Question about led flasher circuit

The 7404 is a logic gate, the output will be either a "high" or "low" voltage, depending on the input state, and nothing in between.

That circuit linearizes the gate with that 1megohm resistor, so it can oscillate.

The 7404 that you are using won't work in the circuit. It is internally TTL (a forty year old logic family), and must run off a 5v supply, and has a low input impedance. The large value resistor won't work with it, and the IC will not work off a lower voltage supply.

The 74HC04 came later, and uses CMOS internally instead of the bipolar transistors of the original. This has a very high input impedance, meaning that 1 megohm resistor will linearize the gate enough, and it will work as intended (at least, if the original does work, I've never tried it).

CMOS logic is less restrictive about power supply voltage. TTL wouldn't work at a much different voltage than 5v, CMOS will. I can never remember the exact intend of the "HC" in the middle of the number, but the original CMOS variant of the 7404, the 74C04, allowed for operation up to about fifteen volts.

A 1.5v supply would seem to be a bit low for the 74HC04, but in this sort of thing likely it can be done.

If this was just using the IC as logic, ie that "high" or "low" state, likely you could get away with the 7404, if that was the only IC being used. (You can do some mixing of the logic families, but there are limits.) But this is an odd use of the IC, and in this case you really do need the 74HC04 (not just because it can be the straight 7404, but because when someone cooks up a use for an IC that is or may be out of spec, they are counting on some vagary of the specific IC)

Michael

The 74hcxx series is CMOS, so even though it's specified at 5V, it will work at lower voltages. This is not necessarily true of other families.

I have just tried the one with just one inverter with an 74hct00b1, tieing the inputs together gives us invertes. It doesn't work with