deduction of output voltage relation

4electros

New Member

• Jun 25, 2006

• #1

in the circuit shown :

assume that reset switch is closed for t<0, then opened for t=0

it's needed to deduct the relation of output voltage vo(t) for t>=0

firstly i think i have to use superposition theorem.

 

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dknguyen

Well-Known Member

Most Helpful Member

• Jun 25, 2006

• #2

Transient analysis is what you need to use, not superposition. Superpositions is when different frequency signals are in the same circuit and you analyze the circuit separately for each frequency.

For initial conditions (t<0) you assume that the switch has been closed for a very long time and so C1 is storing no charge. Then you do transient analysis.

 

Last edited: Jun 25, 2006

4electros

New Member

• Jun 25, 2006

• #3

concerning t<0 , i think that v-=v+=Vos nearly due op-amp characteristic A=infinity so v(t=0)=vos is to be after very long charge

anyway what would i do to apply transient analysis? where should i start?

 

Last edited: Jun 25, 2006

R

Russlk

New Member

• Jun 25, 2006

• #4

First you need to know whether Ib*R1 is more or less than Vos. The sum of the currents at a junction must equal zero.

 

4electros

New Member

• Jun 26, 2006

• #5

Russlk said:

First you need to know whether Ib*R1 is more or less than Vos. The sum of the currents at a junction must equal zero.

Click to expand...

yeah .v+=v- due to opamp characteristic A=infinity where vo=A(v+ - v-)

so Vos=IB1*R1

 

R

Russlk

New Member

• Jun 26, 2006

• #6

"so Vos=IB1*R1"

No, if it were so, the solution would be trivial. You have to assume that IB1*R1 is not equal to Vos. But v+ and v- of the opamp do have to be equal. Where does the excess current go?