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negromonte1
Guest
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 Posted:
Mon Dec 12, 2005 5:35 pm Post subject:
Op.Amp:integrator and offset/Ibias errors |
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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.
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John Popelish
Guest
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| Posted:
Mon Dec 12, 2005 5:35 pm Post subject:
Re: Op.Amp:integrator and offset/Ibias errors |
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negromonte1 wrote:
| Quote: | 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.
| Quote: | 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.
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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.
> Thanks in advance for the help. |
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Fred Bloggs
Guest
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| Posted:
Mon Dec 12, 2005 5:35 pm Post subject:
Re: Op.Amp:integrator and offset/Ibias errors |
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negromonte1 wrote:
| Quote: | 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.
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neon
Joined: 25 Feb 2006
Posts: 594
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| Posted:
Fri Mar 31, 2006 1:19 pm Post subject:
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| with the switch across the amp open or close the amp is saturated to whatever the it goes to open loop depends on A IT SHOULD SATURATE both way. |
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