Transimpedance Operational Amplifier Simulation circuit help

[CNC101]

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

 

[gneill]

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

In post #28 you wrote:

Ok, (-I*Rf-Vout)/R1+(-I*Rf-0)/R2-I= 0 transposes as Vout=-I*Rf(R1/R2+1) to which I= 1nA.

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.

The exact expression is:

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

 

[CNC101]

Well my exact expression was -I(R1(Rf/R2)+1)+R)=V Which gave me 0.9991nA which on the simulation, Vo is shy by just 0.002mV compared to 0.09mV!

 

[Rafeng404]

Hi guys,

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 = 1*10^9. Then if you transpose the Vout= -Rf/Rin*Vin formula and substitute Rf*(R1+R2)/Rf+(R1+R2) for Rf and put Rin as 1 (this was me using half educated reason because 0 would bugger the whole thing up) you get 1.01*10^-6 V for the input voltage which I think was mentioned earlier.
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

 

[gneill]

We know there is -0.1V across R1 and R2 so R1 must drop -0.09V and R2 must drop -0.01V.

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.

 

[Rafeng404]

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.

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?

 

[gneill]

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?

Yes, just be sure to state clearly why your assumption/approximation is valid.

 

[Rafeng404]

Will do, thank you.

 

[dave pallamino]

Hi All,

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!

 

[Michael Neo]

Is I = +1nF or -1nF?

 

[Michael Neo]

The exact expression is:

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

When I substitute 0.1 V for V_out and solve for I.
The result is, I = -1/1,900,000 A

I do not think this is the correct answer.

 

[The Electrician]

Hi All,

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!

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

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.

 

[Gremlin]

The exact expression is:

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

Hi, would it be possible to run through the steps of getting from

(((-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 textbook as a reference but I don't have it with me.

Thanks.