While reading microcontroller data sheet i came across i/o types like

i) TTL

ii) schmitt trigger.

I came to know that TTL i/o logic means 0v-0.8v -logic state 0, 2-Vcc means logic state 1 (as per wiki [1] http://en.wikipedia.org/wiki/Digital_signal#Logic_voltage_levels ).

my questions are

i) what will be the logic state when my input voltage is 1-2 in TTL in i/o type

ii) as per definition schmitt trigger is a circuit with positive feedback, but i didnt understood how it works, advantage of using schmitt trigger type input over TTL


Regarding your first question, when the input voltage is between the valid regions for a logic 0 and a logic 1 then the output cannot be specified. It may be a 1, it may be a 0, it may be at some voltage in between a valid 1 and a valid 0.

For the second question, the voltage switching points at the input to a Schmitt trigger are different depending on whether the output is currently a 1 or 0 (for the non-inverting case). When the output switches from a 0 to a 1 the voltage thresholds shift downward, favoring the 1 state and making it more difficult to switch back to a 0. The reverse happens when the output switches from a 1 to a 0. The benefit is that noise on the input signal won't cause glitches on the output. A Schmitt trigger is also often used to "sharpen" a slowly changing input signal.

  • \$\begingroup\$ if possible can i expect some more detailed explaination on schmitt trigger. \$\endgroup\$ – tamil_innov Jul 20 '13 at 18:43
  • \$\begingroup\$ Suppose the input is low and the output is low. There is some specified level for the input voltage such that the output is guaranteed to be high when the input is above that voltage. This voltage level is called \$V_{IH}\$ and let's say it's 2.4V. Once the input has risen above that level and the output changes to a high voltage (a logic 1) then internal positive feedback causes \$V_{IH}\$ to be reduced, let's say to 2.0V. This means that any noise riding on the input signal that is less than 0.4V cannot cause the output to switch back to a 0. \$\endgroup\$ – Joe Hass Jul 20 '13 at 19:12

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