Could someone explain the concept of 'overdrive' to me?
I'll try.
A comparator is really just a high-gain amplifier. With low overdrive, the low bandwidth of the "amplifier" begins to rear its ugly head, so to speak. Let's say you have a gain of 10,000, and Vcc=10V, Vee=0. Oversimplifying, it will then only take an input differential of 1mV to drive the output rail-to-rail. Let's say you have an open loop bandwidth of 1kHz (sounds low, but it's realistic). That corresponds to a time constant of about 159us. OK, so if I drive the input differentially from -1mV to +1mV in, say, 10ns (fast, in other words), the output (assuming a linear amplifier) will try to swing exactly -10V to +10V, but the output is limited initially to zero volts.
Vout=Vf*(1-e^(-t/RC))
To reach Vout=5V (the 50% point),
5=10*(1-e^(-t/RC))
1-e^(-t/RC)=0.5
e^(-t/RC)=0.5
t/RC=-ln(0.5)
t=0.693*RC
t=110us to reach 5V with 1mV of overdrive.
Now let's drive the input differentially from -50mV to +50mV, still in 10ns. With 50mV of overdrive, the output starts at GND and heads for (50mV*10,000) 500V. The -50mV just served to keep the output at zero. Ok, how long does it take for the output to reach 5V?
V=Vf*(1-e^(-t/RC)).
To reach Vout=5V (the 50% point),
5=500*(1-e^(-t/RC))
1-e^(-t/RC)=.01
e^(-t/RC)=-0.99
t/RC=-ln(0.99)
t=-RC*ln(0.99)
t=159us*0.01
t=1.6us to reach Vout=5V with 50mV of overdrive.
If you calculate the time to reach 10V, it will be about 3.2us.
(I hope I got all the parentheses paired up, etc.)
This is what overdrive does.
A real comparator is not a linear amplifier. It generally has at least a couple of limiting amplifiers in cascade, so the response time will not be linearly proportional to the overdrive as it is in this case, but the principle is the same.
Below is a schematic of the circuit described above, and the .ASC file, in case anyone wants to simulate it in LTspice.