3V coin cell batteries, and when connected directly to the LEDs they are extremely bright.

So my questions are: 1. Is there a configuration change I can make to get more current through the transistors? (more base current maybe?)

  1. Is there a better or more efficient setup to perform essentially the same function?

Thanks! Imgur Schematic

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  • \$\begingroup\$ 750mA out of coin cells?! That is a very short-lasting, expensive solution, and coin cells usually aren't even designed for such extremely high currents! \$\endgroup\$
    – Marcus Müller
    Jul 3, 2018 at 1:06
  • \$\begingroup\$ And aside from the LEDs, the NE555 would already be a waste of energy that I wouldn't recommend running off coin cells, as it would eat through these in no time. \$\endgroup\$
    – Marcus Müller
    Jul 3, 2018 at 1:08
  • \$\begingroup\$ "npn and pnp transistors" - do you think you could be just a little more specific? \$\endgroup\$
    – brhans
    Jul 3, 2018 at 1:13
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    \$\begingroup\$ I give you minutes at most - let's say a coin cell has a rated capacity of 200 mAh at small discharge currents until it drops to 2V. At 750 mA, my rough estimate for an upper limit of how much you can actually get from that coin cell is 10% - so, 20 mAh, or 20/750 h worth of operation; that's but 16 minutes. I doubt it's that much - internal resistance will cut you off much earlier. \$\endgroup\$
    – Marcus Müller
    Jul 3, 2018 at 1:29
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    \$\begingroup\$ I mean, it's your money, and whilst I do have environmental concerns about this, there's probably good reasons you want these coin cells – problem really is: these coin cells really aren't designed to deliver 750 mA, and that means their internal resistance will eat up most energy. If you want a more effective circuit, replace with a different type of battery. (measure the voltage across the coin cells when connecting LED + R across them; it'll be below 6V. How far below is defined by the internal resistance) \$\endgroup\$
    – Marcus Müller
    Jul 3, 2018 at 1:50

2 Answers 2

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Coin cells or no coin cells, the base resistors R3 and R4 are much to big. You will need to chose transistors that can sink more current and feed them with a base current of 15mA or so (divide your desired LED current by 50, typically). Your base resistors need to be a few hundred ohms.

Also, you shouldn't use a PNP as a low side switch. It will only act as an emitter follower and never reach saturation. Move the PNP to the high side...that is, between the anode of D2 and the supply voltage.

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In your currently drawn circuit, I don't see how Q2 would turn on all the way. It should be on the high side.

Here is a sketch of something that might work for you. You will need to pick FET's with low Rds on (maybe around 300 mOhm or less). Rds must be specified at 4.5V or lower.

schematic

simulate this circuit – Schematic created using CircuitLab

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