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I am building a LED spinner circuit and I am at the point of optimizing it. The whole circuit itself only draws about 10-20mA max. I was today looking at this part of the circuit: LED spinner on/off transistor

Now as you can see, when my switch is at position 5, it turns the circuit off. But, now when my circuit is off, there is still current flowing through the pull down resistor, draining the battery. I know this is a very small current, but I was wondering if there was a way to make this switch so that it does not draw any current when switched off.

Edit: I should have maybe put the whole circuit in. Full Circuit

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    \$\begingroup\$ There will always be some kind of leakage in "off" switches. You could use bigger resistors, or a FET in place of a resistor with an extremely high open/off resistance, but you will always have some leakage. \$\endgroup\$ – schadjo Sep 30 '19 at 17:30
  • \$\begingroup\$ I understand that with most solutions there will be leakage as well as during my on time there is wasted current going through that transistor to the pulldown resistor. I was just curious if there was a way to completely stop the current when the circuit is off and I must thank Dave for the answer to my question. \$\endgroup\$ – Francois landry Sep 30 '19 at 20:11
  • \$\begingroup\$ you can save 9 resistors by putting the resistor after the LEDs instead of before, you can also get a kind of dual brightness effect by putting a resistor in series with the 4017 VCC \$\endgroup\$ – Jasen Oct 1 '19 at 4:15
  • \$\begingroup\$ I didn't think of that, thanks! that will make room for the diodes i'm adding. Also, could you explain a bit more of the dual brightness effect, I'm not seeing how adding a resistor there would do that. \$\endgroup\$ – Francois landry Oct 1 '19 at 12:34
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    \$\begingroup\$ @Francoislandry it's magic! actually the 4017 can accept power threough the protection diodes on the clock input, so with a resistor in the main supply it gets a lower voltage and can receive a relative voltage boost when the 555 output is high. \$\endgroup\$ – Jasen Oct 1 '19 at 20:14
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Note that the current is wasted regardless of whether the circuit is "on" or "off" — when it is "on" the voltage drop across R11 is only slightly less than when it is "off".

Using a PMOS transistor instead of the PNP would mean that the pulldown resistor could be on the order of megohms, reducing the "leakage" current to microamps.

Or you could use a different strategy altogether, eliminating the off-state current entirely:

schematic

simulate this circuit – Schematic created using CircuitLab

Better still, combine both ideas and get minimal wasted current in the on-state, too:

schematic

simulate this circuit

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  • \$\begingroup\$ I think you'll find that this circuit will be slow to turn off. because C1 will back-feed your Q1. but at 20mA that should be mostly harmless. \$\endgroup\$ – Jasen Oct 1 '19 at 4:20
  • \$\begingroup\$ @Jasen: Slow only in the sense that the circuit won't turn off until the current timing cycle completes and the 555 pulls pin 7 low. Hmmm -- however, once power is removed, pin 7 will no longer be active, and the residual charge on C1 may cause the circuit to re-power briefly, and there may be a series of such oscillations until the charge on C1 is completely gone. \$\endgroup\$ – Dave Tweed Oct 1 '19 at 14:57
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    \$\begingroup\$ First: where did my other comments go?? Second: As long as it shuts off completely, even if its after a few seconds(without damaging anything of course) it should be fine. Since the charging for the capacitor comes from the switch being in one of the three ON positions, I dont see it coming back on fully. \$\endgroup\$ – Francois landry Oct 1 '19 at 15:08
  • \$\begingroup\$ I deleted your other comments because they served their purpose of causing me to revise my answer (again). If you're fine with the odd circuit behavior that I described, then go for it. It won't damage anything, and you still get the zero off-state current. \$\endgroup\$ – Dave Tweed Oct 1 '19 at 15:11
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    \$\begingroup\$ That's easy -- the NE555 isn't specified for operation below 5.0V. \$\endgroup\$ – Dave Tweed Oct 1 '19 at 21:14
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  • You could use a PMOS FET in place of Q1. Then R11 could be 50k or 100k instead of 10k, reducing leakage in the off position.

  • You could use a separate "off" switch, or a special rotary switch with a special "off" position that disconnects VCC from the transistor altogether.

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You could use three Schottky rectifiers in place of the transistor and pull-down. Place anodes to switch pins 1, 2, 4, cathodes tied together to "feed main circuit." Disconnect pin 5 so it becomes "true off." The "feed main circuit" will be about 0.25v lower than Vcc.

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You could replace all of the parts in this design except for the switch, battery, and LEDs with a microcontroller and it would have lower off power, lower running power, and likely even lower cost.

The off power savings are thanks to the fact that a modern microcontrollers (like AVR) can use as little as 0.1uA while sleeping, and can wake on a change on one of their input pins.

You connect the micro directly to the power source and then attach the active switch contacts to IO pins. You can enable internal pull-ups on these pins and then use a pin change interrupt to wake from low power sleep. The "off" position need not be connected to any pin - the MCU knows that if none of the other pins are active for more than a certain timeout that the switch is in the off position and it goes to sleep until the switch is moved. The pull-ups do not use any power when the switch is in the off position.

That is the basic idea. There are also refinements you can add like having the off switch attached to a pin with a pull-up so you can instantly detect it - but then the software disables the pull-up on that pin before going to sleep so again no power drain.

Note also that you can directly drive the LEDs from the MCU pins using PWM. This saves avoids the resistors and also gives you the opportunity to overdrive the LEDs for more brightness, which could make sense for a fidget spinner since you are likely going to have less than 100% duty cycle on those LEDs.

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