Op.Amp:integrator and offset/Ibias errors

[negromonte1]

I have an home work to do:

http://img372.imageshack.us/my.php?image=integrator12ww.gif

At t=0+ the switch is open; at t=0- switch
is closed (vc=0). Calculate vo(t).

According the book the result should be:

vo(t) = VOS + (VOS/RC)*t + (IB2/C)*t (when t>=0)

But my guess is:

vo(t) = ± (VOS/RC)*t ± (IB2/C)*t (when t>=0)

(± sign, since neither VOS nor IB2 sign are predicible)

I don't understand the result of the book;
how that VOS (first term) can be in vo(t);
since amp.op. is in common mode range I used
"sum of effects" to calculate vo(t) and
according to me the contribute to Vout of
VOS is just to charge capacitor.

Thanks in advance for the help.

 

[Fred Bloggs]

I have an home work to do:

http://img372.imageshack.us/my.php?image=integrator12ww.gif

At t=0+ the switch is open; at t=0- switch
is closed (vc=0). Calculate vo(t).

According the book the result should be:

vo(t) = VOS + (VOS/RC)*t + (IB2/C)*t (when t>=0)

But my guess is:

vo(t) = ± (VOS/RC)*t ± (IB2/C)*t (when t>=0)

(± sign, since neither VOS nor IB2 sign are predicible)

I don't understand the result of the book;
how that VOS (first term) can be in vo(t);
since amp.op. is in common mode range I used
"sum of effects" to calculate vo(t) and
according to me the contribute to Vout of
VOS is just to charge capacitor.

Thanks in advance for the help.

When the switch is closed vo=v(-) which must be at VOS for v(+)=v(-).
And when the switch opens this condition still holds for v(+)=v(-) so
that vo= VOS + q/C, at each instant by simple sum of voltages around the
loop, where q is the time integral of current from vo to v(-) node
through C. The current required to maintain v(-) at VOS is just IB2 +
VOS/R1, a constant supplied through C, so the integral is then
q=(IB2+VOS/R1)*t making vo=VOS + IB2/C*t + VOS/(C*R1)*t.

 

[John Popelish]

I have an home work to do:

http://img372.imageshack.us/my.php?image=integrator12ww.gif

At t=0+ the switch is open; at t=0- switch
is closed (vc=0). Calculate vo(t).

According the book the result should be:

vo(t) = VOS + (VOS/RC)*t + (IB2/C)*t (when t>=0)

But my guess is:

vo(t) = ± (VOS/RC)*t ± (IB2/C)*t (when t>=0)

(± sign, since neither VOS nor IB2 sign are predicible)

The sign of VOS and IB2 are included in the symbol. In other words,
those variables include both a magnitude and a sign, just like any
other algebraic variable. Your method is more confusing than helpful.

I don't understand the result of the book;
how that VOS (first term) can be in vo(t);
since amp.op. is in common mode range I used
"sum of effects" to calculate vo(t) and
according to me the contribute to Vout of
VOS is just to charge capacitor.

Solve for the case when the switch is closed. That is what you start
with at t=0, before the capacitor has a chance to change its voltage
(integrate) at all.