Understanding "Voltage - Input Offset" Parameter - diyAudio
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Old 11th February 2013, 02:02 PM   #1
Loren42 is offline Loren42  United States
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Default Understanding "Voltage - Input Offset" Parameter

What is the significance of the "Voltage - Input Offset" parameter for an op amp when used as a line stage or buffer amp?

For instance, the OPA134 is rated 500 V and the LM4562 is rated at 100 V. However, what does that mean for the typical topologies used for line amps and buffers?
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Old 11th February 2013, 03:35 PM   #2
rsavas is offline rsavas  Canada
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Input offset voltage is important for an amplifier that operates at or provides gain at DC. If you AC couple signals into & out of the OPA, it really does not mean much.
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Old 11th February 2013, 03:44 PM   #3
DF96 is offline DF96  England
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Input offset voltage is the voltage you may have to provide (from a zero impedance source) to ensure that the opamp output voltage is zero. It could be argued that it is a measure of how well the input differential stage is balanced; this may or may not be of any real significance for audio purposes.

An opamp intended for DC use (e.g. in instrumentation) may have very low input offset voltage. This may, or may not, indicate that other aspects of behaviour more important for AC use have been compromised in order to achieve good DC performance.
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Old 11th February 2013, 03:47 PM   #4
Loren42 is offline Loren42  United States
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Okay. If one op amp feeds a second op amp as a two-stage design and there are no caps used (DC coupled), does it matter?
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Old 11th February 2013, 03:56 PM   #5
DF96 is offline DF96  England
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That would depend on how the second opamp gets its 'ground' reference, and how much DC gain the first stage has. Show us a circuit.
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Old 11th February 2013, 03:56 PM   #6
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Quote:
Originally Posted by Loren42 View Post
For instance, the OPA134 is rated 500 V
This means that if configured for a gain of 1 and DC coupled the output will have a worst case offset of 500uV. If configured with a gain of 10, the output offset will be 5000uV or 5mV.

Opamp offsets are specified as input referred. This is an irreducible apparent voltage between the input terminals. It's actually defined as the voltage you would need to apply to bring the output to zero. It enables you to calculate the likely offset at the output on the basis of the (DC) gain.
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Old 11th February 2013, 04:12 PM   #7
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You have to remember that there are also input offset currents to be taken into consideration. Since these are generally supplied via the biassing and other resistors connected to the opamps inputs they develop a voltage across these resistances, and hence they will also contribute to any output offset voltage. If the resistances are very low, or the currents very low, as in the case of jfet input opamps, their contribution may be negligible, but they may also be significant.

Last edited by counter culture; 11th February 2013 at 04:15 PM.
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Old 11th February 2013, 04:19 PM   #8
Loren42 is offline Loren42  United States
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Quote:
Originally Posted by DF96 View Post
That would depend on how the second opamp gets its 'ground' reference, and how much DC gain the first stage has. Show us a circuit.
This is one example:

Click the image to open in full size.
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Old 11th February 2013, 04:44 PM   #9
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That's a bit complicated.

There's a DC path from input to output. Put the voltages into the circuit and calculate the DC operating point.

Op Amp Offset Voltage

Op Amp Bias Current

Last edited by counter culture; 11th February 2013 at 04:48 PM.
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Old 11th February 2013, 04:46 PM   #10
DF96 is offline DF96  England
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First stage has gain x2. Second stage up to x4 (ish?). So output offset could be up to 8 times input offset plus 4 times offset for second opamp. Not enough to do any harm, apart from clicks if any switching done. You would need to add on any DC gain in whatever follows.
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