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I'm hoping someone could guide me in the right direction. I have an ongoing project that seemed simple: to utilize a push button. When you press it once, it toggles the LEDs on; when you press it again, it toggles the LEDs off. The entire system operates on a coin cell battery and needs to be a very small PCB.

I managed to get the circuit to function with both an ATtiny10 and a 555 design. However, the power consumption in the off state is too high. I am aiming for the lowest possible power usage when it's in the off state.

I purchased some bistable modules from AliExpress to experiment with and understand their operation. I was astonished by their low power consumption: it's 0.5 uA, but I am struggling to work out how it works. My best guess is a D-type flip-flop?

I've attempted to search for the part number, but I couldn't find anything. Does anyone have knowledge about how something like the board below would work?

Or am I pursuing the wrong approach in designing something like this?

My objectives are:

  • Low power consumption
  • Cost-effective assembly
  • Minimal part count
  • Utilization of the smallest possible component size 0603/SOT23-6

Any guidance or insights would be greatly appreciated. Thank you!

enter image description here

Update!

I'm utterly amazed at how knowledgeable you all are! I spent a significant amount of time searching for that IC wiht no luck but some how you found it!, the explanation provided by "Kuba hasn't forgotten Monica" is absolutely fantastic. It's a shame that I can't fully grasp the circuit simulation, I have just been learning Kicad and doing some simple pcb layouts :( I still got a lot to learn.

Here's the Aliexpress link

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  • \$\begingroup\$ whete does the picture come from? ... please add a reference to the picture into your post ... do not add into a comment \$\endgroup\$
    – jsotola
    Commented Aug 10, 2023 at 2:08
  • \$\begingroup\$ My best guess is a D-type flip-flop ... start thinking in terms of microcontrollers \$\endgroup\$
    – jsotola
    Commented Aug 10, 2023 at 2:09

3 Answers 3

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The board in your picture (YR-1006) is probably based on the Elitechip (丽晶微电子) CS2130 switch IC.

enter image description here

(google translated datasheet excerpt)

I think it's somewhat similar to this open design, but only uses the low level output (OUT2) with a P-channel MOSFET.

enter image description here

My main clue that it's a switch IC from this manufacturer is the Elitechip EC190707 which has a very similar marking style. I think it's the CS2130, because it's the only switch from Elitechip with a compatible pinout and package that I could find.

Small addendum: According to this article the CS2130 was replaced by the EJ2130E which is pin identical so it could be either.

enter image description here

(Source: taobao.com)

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Q1 is the LED driver. Since it is on only when the LED is on, it could be either a FET or BJT and not affect the off-state current.

At that current drain, U1 must a CMOS part. There are CMOS 555 variants in that package, but the package ident text is no help. The problem with this is that a toggle circuit uses at least one capacitor to hold the current state, and the board has none.

Thus, it looks like the board has a uC with no power supply decoupling capacitor.

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  • \$\begingroup\$ Link to the product page - ? \$\endgroup\$
    – AnalogKid
    Commented Aug 9, 2023 at 23:29
  • \$\begingroup\$ I don't recall, but isn't the resister divider going to exceed any chance of using even the 555 CMOS version for very low currents? \$\endgroup\$ Commented Aug 10, 2023 at 0:08
  • \$\begingroup\$ In this case it seems to be a specialized IC and not a μC. See my answer. \$\endgroup\$
    – nanash1
    Commented Aug 10, 2023 at 9:45
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The IC has a flip-flop and debounce circuitry inside. You could design it from first principles. For example:

schematic

simulate this circuit – Schematic created using CircuitLab

The feedback circuit around the buffer NOT1-NOT2 adds hysteresis, so that small changes in voltage on the capacitor won't change the output state. The time constant R1-C1 is about 17ms and should be sufficient to debounce the switch. To be conservative in face of worn-out switches etc., C1 can be increased to, say 220nF or even 470nF. The trade-off is reaction speed to pressing a button vs. susceptibility to bounce.

Inverters could be any Schmitt-trigger input inverting gates like 4093, 40106. The D F-F can be 4013, 4042, etc. If using the 4000 family from 3V, R3 is likely unnecessary and the LED can be connected directly to the Q output. Of course you can also use the 74HC and similar LV CMOS logic families, especially the 74xx1Gxx, e.g. 74LVC1Gxx that has single or double gates in tiny packages. NOT1 and NOT2 could be merged into a single non-inverting Schmitt-input gate, or one that can be pin-strapped to act as a non-inverting buffer.

The 4000 family has some benefits in general. It works from 3V up to 15V, so having some on hand lends flexibility to your designs. It's not very fast, but that's of no consequence for its common applications. Of course the packages are larger, but at least you can get all the 14- and 16-pin chips in the series in DIP as well as various surface-mount packages.

The quiescent current consumption at room temperature should be <<1μA. If it's not, you probably got fake/reject chips. I buy those straight from TI and never had a problem - most of the time they are orders of magnitude better in current consumption and leakage than the specification limits.

The lowest-power 555 variant is LMC555, with idle current of 50μA typical. That's still orders of magnitude higher than CMOS logic.

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  • \$\begingroup\$ Don't think so. 17 ms after SW1 is pressed, the NOT2 output and the NOT1 input are both sitting high. When SW1 is released, they both stay high. When SW1 is pressed again, nothing happens. \$\endgroup\$
    – AnalogKid
    Commented Aug 10, 2023 at 21:20

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