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I have a microcontroller with an active low reset pin. This pin is connected to a timer that periodically triggers a low edge and resets the microcontroller. What would be the best circuit that would allow me to also manually trigger the reset pin through an external circuit while the pin is connected to the timer? Shorting the two together would obviously not work.

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  • \$\begingroup\$ What kind of timer? Post the link to the datasheet. \$\endgroup\$ – Golaž Jul 5 '15 at 9:42
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When either the timer output or the GPIO goes low, \$ \overline{\text{INT}}\$ goes low, so the gate is a positive true AND, 1/4 of an HC08 or somesuch.

enter image description here

or, better yet,

enter image description here

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I assume your timer output is a push-pull circuit?

If you can configure it to be open-drain, then you just need a pull-up resistor on the reset line, and use a switch to short the reset input to ground.

If you can't configure the Timer output to be open-drain, then you just need a series resistor between the timer output (say 1k) and the reset input. Again, the reset switch shorts the reset input to ground.

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SW1 can be pressed in order to generate the reset independent of timer. Whenever the SW1 is pressed first transistor turns on and RST# net will be pulled low to ground. Whenever the Timer is driving low, the first transistor will turn on and RST# line will go low and a low level voltage will be presented to the MCU RESET# Pin.

When either switch or TimerOutput is low, Q3 will have a base current of about 8 mA (assuming VCC as 5 V, Vbe as 0.8 and diode drop of 0.4 V), which will sufficiently to saturate Q3. Q4 too will be saturated so that, the Vol will be less than 0.1 V.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What if the switch is pulled up? I expect the final design to be a gpio instead of a switch. The gpio will be pulled down to trigger the reset. \$\endgroup\$ – Malfunction Jul 6 '15 at 9:40
  • \$\begingroup\$ @Malfunction Sorry for the silly mistake. \$\endgroup\$ – Umar Jul 7 '15 at 3:54
  • \$\begingroup\$ Doesn't work: lowest possible level of ouput will be ca. 1.4V, which is to high for L. \$\endgroup\$ – Curd Jul 7 '15 at 9:07
  • \$\begingroup\$ @Curd pls check now. Sorry for ignoring \$\endgroup\$ – Umar Jul 7 '15 at 9:34
  • \$\begingroup\$ Now logic output level is OK, but still another problem: base resistor R9 is only 100 Ohm, i.e. ca. 36mA at Vcc=5V; might be too much load for TimerOutput. BTW R8 and R9 can be shared. \$\endgroup\$ – Curd Jul 7 '15 at 10:28

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