Output level of LM311 comparator

Question

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.

Answer 1

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.

Answer 2

A transistor is not a perfect switch, so you shouldn’t expect the voltage to be 0. The transistor specification is Vce. This is dependent on the collector current, gain of the transistor, base current and other factors.

For comparison, a bjt is effectively a forward biased diode when conducting whereas a mosfet has a resistance (Rdson). So, if there’s current flowing, you get a voltage drop.

Answer 3

The OTA is meant to be operated in the linear current source mode. Your question violates this requirement.

Like all grounded common emitters Vce is a function of Vbe and Ic. When saturated with low current such as Ic/Ib=1 both BE and CB have about the same forward voltage and thus differences in doping and bulk resistance based on power ratings dominate the residual quadratic differences to <=30 mV. But this might be Rc>>100k to 1M, so tHe load pF will make the risetime slower than with 50 mA or even 1mA drive instead of say 10uA output current.

The incremental output impedance depends on Delta Vce/Ic when saturated and Early leakage R when not . Both being high resistance, unless you specify Tr , Vce(sat) @ Ic, you are wasting your time with this question. (Other than learning the tradeoffs for Vce vs Ic in saturation.

The datasheet plots are misleading on a linear scale near 0 as the Ic must approach Ib to get near lowest practical values.

Better results may be obtained with 74HC logic or 6V CMOS R2R output comparator or OA being around 50 ohms Ro +/-50% with a 1M load.