# Most efficient delay circuit

I am currently trying to design a basic time delay circuit. The circuit's goal is to switch on a BLE beacon (requires 1.8V-3.6V at 1mA) for 3-5 seconds and then switch it off again. My first option was to use a 555 timer, however, due to the fact that the circuit is battery powered (coin cell CR2032) I require something which uses the least amount of power. I thus opted for an n-channel mosfet with a low-side transistor configuration.

• Is this the best, most efficient manner in which I could build a delay circuit?
• If not, what is and why is it more efficient?
• I'd be curious if a ultra low-power microcontroller running on like a 32khz clock and sleeping would be the lowest... – MadHatter May 5 at 21:24
• How often is your device switched on? – MadHatter May 5 at 21:25
• MSP430 can be put to sleep (waking up on a timer on a 32.768kHz clock) on well under 1 uA; that's about 100x less than even a CMOS 555. – Brian Drummond May 5 at 21:25
• @MadHatter No more than 10 times a day. – Eckaard May 5 at 21:40
• @BrianDrummond thank you I'll have a look into it straight away. – Eckaard May 5 at 21:46

Your "beacon" is probably way smarter than you realize (and an NE555 would use way more power than a beacon). BLE chips are actually microcontrollers with a radio frontend – and many of them are user-programmable.

So, this has a simple software solution.

• Marcus, Thank you for your reply. I'm busy working with a JDY-19. Currently I am making use of AT commands to set the major, minor, baud rate, UUID and transmission rate. I have yet been able to actually program the chip otherwise. I used the following data sheet; [link] (myosuploads3.banggood.com/products/20181201/…) Sorry for getting off topic but if you know of a way that would be great. – Eckaard May 7 at 20:59

The following circuit should come close to your specifications. But it probably won't work down to $$\1.8\:\text{V}\$$. You'll need a different MOSFET if you want to lower the workable voltage down that far. But this basic topology should get you there, though the time period will be dependent upon the rail voltage. But it has almost no leakage current when OFF, so that's good as well.

simulate this circuit – Schematic created using CircuitLab

I've established the value of $$\R_2\$$ on the basis that your load current is $$\1\:\text{mA}\$$. If you need a lot more, then $$\R_2\$$ (and probably $$\R_3\$$) will need to be adjusted, accordingly.

The selection of the BSS123 is based upon how common the part is and it's low threshold voltage. But it's just an approximation. If you really need lower voltage operation, closer to $$\1.8\:\text{V}\$$, then you will need to find a different device with a lower threshold voltage.

• Thank you very much @jonk! This is great! I really appreciate the help. 1.8V is perfect and will work just fine. – Eckaard May 7 at 21:04