# Understanding Schmitt Trigger from datasheet

I've asked this question before, but I don't think I worded it correctly, as I got answers to a question I didn't intend to ask... So I'm giving it another shot:

I am having some difficultly understanding the values described in this product sheet for the SN74AHCT14 Schmitt Trigger Inverter:

Vcc will be 4.5V.

If I understand correctly, this trigger is an inverter, so a high input results in a low output.

If the input is at 0V, I expect to get a high output. As my input slowly increases from 0V up to 4.5V, what is the lowest voltage that may cause the trigger to begin outputting a low output? And what is this value denoted as in the product spec?

I am not interested in what will guarantee a low output, but instead, what is the earliest value that may cause the trigger to switch to low output.

• datasheet says 0.9 to 1.9 V. it is a band and hence, for 100% guaranteed response all the time, expect a minimum of 1.9 V at the input pin, to get the low on the output pin. Aug 3, 2015 at 6:06
• So if it switches to low output as early as 0.9V, then if I understand the datasheet this means that, conversely, it will switch to high output on the way from 4.5V down to 0V as early as 1.5V... But this sounds wrong to me (switching at 0.9V on the way up, but at 1.5V on the way down??). Maybe you can understand why this is confusing, but I'll draw a picture sometime soon in case... Aug 3, 2015 at 6:34
• Do not mix max of VT- and Min of VT+. Please refer to delta VT which says that a minimum of 0.4 V is always ensured. When the output changes to low with input of 0.9 V, the output will change to high only when the input crosses 0.5 V Aug 3, 2015 at 6:58
• A schmitt trigger has hysteresis by design and this means it switches at one DC level on the way up and a different dc level on the way down - by design! Aug 3, 2015 at 7:44

Short: The hysteresis specification ensures that the negative going transition voltage is always greater than the positive going transition voltage.

Define:

• Vtpos = input transition voltage with Vin increasing
• Vtneg = input transition voltage with Vin decreasing
• Vhyst = Vtpos - Vtneg for a given IC and conditions.

Look at delta-Vt figure in table 7.7. 3rd line on page 5 in datasheet.

This guarantees:
a minimum hysteresis of 0.4V and
a maximum of 1.4V at Vcc=4.5V and 25 degrees C.

Vhyst = 0.4 to 1.4 V

This spec is then read in conjunction with values for Vt positive and negative going.

Vt pos going = 0.9 to 1.9
Vt neg going = 0.5 to 1.5

Without the hysteresis spec this suggests that the negative going level MAY be above the positive going level (which makes no sense) BUT add the hysteresis spec as well and it makes sense.

With Vt neg = 0.5V then Vtpos MUST be >= (Vtneg + hyst min) = 0.5 + 0.4 = 0.9V.
and can be as high as 0.5 + 1.4 = 1.9V.

The following may bend the brain somewhat - it's not strictly necessary to follow it if you just follow the above ranges. [E&OE - if I've accidentally said something wrong plase point it out - easily done in such cases].

ie it is seen that with Vtneg at it's minimum values the addition of the range of possible hysteresis value gives exactly the range of Vtpos values - which is as it should be.

However, Vtneg max + hysteresis range = 1.5 + (0.4 to 1.4) = 1.9 to 2.9 which is from the max level of vtpos or above - ie if the negative value is at a maximum of 1.5V then Vtpos will be at its max value of 1.9V. ie when Vtneg is at it's maqximum value the hysteresis range will be at it's minimum. If this was not so then rather high values of Vtpos could result.

It is easy enough to draw a 'state diagram' from the above showing areas of valid operation. There can never be any overlap. ie for a given situation Vtpos will always be > Vtneg.

• Thanks for the detailed explanation, Russell! I think this all makes sense to me. Looks like the min hysteresis is exactly the value that I'm looking for to ensure that the time between switching is above the threshold I need. Aug 3, 2015 at 23:22