ESP8266 Community Forum

Hi Kolban

I'm here because of two things:

1-
First of all let me say I downloaded your ESP8255 eBook for free to see what it was.
And it was (sorry, it is) an absolutely fantastic work (I'm only at page 37) .
That said I will most probably buy it.

2-
As you asked instead of emailing you, to make technical comments at this post, here is one:

I'm afraid the reset circuit at page 37 could damage the module.

Attached are the simulation screenshots (schematic and graphic from LTspice).




The green line is the pulse that turns ON the switch (S1 at your schematics) .
So, when the green line is at 3.3V it is like you've pressed the S1 button.
When the green line is at 0V it is like you've released the S1 button.
The white line is the Vc : The Reset line, which connects to the module reset.

As you can see, when we press the S1 button, the Vc goes to 0 and then exponentially
goes again to 3.3V (all ok here).

The problem is, when we release the S1 button, Vc goes almost immediately to around 5V
(in my opinion, because of the previous capacitor charge of 3.3V is, for brief instants, somewhat added to the power supply voltage), then exponentially discharges to 3.3V .

This is a problem.

Miguel Garcia

Thanks Miguel, my electronics skills are negligible. Let me study what youve written and, most likely, I'll just pull the circuit diagram from the book rather than risk anyone damaging their units.

Neil

Hi again Kolban

You do not need to remove the circuit because, in my point of view, all other things are correct.

You only need to replace the cureent circuit with a simple RC reset circuit
like this one:




NOTE:
I think the diode is even optional if ESP8266 has internal ones to GND and +V (which I think it has).

Miguel

I have downloaded the book and was just about to reply here to warn about the reset pin! But it seems you have got here before me.

What happens is that when you press the reset button, the capacitor is grounded and starts to charge up. The more you hold the button down the more it charges. With 10K and 10uF it is about 300ms-400ms to achieve almost full charge. Then you release the button and two things happen. First, the negative terminal of the capacitor is removed from the ground. Secondly the negative pin, is now connected to the positive supply through the second 10K resistor. So now we have a charged capacitor with say 3.3V on it, and we raise its "bottom" to the supply rail, at 3.3V. That means that its "top" will be its "bottom", at the 3.3V plus its own charge, another 3.3V, for a total of 6.6V. More or less.

This technique is used on DC-DC circuits and is called "switched capacitor". Basically you connect a capacitor from Vcc to GND, say 5V. After the capacitor has charged up, you connect its "bottom" to Vcc and disconnect the top from Vcc. Now, the "top" of the capacitor is sitting at Vcc + capacitor charge (another Vcc) = total 2*Vcc. So now we have 2*Vcc. Because we do this switching back and forth many times per second (tens or hundreds of thousands) this technique is called "switched capacitor".