How does input DC offset affect the output of opamp?

In open-loop situation, what will happen?
Since DC offset cannot be amplified, what's the difference at the output between with DC offset and without DC offset? any equation to solve this? thanks!

for open loop , the output will be gnd or vdd(no load)
configure the amp as a flower, the difference of output and input is offset(assume the gain is large enough)

I think, in fact, the tempeture is alterable, and when tempeture change, the quiet operation point will change. this change can be amplified by op, so the offset happen

If the DC input offset is very small, it will be amplified, the gain equals to DC open loop gain, right?

pipeline,

DC offset is the DC offset voltage specific to each input terminal when the op-amp is:
1. in open-loop
2. no load at the output
3. no load at the input

Theoretically in principle, DC offset at the input should be 0V with respect to the ground. Output is saturated at Vcc or Vee in Bipolar op-amp, or Vdd or Vss in CMOS op-amp with respect to ground.

DC offset doesn't get amplified in open-loop op-amp. In closed loop, it gets amplified.

Why in closed loop, it gets amplified?

Pipeline said:

If the DC input offset is very small, it will be amplified, the gain equals to DC open loop gain, right?

Click to expand...

YES you are right , but since the offsets are usually not equal 0 and the amplifier gain is extreemly high, the output is - or + supply voltage (depending of the offset sign). In idealised case however when the offset is equal 0 ( I mean real 0,0000000000000000000.......) at the output you should get also 0V.

starcoming said:

Why in closed loop, it gets amplified?

Click to expand...

Because the offset voltage is added to the input signal.
(This addition takes place inside the Op amplifier structure )

One of the method do measure the offset voltage is to apply a 0V at the input (at this stge you will get +V or -V at the output) and then you adjusted the input (+ or -) to get 0V at the output. The input volltage at this condition is the offset value.

I can also add, that the offset can be split to two components:
the voltage offset and the current offset.

SkyHigh said:

DC offset doesn't get amplified

Click to expand...

Are you sure about this? Can you describe it more precisely?

In my opinion, DC offset is amplified either in closed loop or in open loop, either it is small or large. But if the offset is quite large, the output may be saturated to vdd or vss due the amplification of the input offset voltage.

in closed loop the front-stage's out is isolated from the later-stage's in by capacitor .

microwavefly said:

in closed loop the front-stage's out is isolated from the later-stage's in by capacitor .

Click to expand...

By DEFINITION the operationl amplifier is an amplifier with DC coupling.
Hence there is no coupling capacitances between the stages.

By addind the external resistors and capacitors one can get different gain for DC and AC components. In simplest case you can connect the capacitor at the output isolating the DC component but in this case you can not analize the circuit as op. amp.

There are compensation capacitors in the op amp. structure to ensure the amplifier stability when feedback is applied, but they shape the frequency responce at the upper frequences only.

If the OPAMP deviced are mismatched, then the DC offset present at the i/p will get amplified by the common mode gain.

Again, the presence of DC offset wiill limit the o/p voltage swing.