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How to simulate offset voltages?

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lics

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Hello everyone,
I use cadence to simulate an amplifiers offset voltage, but it shows the offset is 170mV. I think it's too big. My method is as following: First, I connect the negative input and the output of the amplifier. Then, I add a DC signal at the positive input and sweep DC voltage. I get output voltages changing with the DC signal. When output is zero, the positive input is about 170mV. So this voltage is offset voltage. I don't know whether this method is corret. Can anyone tell me?
Thank you in advance!
 

invent

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attached is a circuit commonly used. However other considerations are needed, such as if it is an open drain configuration, a resistor is needed at the output. Also what is the operating voltage? Try not to operate at the minimum voltage.

Another consideration is that if you are operating the circuit on a single supply (i.e. vdd=3V, vee=gnd) then the vio is not at 0v. Rather, it is usually taken at the middle point (1.5V).

hope this helps.
 

pfd001

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lics said:
Hello everyone,
I use cadence to simulate an amplifiers offset voltage, but it shows the offset is 170mV. I think it's too big. My method is as following: First, I connect the negative input and the output of the amplifier. Then, I add a DC signal at the positive input and sweep DC voltage. I get output voltages changing with the DC signal. When output is zero, the positive input is about 170mV. So this voltage is offset voltage. I don't know whether this method is corret. Can anyone tell me?
Thank you in advance!

You don't need to connect the negative input and output of the amp. You can leave the amplifier open. Then you can give a correct common mode voltage at the negative input, sweep the DC voltage at the positive input, the difference between the negative and positive is the offset voltage when the output is zero (maybe at the common mode voltage)
 

yutian

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thans ,i know,but i think this method is the same as the Allen's book.
 

timo67

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As you still are in simulation and not in test
I think you need only to simulate your op amp in open loop, that is :
negative input : grounded ( if your op amp has a symetric power supply ie vdd=-vss)
positive input swept from -10mV to 10mV or less

Then yo note the value of the positive input that gives zero volt at output
So you have mesured the output offset voltage
Yo need to divide this output offset voltage by the DC open loop gain of your op amp you find the input offset voltage of your operational amplifier
PS: if your op amp does not have a symmetrical power supply you have to replace the ground by an adequate DC bias voltage.
good luck!
 

lics

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I try another methode. First, let the negative input is the common voltage. Then, sweep the positive input. See the output voltage when the positive is at the common voltage. This output voltage is offset output voltage. Using the voltage divided by the amplifier's open gain is input offset voltage.
Is this methode right?
 

jwfan

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If you use this method, if the gain is large, you may see 0 or VDD at output when the positive terminal is also common voltage.

I think pfd001's method is right.

lics said:
I try another methode. First, let the negative input is the common voltage. Then, sweep the positive input. See the output voltage when the positive is at the common voltage. This output voltage is offset output voltage. Using the voltage divided by the amplifier's open gain is input offset voltage.
Is this methode right?
 

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