I am simulating the circuit above in LT Spice, where a LM311 (a LT1011 is simulated, but should be largely equivalent) is used to transform a (noisy) sine wave signal into a square wave. Value of R5 controls hysteresis as seen in red waveform.
I understand that the comparator has an open-collector output, pull-up resistor R1 is used to obtain 5V on high output.
My question is about the low output level. Even though V- is grounded, the voltage at the output doesn't quite reach 0V, rather it stays at 200.3mV; changing the value of R1 from 1kOhm to 10kOhm decreases this level to 155mV. How does this voltage behave and how can it be selected by the circuit designer? Is it any of the specs in the datasheet?
Looking at datasheets for LM311/LT1011 it seems pin 1/EMIT OUT is connected to the rest of the circuit through a 4Ohm resistor. If this was the only factor, I'd expect a 1kOhm by 4Ohm +5V to 0V voltage divider to provide a much lower 19.9mV. Instead, it seems to behave as a 42Ohm resistor with a 1kOhm R1, and it even changes to a 320Ohm resistor with a 10kOhm R1.
Best Answer
The LM311 and LT1011 are bipolar chips- the LT1011 output is an NPN transistor with what looks like some current-limiting circuitry (including the 4 ohm resistor in the emitter). So the output will not behave in a “resistive” manner.
The typical and worst-case voltage drop is specified in the datasheet. For the LT1011 if you call on it to sink less than 8mA you can count on the low voltage voltage being less than 400mV given some overdrive and Tj < 100 degrees C. For 8mA sink and 3mV of overdrive it will typically be around 260mV at 25 degrees C.
If you are using the LM311 refer to the relevant datasheet, of course, and preferably simulate with the correct model.