Tool/software:
Dear Technical Support Team,
I am desgnin a reset circuit for my microcontroller with your product SN74LVC1G10DCKR.
What I am intending to do is, when the three buttons pressed in the device, a reset is performed to the microcontroller.
These are the connections of the membrane keypad:
BTN_EXT is not currently used.
This are the connections to the SN74LVC1G10DCKR chip and the microcontroller:
Appart from taking into consideration the pull-downs for the membrane keypad, and considering that VBAT will never go anywhere under 3.0 V, is there any other consideration I am missing for the design?
Any other capacitors, or extra design stuff that I need to take into account for this to work?
Thank you.
The NAND gate's output is connected directly to the R-C reset circuit. So when the gate powers up, it will override the reset pulse. Futhermore, the output might be overloaded because of the large capacitor.
Insert a diode implement a wired AND, and a resistor to limit the current.
Dear Clemens Ladisch,
Thank you for your answer and sorry for my little knowledge in this area.
I partly understood what you said, but I still have a few questions:
So when the gate powers up, it will override the reset pulse
I understand that here you mean that, the R-C reset circuit should provide a reset at the beggining of the power-up sequence of the device, so performing a reset to the microcontroller. Because the output of the reset is directly connected to that R-C reset circuit, the RST signal may force a '1' in that pin, so never performing a reset on the microcontroller, is that correct?
Futhermore, the output might be overloaded because of the large capacitor.
This, I don't understand. Could you explain it please?
Insert a diode implement a wired AND
Here, are you suggesting I place a diode like this?:
and a resistor to limit the current.
Here I am a bit lost, where should I put the resistor? Before the diode?
Thank you.
You want to pull down the reset signal, but not up. So the diode must go in the opposite direction.
A capacitor acts like a short. The absolute maximum for the output current is 50 mA; more can damage the gate. Taking the diode's forward voltage into account, you need a series resistor of at least (3.3 V − 0.6 V) / 50 mA = 54 Ω; use 100 Ω to be safe.
Hello,
Okay, I will place the dioe in the opposite direction.
I still don't understand (sorry for my unexperience) where should the resistor be placed. From your answer I understand it needs to be next to the diode: like this maybe?
Thank you
Okay, thank you Clemens, you've really helped me a lot.
I will try to test this in LTSpice and then in my circuit.