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Comparator doesn't work as expected? Answered


I'm having a strange problem with the LT1016

When the -input is at 0.630V, and the +input is at 0.555V, and the comparator is fed with 5V,
the outputs are because of some reason Q(inverted) = 2.10V and Q = 2.67V
2.67V for Q is just out of specifications (minimum High voltage of the pin should be 2.7V).
2.10V for Q(inverted) is not good as it should be less than 0.3V in it's low state.

The "latch enable" pin is grounded.
Both output pins are not connected.

Why doesn't Q(inverted) go to 0.3V and Q to 3V, as the datasheet suggests?
I tested this with another (also new) comparator; also an LT1016, and it behaves exactly the same.

1 V+ 5V
2 +IN 0.555V
3 -IN 0.630V
4 V- 0V
5 Latch Enable 0V
6 GND 0V
7 Qout 2.67V
8 Qout(inverted) 2.10V

Any ideas?


In Appendix A of the data sheet you linked to ,there is a sort-of "artist's interpretation" (figure A-1) of what the outputs of this comparator look like inside. It looks like two NPN transistors stacked on top of each other, with the emitter of the top one connected to the collector of the bottom one.

I don't know if that makes everything clear or not. I am guessing your IC is in fact working as designed. That is to say, those little transistors in its output stage really are turning on and off, but with no load attached, the voltage there is kinda floaty.

I like IrishJim's suggestion of connecting a 1K resistor from the output to ground, and then observing what this does to the output voltage.

I also suggest trying a 1K resistor connected from output to V+ = 5V, and see what that does.

Also try it with both the "pull-up" and "pull-down" resistors together: i.,e with a 1K resistor connected from +V to output and another 1K resistor connected from output to ground.

I am hopeful some arrangement of these pull-up, pull-down, resistors will make your comparator's outputs behave better.

Real hardware design is really hard !

Its a nice comparator for dual supplies, but as the late, great Jim Williams from Linear points out in another app note for another chip, it wasn't great at single ended uses.

OK, I guess it is not a problem with the output stage, since different pull-up, or pull-down, resistors do not help. Sorry about that.

I am starting to think it is an input thing, specifically this op-amp does not like input voltages within about 1.5 volts of either supply rail.

Also I found some scribbles on the datasheet which seem to confirm this hypothesis. In the table on page 3, titled "Electrical Characteristics", on line 5, there's a parameter called "Input Voltage Range". For what it calls "Single 5V Supply" (which I guess means negative rail at V-=0 and positive rail at V+=5V) the input voltage range is only +1.25 V to 3.5 V.

Obviously this is going to be a huge problem if the voltages you want to compare are, like 0.02V and 0.50V in your tests, in the range from 0 to +1.25V.

One solution might be to use a double-sided power supply, i.e. one with negative rail at -5V, and positive rail at +5V, and ground rail at 0V. Then your usable input range should be from -3.75V to 3.5V

Or maybe you'll want to go shopping for a different comparator, one that does not have this funny input range problem.

I'm sorry if it comes to the second option.

It seems like I've had this happen to me before. The ICs look totally sexy when they're in the catalog, but it's only after getting them into a circuit when I discover they're not working the way I imagined they would. Often the ugly truth is there in datasheet somewhere, in the fine print.

You need 4,5,6 to ground. If you're showing your circuit, put ALL the pin numbers on too ;-)

From the data sheet "

A more general consideration is that the common mode range
is 1.25V above the negative supply and 1.5V below the
positive supply, independent of the actual supply voltage."

...which means, on a single supply system, without other bits and pieces, you are limited to +1.25 and 3.5 V inputs

Show your exact circuit.

Try putting a 1Kohm resistor on the outputs to ground, and then measure at the output pin. steveastrouk is correct that open collector outputs must have a load.

The outputs are open-collector, they NEED a load.