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Why is the offset voltage present on the non inverting terminal? What would happen on placing it on inverting terminal?

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    \$\begingroup\$ It is the same as a piece of string being too short at only one end. \$\endgroup\$ – jippie Feb 2 '16 at 17:34
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Offset voltage is unavoidable error in the construction of a opamp. Nobody "places" it anywhere. Manufacturers go to great lengths to reduce it, but of course, can not make it zero.

It also makes no sense to ask whether the offset voltage appears on the positive or negative input. The offset is between the two inputs. A opamp ideally does:

  Out = Gain(Pos - Neg)

In reality, it does:

  Out = Gain(Pos - Neg + Ofs)

The offset error is not specifically added to either input. It is a error resulting from finding the difference between the two inputs. You therefore can't say whether it appears on just the positive or just the negative input, and you wouldn't be able to tell the difference anyway.

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  • \$\begingroup\$ Is it constant if we take single op-amp or depends on other factors like temp,VCC,Input to OP-AMP etc ? My question is if I buy a op-amp from market, so offset will be constant to that specific op-amp due to some mismatch on input pins while IC fabrication? \$\endgroup\$ – Electroholic Feb 3 '16 at 6:18
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    \$\begingroup\$ @Elec: Some part of the offset of a particular opamp will be constant, other parts will drift over temperature and time. Sometimes datasheets will tell you how much the offset can drift. \$\endgroup\$ – Olin Lathrop Feb 3 '16 at 11:35
  • \$\begingroup\$ Ok then which is the parameter which will tell us about offset drift? \$\endgroup\$ – Electroholic Feb 3 '16 at 11:37
  • \$\begingroup\$ @Elec: It's usually cleverly disguised as "offset drift". \$\endgroup\$ – Olin Lathrop Feb 3 '16 at 14:13
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The offset is mostly due to mismatch in the 2 devices forming the differential pair at the input to the op-amp. The offset is differential, meaning it applies to the difference in voltage between the inverting and non-inverting terminals. For convenience, the offset is often referred to the non-inverting terminal and modeled as a voltage source in series with the (ideal) differential input.

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To turn on a transistor takes a few hundred milli volts. There are two transistors that are on the inputs of an op-amp; one connected to the non-inverting input and one on the inverting input. If these transistors were identical they would have identical input voltage characteristics and the input offset voltage (i.e. the difference between the two characteristics) would be zero.

So, input offset voltage is a measure of the difference between the two transistors' characteristics. This means it is arbitrary to think of that "error" voltage being on one terminal or the other.

The same is true of input bias currents - ideally, both input transistors would need the same amount of bias current to turn on a set amount (I'm particularly thinking of BJTs here because they do need a small base-bias current).

But there is usually always a difference and this is called "offset" current and, again, it's arbitrary as to which input you choose to regard this offset current as emanating from/to.

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Let`s take as an example the classical inverting amplifier configuration with the closed-loop gain of (-R2/R1).

Case (a): Connect a small dc voltage Voff=1µV between the non-inv. input and ground; Case (b): Connct the same voltage source between the inv. input terminal and the common node of both feedback resistors.

Calculate for both cases the resulting DC output voltage (for ideal open loop gain Aol approaching infinity. In both cases, the result will be identical (noise gain):

Vout=Vo(1+R2/R1)

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In the model of the op-amp as an ideal op-amp with a voltage Vos placed in series with the non-inverting input, you have something like this:

http://images.elektroda.net/69_1236538060.jpg

The Vos in this case adds to the input on the non-inverting input, so a positive Vos drives the output of the amplifier more positive. If you move it to the non-inverting input then the result will be identical from all outward appearances, however you'd need to draw it 'flipped' so that it subtracted from the inverting input voltage to keep the sign the same.

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