How to use the 555 in bistable mode as a voltage level detector? I want to use a narrow range 3.5 - 4v input from a Hall detector

The 555 gets a lot of hate for being old and having no place in a modern system, but it has a redeeming feature: a ridiculously fast and strong output (compared to anything you could easily build yourself with discrete components) - strong enough to be considered a half-decent MOSFET gate driver. It's a simple, modular and easy to understand integrated circuit embodying a number of key concepts, which even your 9-year old grandson can grasp, with the right instruction, patience, tinkering and magic smoke.

This is how you would wire up the 555 to implement hysteresis:

^{simulate this circuit – Schematic created using CircuitLab}

Literally no other components necessary, except a power supply! Although you should probably put the recommended 10nF capacitor from pin 5 (CONT) to ground. Since there's no timing capacitor to discharge, pin 7 "DISCHARGE" is not used or connected to anything.

Using a supply of \$V_{CC}=+12{\rm V}\$, resistors R1, R2 and R3 all develop the same voltage \$\frac{V_{CC}}{3}\$, since they all have the same value 5kΩ. Therefore absolute potentials at HI and LO are \$V_{HI}=+8{\rm V}\$ and \$V_{LO}=+4{\rm V}\$. These define the upper and lower switching thresholds for hysteresis. The comparators will both be comparing the same input \$V_{IN}\$ against these two thresholds, so pins "THRESHOLD" and "TRIGGER" are joined to the same IN.

The SR flip-flop FF1 has the role of reacting to changes of comparator output. When CMP1 output (R - reset) goes high, Q goes low and stays there even when CMP1 output falls low again, effectively "latching" that Q state until ... when When CMP2 output (S - set) goes high, Q also goes high and stays there even when CMP2 output falls again, effectively holding that Q state until ... and so on.

CMP1 of course produces a high output when \$V_{IN} > V_{HI}\$. CMP2 output goes high when \$V_{IN} < V_{LO}\$. Hopefully you can see that a "slowly" rising or falling input \$V_{IN}\$ (blue below) will result in \$V_{OUT}\$ (orange) going to \$V_{OUT}=+12V\$ when \$V_{IN}\$ exceeds the upper +8V threshold, but only going low again when \$V_{IN}\$ falls below the lower +4V threshold:

_{Note: I had to scale \$V_{OUT}\$ to obtain the real-life 0V/+12V output, since the simulator's SR flip-flop model produces only 0V/+5V output - very frustrating.}

The next thing to address is how to change those thresholds. Sadly, you can only set the upper threshold directly, which is "control voltage" \$V_{HI}\$ at pin 5. You'll notice that R2 and R3 divide by two, so that we are constrained always to have \$V_{LO}=\frac{1}{2}V_{HI}\$. We must also be aware that the internal comparators are unlikely to work with \$V_{HI}\$ too close to either supply rail, 0V or +12V here. Strangely, the 555 datasheet does not seem to document what the functional limits of pin 5 potential are, so you'd have to experiment to find those limits.

Setting pin 5 to +6V would yield thresholds of \$V_{HI}=+6{\rm V}\$ and \$V_{LO}=+3{\rm V}\$ respectively:

^{simulate this circuit}

I used a voltage source to apply +6V directly to CONT (pin 5), but the same effect can be achieved with a resistor R4 from CONT to ground, which effectively places it in parallel with R2 and R3, below left:

^{simulate this circuit}

With R4 at 2kΩ, this is sufficient to "pull down" the potential at CONT to exactly +3V (assuming \$V_{CC}=+12V\$), which will probably be suitable for your own tachometer signal \$V_{IN}\$, which slightly exceeds that.

Above right I "pull up" the potential at CONT by connecting R4 to \$V_{CC}\$ instead. I won't go into the algebra for this, but it's not complicated, though perhaps a little advanced for a 9-year-old.

Can you use this to drive a MOSFET? Sure. Is it as quick as a purpose-built gate driver IC? No. Is it quicker than a reed switch? Yes, a thousand times quicker. Strictly speaking, the "bistable" part of this is the SR flip-flop, and this circuit as a whole is better described as a "schmitt trigger".

Choose a power supply voltage for the 555, (which determines \$V_{OUT}\$) suitable for the gate of your MOSFET, and set \$V_{HI}\$ (called \$V_{CONT}\$ in the datasheet) to somewhat lower than your analogue control signal \$V_{IN}\$, and you should be good to go.

There are a few potential gotchas to be aware of:

1. I don't know the lowest permissible potential at CMP2's inputs. Since the datasheet doesn't elaborate, you'll have to experiment.

2. This is an inverting schmitt trigger - when the input is high, the output is low (I might be wrong here, I'm working from memory) . This might the opposite of your actual requirements, in which case you'll need to "invert" \$V_{IN}\$ somehow. Don't be tempted to invert \$V_{OUT}\$, because that would defeat one of the main reasons for using the 555 in the first place, its gate-driving ability.

3. When charging and discharging a MOSFET's gate, OUT (555 pin 3) can source and sink well over 100mA of current, momentarily. Also, the motor coil demands a sudden burst of current. This can murder a smooth \$V_{CC}\$, so be sure to decouple the supply very close to the 555's supply pins, with capacitors between them of 10μF electrolytic and 100nF ceramic. This is not optional.

A 555 isn’t much suitable for handling an analog signals.

Here is a solution that ensures fast enough edges for driving the mosfet.

The P1 sets the desired trip-point of hall sensor signal.