- #26

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(I*Rf-Vout)/R1+(I*Rf-0)/R2-I= 0 , then im stuck at a dead end :(

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- #26

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(I*Rf-Vout)/R1+(I*Rf-0)/R2-I= 0 , then im stuck at a dead end :(

- #27

gneill

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Why?

(I*Rf-Vout)/R1+(I*Rf-0)/R2-I= 0 , then im stuck at a dead end :(

First of all, you seem to be stuck on getting the sign right for the relationship between ##I## and V1. I flows from left to right through ##R_f## causing a drop in potential. The left end of ##R_f## is at zero potential thanks to the input of the opamp being held at zero potential. That means V1 must be: ##V1 = - I R_f##. Note the negative sign!

Second, you end up with a single equation containing both ##I## and ##V_{out}##. Isn't that ideal? Just rearrange it to solve for ##V_{out}## in terms of ##I##, or vice versa.

- #28

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To get the Simetrix schematic Im having trouble running a simulation as I keep getting errors (as seen in the command shell) and I have no idea, as a novice, how to rectify it. I need to probe "Vout" using the current generator featuring current 1n amps.

- #29

gneill

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Presumably you've made a simplification based on a justified numerical approximation along the way?Ok, (-I*Rf-Vout)/R1+(-I*Rf-0)/R2-I= 0 transposes as Vout=-I*Rf(R1/R2+1) to which I= 1nA.

I'm not familiar with that simulation software. Looking at your schematic, I think you should turn your current source around so that the current is injected into the input. Check the polarity of the power supply connections to the TL072.To get the Simetrix schematic Im having trouble running a simulation as I keep getting errors (as seen in the command shell) and I have no idea, as a novice, how to rectify it. I need to probe "Vout" using the current generator featuring current 1n amps.

- #30

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Wel the simulation works however the voltage outcome is disappointing as you can see:

- #31

gneill

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- #32

NascentOxygen

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With 1nA of input current, what does your formula predict the output to be? Try it with a few μA.

- #33

gneill

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- #34

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- #35

gneill

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That result is almost exactly right if you don't make a numerical approximation simplifications in the expression for K. An exact value for K would yield 100.09 mV for the output voltage.

- #36

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Could you explain what you mean by Numerical approximation simplification, cheers :)

- #37

gneill

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In post #28 you wrote:Could you explain what you mean by Numerical approximation simplification, cheers :)

Which is true if you've made the assumption that ##\frac{(R_1 + R_2)}{R_2} R_f >> R_1##, which is a pretty good approximation in this case.Ok, (-I*Rf-Vout)/R1+(-I*Rf-0)/R2-I= 0 transposes as Vout=-I*Rf(R1/R2+1) to which I= 1nA.

The exact expression is:

##V_{out} = -I \left[ \frac{(R_1 + R_2)}{R_2} R_f + R_1 \right]##

- #38

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- #39

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Just thought I would chime in in case as a semi novice like me is reading this thread and to make sure there is nothing wrong with what I have done or if there is a situation where it would be none sense. Hopefully one of you clever lads will be able to put me right.

I got the 1nA just from doing ohms law. We know there is -0.1V across R1 and R2 so R1 must drop -0.09V and R2 must drop -0.01V. We can also agree that R2 is in parallel with RF which also much have the -0.01V volt drop. There for I=0.01/10*10^6 =

Did I get lucky here? or is that a reasonable way to do it? because If I am honest I haven't done your methods for years and couldn't be arsed having to re learn it haha.

Cheers

- #40

gneill

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How do you know that? There's a junction between R1 and R2 where there's a third connection (to Rf), and you can't make the assumption that it doesn't interfere with the potential drop ratio without justification. In the case of the given problem with the given component values it is possible to make such a justification, but if those values had been different it might not have been the case.We know there is -0.1V across R1 and R2 so R1 must drop -0.09V and R2 must drop -0.01V.

If you submitted your working for marking without providing such a justification, you'd very likely lose points.

- #41

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Thank you for your reply, just so that I understand you correctly, in this case, as we know all the values it is ok to do what I did?How do you know that? There's a junction between R1 and R2 where there's a third connection (to Rf), and you can't make the assumption that it doesn't interfere with the potential drop ratio without justification. In the case of the given problem with the given component values it is possible to make such a justification, but if those values had been different it might not have been the case.

If you submitted your working for marking without providing such a justification, you'd very likely lose points.

- #42

gneill

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Yes, just be sure to state clearly why your assumption/approximation is valid.Thank you for your reply, just so that I understand you correctly, in this case, as we know all the values it is ok to do what I did?

- #43

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Will do, thank you.

- #44

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My attempt is a little different, could any of you guys confirm I've gone about this one the right way?

Using KCL at V-

I=(0-Vx)/Rf

Vx=-I*Rf

Using KCL at Vx

((Vx-Vout)/R1)+(Vx/R2)=0

(R2*Vx)-(R2*Vout)+(R1*Vx)=0

Vout=(R1+R2/R2)Vx

Vout=-(R1+R2/R2)I*Rf

Vout=-kI

k=(R1+R2/R2)Rf

I=-((Vout*R2)/((R1+R2)Rf))

I=-((0.1*(10k))/((90k+10k)10M))=-1nF

Any advice would be greatly appreciated...Thanks!

- #45

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Is I = +1nF or -1nF?

- #46

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When I substitute 0.1 V for VThe exact expression is:

##V_{out} = -I \left[ \frac{(R_1 + R_2)}{R_2} R_f + R_1 \right]##

The result is, I = -1/1,900,000 A

I do not think this is the correct answer.

- #47

The Electrician

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Your KCL equation at Vx: ((Vx-Vout)/R1)+(Vx/R2)=0

My attempt is a little different, could any of you guys confirm I've gone about this one the right way?

Using KCL at V-

I=(0-Vx)/Rf

Vx=-I*Rf

Using KCL at Vx

((Vx-Vout)/R1)+(Vx/R2)=0

(R2*Vx)-(R2*Vout)+(R1*Vx)=0

Vout=(R1+R2/R2)Vx

Vout=-(R1+R2/R2)I*Rf

Vout=-kI

k=(R1+R2/R2)Rf

I=-((Vout*R2)/((R1+R2)Rf))

I=-((0.1*(10k))/((90k+10k)10M))=-1nF

Any advice would be greatly appreciated...Thanks!

is incomplete. It should be: ((Vx-0)/Rf)+((Vx-Vout)/R1)+(Vx/R2)=0

Also, the units in your final answer should be in nA, not nF.

- #48

Gremlin

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Hi, would it be possible to run through the steps of getting fromThe exact expression is:

##V_{out} = -I \left[ \frac{(R_1 + R_2)}{R_2} R_f + R_1 \right]##

(((-IRf)-0) / R2) + (((-IRf) - V) / R1) - I = 0 through to the above?

Initially I would tend to x R1 to remove one of the denominators and get to:

(-IRfR1/R2) + (-IRf - V) - IR1 = 0 but then I forget. I usually use my Stroud engineering mathemtics text book as a reference but I dont have it with me.

Thanks.

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