cmos schmitt trigger design

cmos schmitt trigger

How do I design a cmos schmitt trigger (single input and single output)? In fact I am not even sure how the cicuit looks like? Any help will be appreciated.

Thanks

schmitt trigger cmos

i think in some MCU the schmitt triger already there you don't have to design or just use 74ls14 ic

schmitt trigger design

Dear ccw27,
Look through chapter 18 of baker, li, boyce.
It gives complete explanation of it.
hope it helps

Can anyone post the book "CMOS Circuit Design,Layout and Simulation" by baker,li,boyce?

Thanks

design of schmitt trigger

There is good explanation in book: digial circuit design-a perspective.

cmos schmitt trigger circuit

Hi ccw27,

The basic principle is very simple: it adjust the triggering threshold using positive feedback according to the input,thus hysteresis effect forms.

Hope it helps.

regards,
jordan76

design of cmos schmitt trigger

You can use a non inverting input like one from CD4050.
Let's see how it works.
For begining let's suppose that the output of the gate is at low level, near ground.
The voltage V2 = ( R2 x V1 ) / ( R1 + R2 ).
Now raising the input voltage V1 from immediately bellow VT+.
When voltage V2 will cross beyond the transition voltage Vtr of the gate, the positive feedback that appears from the output to the input forced the output swinging to high level.
This happens very quickly due to the positive feedback.
Vtr = (( VT+ ) x R2 ) / ( R1 + R2 ) -> ( VT+ ) = (Vtr x ( R1 + R2)) / R2
When the input voltage V1 decrease, the output V3 goes to low level when
V2 = Vtr = (VT-) + ((VDD - (VT-)) x R2) / ( R1 + R2 )
We can get now
(VT-) = Vtr - ((VDD - Vtr) x R1 ) / R2
The hysteresis will be:
H = (VT+) - (VT-) = (VDD x R1) / R2
If Vtr = VDD / 2 then
(VT+) = Vtr x ( R1 + R2 ) / R2
(VT-) = Vtr x ( R2 - R1 ) / R2
The histeresis not depends upon transition voltage Vtr.
The resistance of the source signal is included on R1, too.