Transimpedance Operational Amplifier Simulation circuit help

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  • Thread starter CNC101
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  • #26
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If you mean to replace V1 in the nodal equation with I*Rf i.e
(I*Rf-Vout)/R1+(I*Rf-0)/R2-I= 0 , then im stuck at a dead end :(
 
  • #27
gneill
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If you mean to replace V1 in the nodal equation with I*Rf i.e
(I*Rf-Vout)/R1+(I*Rf-0)/R2-I= 0 , then im stuck at a dead end :(
Why?

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!

upload_2016-2-1_19-14-28.png


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

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.

schem1.png
 
  • #29
gneill
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Ok, (-I*Rf-Vout)/R1+(-I*Rf-0)/R2-I= 0 transposes as Vout=-I*Rf(R1/R2+1) to which I= 1nA.
Presumably you've made a simplification based on a justified numerical approximation along the way?
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.
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.
 
  • #30
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Wel the simulation works however the voltage outcome is disappointing as you can see:

schem1.2.png
 
  • #31
gneill
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Try changing the op amp type to an ideal op amp. The software should have an ideal op amp in the parts list.
 
  • #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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FYI, I set up and ran a simulation under LTSpice in about two minutes and obtained the predicted results.
 
  • #34
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With LTspice and a precision operational amplifier model, I have -100.088mV which I think is close enough. Simetrix required a voltage controlled voltage source to create ideal opamp conditions which just got confusing.

op amp.png
 
  • #35
gneill
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With LTspice and a precision operational amplifier model, I have -100.088mV which I think is close enough. Simetrix required a voltage controlled voltage source to create ideal opamp conditions which just got confusing.
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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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]##
 
  • #38
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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!
 
  • #39
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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
 
  • #40
gneill
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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.
 
  • #41
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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.

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?
 
  • #42
gneill
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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?
Yes, just be sure to state clearly why your assumption/approximation is valid.
 
  • #43
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Will do, thank you.
 
  • #44
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!
 
  • #45
Is I = +1nF or -1nF?
 
  • #46
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 Vout and solve for I.
The result is, I = -1/1,900,000 A

I do not think this is the correct answer.
 
  • #47
The Electrician
Gold Member
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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.
 
  • #48
Gremlin
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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 text book as a reference but I dont have it with me.

Thanks.
 

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