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I've seen this in a bunch of ATMega / Arduino reference designs:

enter image description here

The RESET pin is pulled up high with R10, pulled low with a switch (S1), but then tied to the DTR line from a UART breakout via C27 - a small capacitor. This last bit is how the Arduino bootloader triggers the reset, but what's actually happening here? How does DTR going low translate to the pin triggering?

kolosy
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  • The low signal on the DTR causes a brief low pulse on the reset pin. That resets the chip. – Majenko Sep 29 '14 at 15:16
  • well, I gather as much. My question is why does that particular construct (line through a cap) cause it? – kolosy Sep 29 '14 at 15:18
  • The new difference in the potential of the two sides of the capacitors allows the capacitor to charge up. It does that through the 10KΩ resistor. The capacitor starts empty, and rapidly fills up, so the voltage at the reset side of the capacitor drops to 0V then rises back to 5V as it fills up again. – Majenko Sep 29 '14 at 15:22
  • Ok, that makes sense. If only this were an answer I could mark as accepted :) – kolosy Sep 29 '14 at 15:23
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    I was drawing pictures ;) – Majenko Sep 29 '14 at 15:39
  • you found this diagram where? where can I find the full version of it? – Renato Tavares Jun 14 '16 at 15:26

1 Answers1

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The LOW signal on the DTR creates a potential difference of around 5V between the two sides of the capacitor. This allows the capacitor to charge up. It does so through the 10KΩ resistor.

The capacitor starts empty, so the voltage is seen as 0V on the RESET pin. That rapidly rises back up to 5V at a rate of \$R×C\$ (that is the time taken to charge up to 63.2% of full), so \$10000×0.0000001 = 0.001s\$ or 1ms.

When the DTR line goes HIGH again it then creates another pulse in the opposite direction (going higher than high) which gets absorbed by the internal ESD diodes in the chip.

Here's a simulation: enter image description here

Majenko
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  • You can think of the series cap as differentiating the square wave. When the incoming slope is high (at an edge) you get a pulse on the other side of the cap. – mixed_signal Sep 30 '14 at 03:27