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I put together the basic 555 monostable circuit, but found it drew a bit over 2mA when "off".

To an extent, this makes sense as the 555 effectively has a 15k resistor between VCC and GND. I thought that I could cut power to the 555 when it's off to drop to a really low quiescent current, so I put in the two transistors and associated components up above the IC in the schematic.

While technically a success in that this dropped the quiescent current draw, it only dropped down to 1.76mA. I took a look at a similar question and answer, and the only difference that I think I have in the power setup is that I'm using BJTs instead of MOSFETs. I looked up the datasheet on the PNP (a PN2907, but KiCad didn't have that) and saw the leakage current to be far less than what I'm seeing.

For testing purposes, I replaced the output sub-circuit on the right with just a resistor and LED for an indicator/dummy load. The test setup ran from a 5V breadboard power supply, though the final version is intended to run from a nominal 4.5V (three AA batteries).

Why is this still drawing milli-amps of current and what is the minimal change needed to correct this? (General critique is also welcome, though I intend to stick with through-hole and the 555 for the project at hand.)

Problematic Schematic

Update: Adding the corrected schematic for the sake of others who may come looking.

Fixed Schematic

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The LMC555 shown in your schematic is a CMOS type and has 3x 100K resistors rather than 3x 5K resistors in the bipolar 555.

It typically draws around 100uA. The old-fashioned bipolar type draws a couple mA typically (there is current draw in both cases that is in excess of the divider chain).

The leakage in Q3 will be multiplied by the hFE of Q2 so you should have resistor E-B on Q2, but that's not likely to be your problem (it would tend to show up as a latent design error at high temperatures).

I'm thinking that the problem is not shown on your schematic and you have an issue with switching the load with a low-side switch, which is causing sneak paths in your load to ground drawing current. Use a high-side switch and ensure that no inputs to your load are more than few hundred mV above ground.

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  • \$\begingroup\$ "the leakage in Q3 will be multiplied by the hFE of Q2" Good point. "you should have resistor E-B on Q2" I have no idea what this would do for the circuit. Is there some key term I should look up to enlighten myself on it? "For testing purposes, I replaced the output sub-circuit on the right with just a resistor and LED for an indicator/dummy load." So it's really just Output -> LED -> Resistor -> GND for that. \$\endgroup\$ Dec 28, 2020 at 2:21
  • \$\begingroup\$ You want a resistor low enough that it prevents the E-B voltage on Q2 from exceeding a couple hundred mV at worst-case Q3 leakage, yet doesn't steal too much base current. In this case you have a very low base drive resistor, so something like 10K is fine. Measure the voltage on the Vcc of the 555 to see what it is. \$\endgroup\$ Dec 28, 2020 at 2:24
  • \$\begingroup\$ OK, so it's basically a pull-up resistor. \$\endgroup\$ Dec 28, 2020 at 2:28
  • \$\begingroup\$ VCC to GND on the 555 is 3.3v. E-C on the PNP is 1.7v. Adding your pull-up resistor cut the current draw half of a milli-amp to 1.2mA. \$\endgroup\$ Dec 28, 2020 at 3:04
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    \$\begingroup\$ Aha. That did it. I dropped an extra diode in series with R1 to ensure that the pull-up resistor can't become a back channel for current to reach U1. Doing that dropped the current draw to tenths of a milli-amp. Doing that with the pull-up on the base of Q2 cut it down below what my multi-meter can read. Thanks for helping me debug! \$\endgroup\$ Dec 28, 2020 at 3:39

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