Engineering Transimpedance Operational Amplifier Simulation circuit help

[CNC101]

Homework Statement

(a) Derive the relationship
between the output voltage V and the input current I; i.e. if V = kI
find k in terms of R1, R2 and Rf.

(b) Calculate the current I if Rf = 10 MΩ, R1 = 90 kΩ, R2 = 10 kΩ
and V –0.1 V.

(c) Model the circuit using PSPICE [using an ideal opamp]
and use the model to confirm the calculation made in (b).

Homework Equations

The Attempt at a Solution

(a) V= KI, K= R2/ R2+ (Rf*R1/Rf+R1)

(b) I= V/K , I= -0.1/(10k/10k+(90k*10M/90k+10M), I= -0.992A (too big a current?)

(c)

Im not sure how to complete this schematic, I need to check current calculation in (b). I don't kow what "Vin" is. Any help is greatly appreciated as I have no training on this software.

 

[gneill]

In part (a) the syntax in your algebraic expression for K is ambiguous: the order of operations is not clear. Use parentheses to make the order of operations unambiguous. It also seems unlikely that the output voltage would be positive if $I$ is positive. Shouldn't K have a negative value? The input current $I$ should impose a potential drop across the feedback resistor, pushing the junction of R1 and R2 negative.

Can you show your derivation of K?

For part (b), yes, that current looks to be too big and of the wrong sign. Note that the given output voltage is negative.

For part (c) you should be able to use a current source at the input. Set it to the value you calculate in part (b) and check the resulting output voltage V.

 

[CNC101]

Ok, for (a) my derivation of "K" is that Rf (feedback resistor) is in parallel with R2, so I have (Rf*R2)/(Rf+R2) = (10*10^6*10*10^3)/(10*10^6+10*10^3) = 1. Using the voltage divider rule I have -(R1/(R1+1)) = K (where 1= (Rf*R2)/(Rf+R2)). Thats my relationship for K, for (b) I have V=KI, I=V/K therefore -0.1/-(90*10^3/(90*10^3+1)) =0.1A which is positive but still too large considering it is a transimpedance circuit that measures very small currents.

 

[gneill]

You should avoid plugging in numbers and rounding before the very end when you have values where small differences can be important to the behavior of the system. A purely symbolic derivation is safer. Also, I think you've lost several orders of magnitude in calculating the value of Rf || R2. It should be closer to 10 kΩ, not 1 Ω .

By using the voltage divider rule you're ignoring the current $I$ that's injected into the junction of R1 and R2.

I get a different value for K, one that is negative and quite large.

Instead, try KCL node analysis at the junction of R1 and R2, and note that the voltage there is fixed by the potential drop across Rf (v1 in the image below).

 

[CNC101]

KCL node analysis at the junction of R1 and R2, I have (V1-Vout)/R1 for I1 and for I2 I have (Vout-V1)/R2.

 

[NascentOxygen]

KCL node analysis at the junction of R1 and R2, I have (V1-Vout)/R1 for I1 and for I2 I have (Vout-V1)/R2.

And what's your related expression for I, the current through R_f?

 

[LvW]

...and for I2 I have (Vout-V1)/R2.

Ohms law for R2 results in I2=-V1/R2.

 

[CNC101]

Well according to KCL, I1+I2=I so V1/R1-V1/R2= I

 

[gneill]

Well according to KCL, I1+I2=I so V1/R1-V1/R2= I

Nope and nope.

Both I and i2 are shown flowing into the node, while only i1 flows out. So i1 = i2 + I. That means i1 - i2 = I. Also, R1 and R2 terminate at different nodes at different potentials (R1 terminates at the output node that has potential V), so you can't write V1/R1 - V1/R2 = I. You need to write the node equation that takes into account the nodes where the branches terminate.

 

[CNC101]

If R1 terminates at the output node then that's Vout/R1 and R2 terminates at the central node so that's -V1/R2, and for Rf is V1/Rf. Since I1=I+I2 then Vout/R1=-V1/R2+V1/Rf.

 

[gneill]

No, in nodal analysis you write the current on a branch between two nodes as the potential difference between the nodes divided by the branch resistance. So for example if there was a resistance R between a node at potential Va and another node at potential Vb, then the current flowing from node A to node B through resistance R would be given by (Va - Vb)/R.

 

[CNC101]

Well for that I have (V1-Vout)/R1 + (V1-Vout)/R2 + (Vin-V1)/Rf assuming the reference node V1.

 

[gneill]

Well for that I have (V1-Vout)/R1 + (V1-Vout)/R2 + (Vin-V1)/Rf assuming the reference node V1.

R2 connects between v1 and ground (0V), not Vout. You already know the current coming from the Vin node as $I$, so you don't need to know Vin. Nodal analysis uses a sum of current terms (it's KCL at a node), and $I$ is the known current in that particular branch.

 

[CNC101]

If I have- (V1-Vout)/R1+(0-V1)/R2 =0 for my nodal equation, where does Rf come in? As I need to find K in terms of all 3 resistors.

 

[gneill]

If I have- (V1-Vout)/R1+(0-V1)/R2 =0 for my nodal equation, where does Rf come in? As I need to find K in terms of all 3 resistors.

There are three branches meeting at node v1. You need three currents... which one's missing in your equation?

 

[CNC101]

Well its "I" flowing through Rf, Which is V1/Rf, So the three currents therefore are (V1-Vout)/R1+(0-V1)/R2 + V1/Rf =0

 

[gneill]

True, but look at the diagram I posted in post #4. That current is a given quantity (well, a given symbolic quantity anyways).

 

[CNC101]

Are we talking about representing V1/Rf as simply "I" in our nodal equation? If so how is Rf represented for my derivation of K? Thanks for your patience.

 

[gneill]

Are we talking about representing V1/Rf as simply "I" in our nodal equation? If so how is Rf represented for my derivation of K? Thanks for your patience.

Yes, you need both an "I" and and "Rf" eventually. Since "I" is a given you can use it as though it were a constant. So plugging it into the node equation you have (summing currents leaving the node):

$\frac{v1 - 0}{R_2} + \frac{v1 - Vout}{R_1} - I = 0$

Now since $I$ is a given quantity you can find a value for v1 separately. Just follow the path from the input to the v1 node. What potential drop can you see?

 

[CNC101]

The p.d is across the Rf resistor from the input to V1 node, and V1 represents the p.d.

 

[gneill]

The p.d is across the Rf resistor from the input to V1 node, and V1 represents the p.d.

Right. So write the corresponding expression for v1. v1 = ?

 

[CNC101]

Well V1=I*Rf. So the nodal equation is (V1-Vout)/R1+(V1-0)/R2-(I*Rf)

 

[gneill]

Well V1=I*Rf. So the nodal equation is (V1-Vout)/R1+(V1-0)/R2-(I*Rf)

Ah, no and no

What's the direction of the potential drop from the input across the resistor Rf? Is it a positive or negative change in potential? In other words, will the change cause v1 to be positive (above the potential at the input) or negative (below the potential at the input)?

For the node equation, don't substitute a voltage for a current! I*Rf would be a voltage. Leave the $I$ alone! It's a given value, so don't complicate it. What you really want to do is get rid of the v1 in the node expression. That's why we're looking at finding v1 via the current $I$ and the drop across Rf.

 

[CNC101]

The p.d has a negative potentional, not too sure how to remove V1. In terms of (V1-Vout)/R1+(V1-0)/R2= 0, I get Vout=V1((R1/R2)+1). If its a negative p.d, does V1 in that equation become Vout= -I*Rf(R1/R2+1)

 

[gneill]

You left out the $I$ current in the node equation; There are three terms as I wrote in post #19.

To remove V1 from the node equation you find it a different way and use that to replace it. That's why you want to find the potential V1 using the current $I$ and the resistance $R_f$.

 

[CNC101]

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 I am stuck at a dead end :(

 

[gneill]

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 I am 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!

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.

 

[CNC101]

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 I am 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.

 

[gneill]

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 I am 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.

 

[CNC101]

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

 

[gneill]

Try changing the op amp type to an ideal op amp. The software should have an ideal op amp in the parts list.

 

[NascentOxygen]

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

 

[gneill]

FYI, I set up and ran a simulation under LTSpice in about two minutes and obtained the predicted results.

 

[CNC101]

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.

 

[gneill]

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.

 

[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.