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
Best Answer
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.