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I am using Comparator (schematic shown below). I have connected output with Vcc via 2k ohm resistor, as it has open collector output.

In this configuration output of the opamp should be 3.3V. But the output is coming 0V. enter image description here

But if I change the supply voltage to 5V, then the output is showing 5V. (schematic below)

enter image description here

Is there any rule "voltage at the terminal (inverting/non-inverting) of the comparator should not be higher than supply voltage of the opamp.

Regards,

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  • \$\begingroup\$ Output of the comparator is connected with Vcc via 2K resistor. \$\endgroup\$ – TapasX Jan 6 '18 at 13:04
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It's a comparator, not an op-amp. The question does not arise with op-amps because they generally do not have open drain outputs.

The answer is maybe. You will have to read the datasheet for the particular comparator (assuming it has open drain/open collector output).

In the case of the LM339 the maximum output voltage is 36V regardless of Vcc.

Some comparators may have protection diodes on the output, so the datasheet would likely state a maximum such as Vcc+0.3V.

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Look up the common mode input range in the datasheet. This most likely doesn't exceed the positive supply. If it doesn't, then your first circuit is running the part out of spec, and you have no guarantee what it will do.

Possible solutions:

  1. Run the comparator from a supply voltage that is high enough so that the expected input signals are within the common mode input range.

  2. Use resistor dividers on the input signals to bring them down to the common mode input range.

As always, you have to actually read the datasheet.

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For almost all ICs, the voltage on the inputs must be somewhere between the supply voltage and Ground for proper operation. For many linear parts, the inputs must be a significant voltage below the positive supply, or above the negative supply, for proper operation.

Read the datasheet for your device to learn its requirements.

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