Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

How does input DC offset affect the output of opamp?

Status
Not open for further replies.

Pipeline

Advanced Member level 4
Joined
Dec 7, 2004
Messages
118
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,296
Activity points
881
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!
 

sunking

Advanced Member level 3
Joined
May 25, 2004
Messages
874
Helped
70
Reputation
140
Reaction score
23
Trophy points
1,298
Activity points
6,283
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)
 

wjxcom

Full Member level 5
Joined
Sep 7, 2005
Messages
278
Helped
4
Reputation
8
Reaction score
3
Trophy points
1,298
Activity points
3,798
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
 

Pipeline

Advanced Member level 4
Joined
Dec 7, 2004
Messages
118
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,296
Activity points
881
If the DC input offset is very small, it will be amplified, the gain equals to DC open loop gain, right?
 

SkyHigh

Advanced Member level 1
Joined
Jan 13, 2005
Messages
432
Helped
79
Reputation
158
Reaction score
27
Trophy points
1,308
Activity points
7,319
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.
 

AndrzejM

Member level 4
Joined
Dec 3, 2004
Messages
71
Helped
12
Reputation
24
Reaction score
3
Trophy points
1,288
Location
Szczecin-Poland
Activity points
1,053
Pipeline said:
If the DC input offset is very small, it will be amplified, the gain equals to DC open loop gain, right?
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?
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
Are you sure about this? Can you describe it more precisely?
 

Hughes

Advanced Member level 3
Joined
Jun 10, 2003
Messages
717
Helped
112
Reputation
224
Reaction score
23
Trophy points
1,298
Activity points
5,986
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.
 

microwavefly

Newbie level 3
Joined
Aug 15, 2005
Messages
4
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,307
in closed loop the front-stage's out is isolated from the later-stage's in by capacitor .
 

AndrzejM

Member level 4
Joined
Dec 3, 2004
Messages
71
Helped
12
Reputation
24
Reaction score
3
Trophy points
1,288
Location
Szczecin-Poland
Activity points
1,053
microwavefly said:
in closed loop the front-stage's out is isolated from the later-stage's in by capacitor .
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.
 

alok_ky

Full Member level 1
Joined
Aug 24, 2005
Messages
96
Helped
2
Reputation
4
Reaction score
1
Trophy points
1,288
Location
bangalore india
Activity points
2,071
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.
 

Status
Not open for further replies.
Toggle Sidebar

Part and Inventory Search


Welcome to EDABoard.com

Sponsor

Sponsor

Design Fast


×
Top