I was reading about Schmitt triggers and something is bothering me. Given that each component is memoryless- shouldn't the circuit be solvable(as in, for any input voltage it should always be immediately at a steady-state)?

Why can't it be described by a set of time independent algebraic equations since the V/I relationships of all components in the circuit are time independent?


1 Answer 1


A Schmitt trigger is not a memoryless circuit. It has positive feedback, which is one way of implementing state (or memory).

Similarly, a flip-flop can be made from only NAND gates, even though the individual gates would be considered "memoryless" when used individually. Again, it's feedback that gives the overall circuit its "memory".

  • \$\begingroup\$ or from NOR gates. that's the easiest, most direct RTL using BJT. \$\endgroup\$ Commented Feb 27, 2017 at 5:14
  • \$\begingroup\$ The saturation of a transistor results in significant base charge storage; it is the base capacitance of the saturated transistor that contains the 'memory' of the two-transistor Schmitt. \$\endgroup\$
    – Whit3rd
    Commented Feb 27, 2017 at 7:38

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