The Designer's Guide Community Forum

The Designer's Guide Community Forum
https://designers-guide.org/forum/YaBB.pl
Design >> Analog Design >> Comparator Negative Input Voltage
https://designers-guide.org/forum/YaBB.pl?num=1408975108

Message started by bki on Aug 25^th, 2014, 6:58am

Title: Comparator Negative Input Voltage
Post by bki on Aug 25^th, 2014, 6:58am

Hi,

i need to compare a negative output voltage with a reference signal.
The reference signal has 2,4V. The output signal is -2.4V or lower.
I thought of putting the Output signal to the negative input of the comparator and turn the reference signal to -2,4V (through turning the source) to put it to the positive comparator input. Then i receive the following formula:
-(-Vout)+(-Vref)=Vout-Vref
But now my advisor says it´s not possible to turn the reference voltage to negative. I am wondering why, can´t i just connect the source opposite?
Maybe can you explain me what the problem is and give an alternative solution?

Thank you!

Title: Re: Comparator Negative Input Voltage
Post by RobG on Aug 25^th, 2014, 1:34pm

It is hard to understand what you are doing. It will be helpful if you can draw and post a schematic of how you are trying to accomplish your goal.

Title: Re: Comparator Negative Input Voltage
Post by aaron_do on Aug 25^th, 2014, 6:31pm

Hi,

I think I understood parts of what you're trying to do. Off the top of my head, if your -ve voltage is reasonably low impedance, you could do a voltage divider between the +2.4 and -2.4 to get 0 V, and then compare the signal with "GROUND". I'm not sure how well that would work. It would most likely be slow. There may also be a switched-capacitor way of doing this. For example, charge cap A to +2.4, and cap B to -2.4, then add the two caps together and compare with "GROUND".

Its also possible to generate a negative voltage with a negative charge-pump, but its not that accurate. Actually I have a sneaking suspicion that you already did that, and you're trying to make it more accurate :P

You are going to have to explain what you mean by "connect the source opposite". What is the source? Also, as RobG mentioned, you should probably describe your application so people know what you're talking about...

cheers,
Aaron

Title: Re: Comparator Negative Input Voltage
Post by Ken Kundert on Aug 25^th, 2014, 9:31pm

A comparator acts like a very high gain amplifier where the output is expected to clip at one of the supply voltages. Thus, it roughly follows this pseudocode:

Code:

Vout = Av*(V(pin) - V(nin));
if (Vout > Vdd)
V(out) = Vdd;
else if (Vout < 0)
V(out) = 0;
else
V(out) = Vout;

where Av is assumed to be very large (>10,000).

Say V(pin) is fixed at 2.4V and V(nin) = -2.3V. Then V(out) will equal Vdd as expected (because Av*(2.4--2.3) = 4.7*Av, which is a very large positive number). Now let V(nin) = -2.5; V(out) will still equal Vdd (because Av*2.4--2.5) = 4.9*Av, which is still a very large positive number).

If you want to compare 2.4V with -2.4V, you should add them together and then compare the result against 0.

-Ken

Title: Re: Comparator Negative Input Voltage
Post by loose-electron on Aug 28^th, 2014, 12:59pm

bunch of different ways of doing this - but the problem needs to be better defined

If the OP can better define the problem?

sounds like a school project of some form.