# A threshold version of a CMOS 4050?

I want a buffer that outputs Vcc (of 5 V) when I give it voltage about a threshold, say 1.5 V. And it outputs Vss (ground) below that threshold.

I have a 4050 CMOS thing but I think it does not have a threshold - it just outputs whatever it sees on the input.

Bonus points if it's pin-compatible with a 4050.

Many thanks Rich

--- edit ---

I'm not after a specific threshold, it's TTL levels. If it looks like an 'ON', output an 'ON'. Else output 'OFF'.

Sorry, could have been clearer. Rich

• You're after a specific threshold, so you need an (adjustable/dynamic) reference and appropriate speed comparators. You won't get the same pinout. – user2943160 Sep 5 '16 at 19:52
• If you want something that responds to TTL-level logic, then you need a TTL logic gate or buffer (or a 74ACT or HCT part, which is CMOS with TTL input levels.). – Peter Bennett Sep 5 '16 at 19:58
• @RichColours - Hi - This question has been very confusing: First you do want to specify the threshold, and give an example of 1.5V; then you say you don't want to specify the threshold; then you say "it's TTL levels" which does imply specific thresholds! This sounds a lot like an XY problem situation. Please can you explain the overall problem you are trying to solve; explain what part this device plays; explain what devices you've already eliminated; and explain why you are talking about TTL levels and a CMOS device? And perhaps, just perhaps, this may start to make sense :-) Thanks. – SamGibson Sep 5 '16 at 20:33
• futurlec.com/74HCT/74HCT4050pr.shtml – Bruce Abbott Sep 5 '16 at 23:38

You should use a comparator for this (or a single rail op-amp).

Most digital logic gates require more than 30% of VCC to trigger reliably and not burn more static power. You could check some datasheets, if you wanted. The key term you are looking for is $V_{IH}$ (for input-high). This represents the lowest possible input which will register as a clean '1', or high logic. For most 4000 series IC's, this value is ~4V. The $V_{IL}$ is likewise the highest input voltage which will register as a clean '0'. For CMOS 4000 series, this is ~1V.

The difference between these values is ~3V. Normally, higher is better, because this means you would require more noise to cause an output error. The situation you described, with $V_{IL} = 1V$ and $V_{IH} = 1.5V$ only leaves 0.5V for noise. So for general purpose gates, most will not have these characteristics.

4000 series datasheet

But, you can use a comparator (or just plain old single rail op-amp) with the V- = 1.5V and the V+ as your input. You can give the comparator IC supply and ground of +5V and 0V, so it will produce your 0-5V output signal. (Like LM293)

Edit: As pointed out by Peter Bennett, the LM293 has an open collector output. So connect the output to the supply voltage with ~3.3kOhm or so resistor. Most op-amps do not require this pullup resistor; just check the datasheets.

• Most comparators have an open-collector output, so you'll need a pull-up resistor between the output and Vcc – Peter Bennett Sep 5 '16 at 19:55

The MC74VHCT50A is one possibility, but I don't think the pinout matches the olde CMOS parts, especially stupid ones like the 4050 with Vcc on pin 1.

Note that the edited version of the question clarifies that it is TTL-compatible levels that the OP is after, so $V_{IL} = 0.8V \text{and} V_{IH} = 2.0V$, so that standard TTL output level limits of 0.4V/2.4V yield a minimum of 400mV noise immunity.

• Level shifting doesn't give specific thresholds. Looking at that datasheet, there's still a >0.5V invalid input region. – user2943160 Sep 5 '16 at 20:17
• @user2943160 According to the question "it's TTL levels". The inputs of that chip are TTL-compatible, so it correctly answers the question. Any arrangement at all will have a region of uncertainty, of course. – Spehro Pefhany Sep 5 '16 at 20:19
• Well, now that I see the question's edit... My vote on your answer got locked, so if you want to edit your answer to cover anything else, I can change it. – user2943160 Sep 5 '16 at 20:30
• Mentioning the content in this comment on the question would certainly be useful. – user2943160 Sep 5 '16 at 20:31
• Spehro , wouldn't applying your logic, with TTL outputs of 0.8~2.0 to same input specs imply TTL has no noise immunity. Actual TTL thresholds are two VBe drops.-~1.3 but HCT is nearer to 1.5Vish – Sunnyskyguy EE75 Sep 5 '16 at 23:29

What you are looking for applies the the 74HCTxx family with T for TTL input thresholds, but the CD4050 was never migrated to HCT.