Comparator output not going to ground?

[r.youden]

I have a simple circuit which is controlled by a comparator into an analogue multiplexer, i.e. when the input voltage at the comparator input is 'high' then the analogue multiplexer output one signal and when the input is 'low' the output is the alternative signal, all very simple.

However I need the output of the comparator to be a solid 0V, currently I am getting around 0.1V which is screwing up the analogue multiplexer. Is there anyway that I can ensure that that output of the comparator goes solidly to ground? I have a 10k pull-up on the output and I am using a LM6511 with single rail supply (I am not able get a negative supply on the board)

Thanks.

 

[ericgibbs]

I have a simple circuit which is controlled by a comparator into an analogue multiplexer, i.e. when the input voltage at the comparator input is 'high' then the analogue multiplexer output one signal and when the input is 'low' the output is the alternative signal, all very simple.

However I need the output of the comparator to be a solid 0V, currently I am getting around 0.1V which is screwing up the analogue multiplexer. Is there anyway that I can ensure that that output of the comparator goes solidly to ground? I have a 10k pull-up on the output and I am using a LM6511 with single rail supply (I am not able get a negative supply on the board)

Thanks.

hi,
Are you using the comparator output to control the analog mux.
Do you have circuit to post.?

 

[r.youden]

I'll try and explain the system as a whole, here goes:

I have a PT100 temperature sensor (2-wire configuration) and the voltage across the sensor (driven by a 1mA constant current source) along with a 100ohm fixed reference is fed into a differential amplifier. That bit works fine and produces a nice 0-1V output signal. However I have come across a problem where by when the sensor is disconnected the output of the differential amplifier saturates at the supply rail, not what I need!

So I had the idea of setting a reference voltage (2v) and the output of the PT100 into a comparator and through an analogue mux which is connected to the output of the differential amplifier, along with the second input being pulled to ground. So when the sensor is removed, the output of the comparator goes low and mux disconnects the differential amplifier and connects the output of the circuit to ground.

Or so I though... The comparator does not produce a perfect ground signal which, for some reason, means the 0v signal coming out of the mux is not 0v, rather 0.04v which is around 4degC on this circuit, again not what I wanted.

I have connected the mux up with a switch connected directly to the supply and ground and this problem goes away, hence my thinking that the output of the comparator not being exactly 0v is causing the problem.

Thanks.

 

[Chippie]

I'll try and explain the system as a whole, here goes:

I have a PT100 temperature sensor (2-wire configuration) and the voltage across the sensor (driven by a 1mA constant current source) along with a 100ohm fixed reference is fed into a differential amplifier. That bit works fine and produces a nice 0-1V output signal. However I have come across a problem where by when the sensor is disconnected the output of the differential amplifier saturates at the supply rail, not what I need!

Swap the connections to the comparator so that the PT100 thermistor is in place of the 100r resistor and the 100r is where the thermistor was, on disconnection the comparator output should then go low..

 

[crutschow]

It is normal for the comparator output to have a slight voltage when low since the output is a bipolar transistor. To get a solid 0V have the comparator drive a CMOS buffer or two inverters. If the CMOS output has a small load, then it will go very close to 0V when low.