Rather than a Zener, I suggest you use 2 red LEDs in series (anode to base) Connect a 270 Ohm resistor to the logic output and a 10k resistor between base and emitter.
Audio, You are missing the point. When the input is at 2.7 V, the voltage across the Zener (or LED) will be 5 - 2.7 = 2.3V.
Thus a 3.3 V Zener would have only a small current through it. As I said in the previous post, low voltage zeners do not have a sharp "knee" so that I why I prefer the LED solution.
I suggest you do a switcherCAD simulation (as I did). With 2 red LEDs or a single blue one, the collector current is <1uA when the input is at 2.7V.
I would vote for the "non-inverting" circuit suggested by Ron as it can provide the near 0-5V drive signal to the P-ch MOSFET, resulting in the least Rds upon turn ON.
I would vote for the "non-inverting" circuit suggested by Ron as it can provide the near 0-5V drive signal to the P-ch MOSFET, resulting in the least Rds upon turn ON.
Yes, I agree provided that mitchjs is happy with a non inverting solution.
However, he specified that he wants the circuit respond to <1V, so Ron's circuit needs a diode or red LED in series with the base in order to ensure this.
And, as I told Ron in a PM, his other solution needs a diode in series with the base as Q1 will not be fully off at the specified 2.7V.
What I said is that low voltage Zeners don't have a sharp "knee" and I can't find a decent data sheet for a either a 2.7V or a 3.3V Zener so I can't calculate whether a Zener is adequate.
Also, my SwitcherCAD simulation software does not have a model for either of these Zener values, so I can't check it by simulation.
You also need to consider thermal effects. If the transisor is partially on at a temperature of say 20C, then it will be even worse at 30C since Vbe decreases with increasing temp.
However, I do have SwCAD models for both a red LED and a blue one. Both of these (ie. 2 red or 1 blue) give satisfactory operation.
That noninverting circuit I posted will work, after a fashion, if you adjust some of the resistor values. The problem is, it doesn't account for the resistance of the logic stage driving the input. It would be better to just use two inverters, as below. There are probably other solutions which are as good or better.