Reset pulse generator for ESP8266

Question

I am trying to generate a reset pulse in order to wake up an ESP8266 device.

The ESP8266 supports a "deep sleep" function, in which it consumes very little power. However, the only way to wake it up is to reset it (by pulling the reset pin low - the device will restart as soon as the pin goes high again).

So I am trying to use an external event (let's say a normally-open switch) to wake up the device. However, there is a snag - the device should wake up immediately after the switch is closed, even though the switch might remain closed for many seconds or minutes afterwards. So I need a circuit that will generate a reset pulse (low, then high after several milliseconds) whenever the switch is closed.

I thought to myself - "no problem, that should be easy!" (yes, I am that much of a newbie! :) ). And I built this circuit - which performs beautifully in the simulator...

But of course real life does not fit the simulation... I am using a Wemos D1 Mini development board - which adds a whole new level of complexity, including a pull-up resistor on RST and a capacitor between RST and ground. So, of course, this simple circuit no longer works :(

My question is - how can I change this circuit so that it does the job? Or, if that is not possible, what type of circuit could I use instead?

I did consider a 555 as well, but a) I am not sure about the power consumption, and b) that is a 5V device, so it might be more difficult to run on batteries.

Some additional challenges:

1. The additional circuitry should use as little power as possible (the whole idea of using deep sleep is getting the device to run for a long time on batteries)
2. Would it be possible to use multiple switches to wake up the device? (I could then run the switches to different GPIO pins, and get the status of the pins on startup to find out exactly which event generated the wake signal)

Thank you!

Answer 1

Using your basic idea, you can do something like this (just adding the one part, the circuit in the dashed box simulates what is on your dev board):

^{simulate this circuit – Schematic created using CircuitLab}

The 2N7000 MOSFET is used to discharge the 100nF power-on reset capacitor on your board.

The simulation shows power-up at t=0 and then the reset switch pressed and held at t = 100ms.

Looks like you could easily go to 10nF and 10M resistors, and even less on the capacitance but take care of the MOSFET input capacitance if you try to go very low or try to substitute a different type of MOSFET (most of which will have much higher gate charge).

Answer 2

A slightly modified version of your solution as pointed by Simon in the comment would work.

The key is that C1 must be much larger than the capacitor inside the board (C2 here) . The pull-up resistor must be lowered by the same order to keep the time constant the same.

Also from the comments I see a false issue with the "voltage doubler", the voltage on Reset pin will be divided between R1 and R3 to 1/10 order which means Reset pin can go at most 0.36V over V1 with a 36uA maximum current which is allowed.

R2 is there to prevent a current spike through SWX, C1 and C2 but if Wemos designers considered is not required you can skip-it to.

^{simulate this circuit – Schematic created using CircuitLab}

For only one switch the diode and the 1M resistor are not needed

The drawback for multiple switches is that you will have to wait for all SW to be steady opened for a time to ensure that you don't loose any data but if they stay mostly opened this simple solution could work well.

Answer 3

Based on our conversation and on your comments, you can take a look at that schematics (Disclaimer: not my schematic).

The capacitor is charged to VCC until you bypass it using the switch. Once the switch is used, it will discharge the capacitor completely. If you want the discharge to be slower, you can put a resistor in series with the switch.

Regarding question 1: the only current that will be wasted is the leakage current that will pass through the reset port and through the capacitor (leakage current). Those are unavoidable unfortunately.

Regarding question 2: You can add many other switches in parallel with the first one. Each switch will be able to reset the ESP8266. However, if you want to route each switch to a GPIO, you will need a different kind of switch and configuration:

^{simulate this circuit – Schematic created using CircuitLab}

R2 and R3 maintain a pull up status to your GPIO to ensure that it can differentiate correctly between a 0 and a 1. You could reverse VCC and ground on that sub circuit if you want the opposite (now that I think about it, it might even be better to pull down and prevent leakage current!) You could create a similar logic using a double throw switch depending on what you have readily in your hand.

Answer 4

I made some modifications in the first circuit to make it work with the RESET signal of the ESP01.

The RESET signal is connected to the node where the Voltimetes is connected (between the 100 ohms resistor and the diode. The 3,3k resistor must be removed, I put it to simulate the internal impedance of the RESET pin. Hope it works for you

Answer 5

This works I use in my project, the second diode is needed to permit flashing with out disconnect the cable