I am using a low power dual comparator (part number ALD2301A). I am trying to take some measurements on the output. I set the inverting to 2.5V and vary the non-inverting between 0V and 5V. When the non-inverting is between 0V and 2.5V, the output should read 0V, correct? When the non-inverting is between 2.6V and 5V, the output should read 5V, correct? Will this is not happening during my simple experiment. I get small voltages (mV) on the output while adjusting the non-inverting between 2.6V and 5V. Why I do not get 5V or something near it on the output? I get 0V when adjusting the non-inverting between 0V and 2.5V.

I set the power supply on the op-amp to 5V. I am not loading the output with any resistor and have not attached any resistore to the inputs of the op-amp.

I am only using one of the comparators on the chip. Nothing is hooked up to the other comparator on the chip..

Why is this happening? Am I missing something? Can anyone help me?

 

Hello:

The spec sheet at http://www.aldinc.com/pdf/ALD2301.pdf. indicates that this device has an OPEN DRAIN output. Try putting a 4.7k 'pull-up' resistor on the output pins 1 or 7.

Should work just fine.

Spoggles

 

Originally posted by Spoggles@May 16 2006, 03:25 PM
Hello:

The spec sheet at http://www.aldinc.com/pdf/ALD2301.pdf. indicates that this device has an OPEN DRAIN output. Try putting a 4.7k 'pull-up' resistor on the output pins 1 or 7.

Should work just fine.

Spoggles

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Thanks for replying. What does open drain mean? Those it mean a path of current must be present to measure a voltage. I will try this in the morning.

 

The Field Effect Transistor(FET) term "open-drain" and it's Bipolar Junction Transistor(BJT) counterpart "open-collector" describe an output structure where the source of a FET or the emitter of a BJT is connected to GROUND and the drain of the FET or the collector of the BJT is open. This means that some piece of circuitry external to the chip must provide a way to pull the output up to a high level when the FET or the BJT is turned off. A pullup resistor to Vdd or Vcc is commonly used.

What value should you use? If you make it too small then when the FET or BJT is on it will draw several milliamps and the resistor may get hot. If you make it too large then the rising edge of the output may get really slow. A value from 1K to 10K should be adequate for most applications.

So an open-drain or open-collector output with no pullup resistor will always be low, and it will never go high.

 

Greetings, dar2525,

To add to the information already posted by papabravo and spoggles, it is probably worth noting that among the reasons for designing comparators with open-drain or open-collector outputs is to permit their outputs to be connected in a "hard-wire" OR'ed configuration. In effect, the output of several comparators can then be connected together and provided with a common pull-up resistor. In this configuration any one of the comparators so connected can drive the common output node low. Conversely, all of the comparators must produce a high level on their outputs for the common node to go high.

I think I noticed in the datasheet for the subject device just such a configuration shown in one of the example figures.

hgmjr

 

Originally posted by Papabravo@May 16 2006, 08:40 PM
The Field Effect Transistor(FET) term "open-drain" and it's Bipolar Junction Transistor(BJT) counterpart "open-collector" describe an output structure where the source of a FET or the emitter of a BJT is connected to GROUND and the drain of the FET or the collector of the BJT is open. This means that some piece of circuitry external to the chip must provide a way to pull the output up to a high level when the FET or the BJT is turned off. A pullup resistor to Vdd or Vcc is commonly used.

What value should you use? If you make it too small then when the FET or BJT is on it will draw several milliamps and the resistor may get hot. If you make it too large then the rising edge of the output may get really slow. A value from 1K to 10K should be adequate for most applications.

So an open-drain or open-collector output with no pullup resistor will always be low, and it will never go high.

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Thanks for replying. I am still not understanding the use of a pull-up resistor. Exscuse my ignorance.

 

Originally posted by dar2525@May 17 2006, 10:34 AM
Thanks for replying. I am still not understanding the use of a pull-up resistor. Exscuse my ignorance.

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If you wire a simple switch with one side to ground and the other side open, then the open side of the switch will be at ground if the switch is closed and it will "float or be at an indeterminate value when the switch is open.

Now wire the simple switch with one side to ground, the other side to a pullup resistor and the other side of the pullup resistor to a positive voltage. Now consider the point between the switch and the resistor.

When the switch is closed the point will be at GROUND. The current flowing through the resistor will by Ohm's Law be V / R. If V is in Volts and R is in Ohms then the computed current will be in amps.

When the switch is open, there is no current flowing. If there is no current flowing in the pullup resistor then there can be no voltage drop across the resistor. If there is no voltage drop across the resistor then the voltage must be the same on both ends of the resistor, which means the point between the switch and the resistor has been "pulled-up" to the battery voltage.

 

Originally posted by Papabravo@May 17 2006, 04:42 PM
If you wire a simple switch with one side to ground and the other side open, then the open side of the switch will be at ground if the switch is closed and it will "float or be at an indeterminate value when the switch is open.

Now wire the simple switch with one side to ground, the other side to a pullup resistor and the other side of the pullup resistor to a positive voltage. Now consider the point between the switch and the resistor.

When the switch is closed the point will be at GROUND. The current flowing through the resistor will by Ohm's Law be V / R. If V is in Volts and R is in Ohms then the computed current will be in amps.

When the switch is open, there is no current flowing. If there is no current flowing in the pullup resistor then there can be no voltage drop across the resistor. If there is no voltage drop across the resistor then the voltage must be the same on both ends of the resistor, which means the point between the switch and the resistor has been "pulled-up" to the battery voltage.

[post=17191]Quoted post[/post]​

I understand,thank you.