# Ideal opamp circuit...

Discussion in 'Homework Help' started by FiReStArtEr21, Mar 13, 2009.

1. ### FiReStArtEr21 Thread Starter New Member

Mar 8, 2009
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Hey again...i'm doing an exercise involving only one ideal opamp, and the goal is to find the current i0 (across the RL resistor), and show that the expression of that current is function of all resistors except RL and of the voltage Vref.

My aproach to this is to define a current across R1 doing Vref/R1 bearing in mind that the voltage on the -ve terminal of the opamp is 0. But then it strikes me: "well, if the expression for the current i0 doesnt depend on RL, there has to be a voltage on the +ve terminal too, just so we can apply KCL in the node between R2 and RL"... Well, i dont know if im thinking right or not, what i know is that im running around in circles with equations that dont make any sense.

Said that, i must confess (although this exercise look very easy to do) i need some guidance in this.

Thanks in advance, and im sorry if my doubts seem noob in some way, but the way my professors teach these elementar notions is (unfortunately) inversely proportional to the difficulty of some of the assingments they ask us to do.

P.S: Btw, How would i do to calculate the voltage Va, function of all the resistors in the circuit (including RL) and Vref??

Cheers

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2. ### steveb Senior Member

Jul 3, 2008
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You are correct that you can't assume Vneg and Vpos terminals are 0. Acutally, you should focus on calculating what that voltage is, and then your other questions will be much easier to answer. There are a few ways to do this, but one way is to notice that the current in both R2 resistors is equal and this current must flow through the upper R1 resistor and the lower R1||RL (parallel) combination. This constraint gives you an easy way to calculate the Vpos voltage in terms of R1, RL and Vref. Then the output current is Vpos/RL. If the current source is truely a current souce, then the RL should cancel out.

Once you have the Vpos voltage it is easy to find Va in terms of Vref because then you have the standard inverting amplifier combined with the noninverting amplifier which has the well known formula, using the superposition principle.

Va=Vpos*(1+R2/R1)-Vref*(R2/R1)

If you don't know that formula, then derive it, memorize it and then don't forget it for the rest of your life.

Last edited: Mar 13, 2009
3. ### steveb Senior Member

Jul 3, 2008
2,433
469
By the way, this circuit may look easy, but there are some definite confusing aspects to it. So don't be discouraged.

Once you solve the problem in the way I've outlined above, I'll mention a peculiar thing, which may be one reason why you had trouble with this.

4. ### Ratch New Member

Mar 20, 2007
1,068
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FiReStArtEr21,

That is an impossible task, and I will show you why.

You appear to be confused. Of course there is going to be some voltage on the V+ terminal if current exists through RL.

It is easy to do, as you will soon see.

Ok, the first thing you should know is that the V+ and V- terminals of the OP amp are going to be equal if the output Va is between the plus and minus voltage rails. With voltage gains of 200,000 to 500,000 and beyond, any slight difference is going to drive the output into one or the other voltage rail. And of course, equality does not necessarily mean zero. That circuit is not very practical because it requires both the upper and lower R2 to be precisely equal. If there is the slightest difference in resistance between the two, the output from the OP amp will not be able to balance its input nodes. So now that we know that the voltage at the input nodes are equal. The OP amp makes the two inputs a virtual node voltagewise. So the current in the top branch is the same as the current in the bottom branch because of the same voltage differences and the same resistances. So now we simply write the equations:

(Vref - V)/R1 = (V/R1) + (V/RL) =====> V = (Vref*RL)/(2*RL + R1) , dividing by RL on both sides gives Io = Vref/2*RL + R1) . So the R2 value is not a factor as long as both R2's are precisely equal, and RL does have an influence on its current. That is why I said what you are trying to do is impossible. Now let's look at the extremes. When RL is a very high value, V = (Vref*RL)/(2*RL + R1) approaches Vref/2, and Io = Vref/(2*RL) . That is because Vref is divided across the 2 identical resistors R1, and of course Io approaches zero due to its high resistance. When RL approaches 0, V approaches zero and again Io approaches zero due to the low voltage across it. You can easily figure out what Va is by knowing the branch currents and V . So there you have it. Just remember to take advantage of the virtual nodes with respect to voltage.

Ratch

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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Ratch, I didn't try to follow your analysis, but I recognize a Howland current pump when I see one.
This is a homework problem. The op amp is ideal, so presumably, the resistors are also ideal. If this is the case, the current through RL is indeed independent of the value of RL, as long as the op amp doesn't saturate.
You seem to be saying that the current will go to zero if RL=0, which is not the case.

EDIT: Perhaps you missed Vref?

6. ### steveb Senior Member

Jul 3, 2008
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469
Ratch, you are falling into the trap that I was going to warn the OP about once he solved the problem the way I recommended. This is a very easy mistake to make, and many (if not most) people fall into it. However, the problem can be solved.

The issue is that the load resistor is needed to bring the circuit away from the marginal stability you are talking about. The load resistor can not be infinite. Also, if the load resistor is very large and the resistors are not perfectly matched, again there can be instability.

7. ### Ratch New Member

Mar 20, 2007
1,068
3
Ron H,

The circuit submitted by the OP is not quite the same as the basic circuit of Fig. 1 of the application note you submitted. The OP's circuit is being driven through a resister on the negative terminal and grounded through a resistor on the positive terminal. The basic circuit in the app note is just the opposite. Also in the OP's circuit, the upper and lower resisters R1 and R2 are identical. That is not true of the basic Howland circuit. Further in the note they modify the circuit extensively. As far as my analysis goes, I stand by it for the circuit submitted by the OP. Try to follow my analysis and see if you agree. It is not complicated.

That means my analysis is wrong. Can you find the mistake?

How so? Vref is included in my derived equations.

Ratch

Last edited: Mar 14, 2009
8. ### Ratch New Member

Mar 20, 2007
1,068
3
steveb,

I am looking forward to your disclosure so as to see if it applies to my analysis.

I pontificated on RL approaching zero or a large value so as to illustrate the validity of my analysis, not as a operational value. Someone, please shoot it down.

Ratch

9. ### steveb Senior Member

Jul 3, 2008
2,433
469
Actually, I agree that large RL is an issue for that circuit.

I'll look at the case of RL=0 more carefully tomorrow. I't's 1:00 AM now. Off the top of my head, I see that V+ approaches zero as RL approaches zero. So we need to look at the stability.

However, there is definitely a wide range of usable resistor values for RL that will work.

I was planning to discuss this once the OP responded with the correct answer, or with a statement that he did not understand my recommended approach. I was trying to not do the work for him so that he could learn.

10. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Guys, read p.2 in AN-1515. Bob Pease explains it much better than I can.

@steveb: you said,
When RL is zero, how can you have a stability issue? It's just an amplifier with negative feedback, and the + input grounded.

@Ratch: Just because RL=0, making V+=0, doesn't mean I0 (load current) is zero. In fact, the current will be independent of the value of RL, unless the op amp saturates. The current comes from the lower R2 (connected to the op amp output).

11. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes, you are right. I meant to say sensitivity issue. It's very late and i need to go to bed. I really don't think there is an issue for low values of RL=0, but I just wanted to be fair to Ratch and say that I would double check at a later time.

Ratch is correct about the issues at high values of RL, but really any current source will have a problem then. High RL means that the voltage gets very large and any current source has a maximum voltage. However, from a mathematical point of view, this circuit also has a marginal stability as RL gets large, which is what is confusing about that circuit.

Basically, there is no real issue here and the OP needs to get the right answer. Let's let him do this, or tell us if he is still having trouble.

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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You've made an error regarding the direction of currents.

Rewriting slightly:

(V - Vref)/R1 = (V/R1) + (V/RL) =====> V = -(Vref*RL)/(R1) , dividing by RL on both sides gives Io = -Vref/R1 .

So, in fact, the current in RL is independent of RL, subject to the limitations of the opamp (supply voltage, frequency response, etc.) and resistor matching.

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13. ### FiReStArtEr21 Thread Starter New Member

Mar 8, 2009
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Hey, i've red your posts and maybe i didnt explain myself properly...but that doesnt matter now. the notion that was evading me when i first started to solve this was that you can turn the input terminals of the opamp into only one node (cos of the virtual short circuit between the terminals). That elementar notion allows me to derive the following:

(VRef-Vx)/R1 = Vx/R1 + Vx/RL + (Vx-Vx)/2R2

VRef/R1 = 2Vx/R1 + Vx/RL

Vx = VRef/2 + (VRef*RL)/R1

but then when i divide with RL, RL still remains in the resulting i0, and that's my main diffilculty now...i cant seem to take RL out of the mix. Do i have to really specify that for a very large RL the statement that io doesnt depend on RL holds?

Thanks a lot

14. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335

You have to start out with a slightly different expression, namely:

(Vx - Vref)/R1 = Vx/R1 + Vx/RL + (Vx-Vx)/2R2

then you'll get the correct result.

15. ### FiReStArtEr21 Thread Starter New Member

Mar 8, 2009
8
0

Ok. Got it...Thank you all

16. ### Ratch New Member

Mar 20, 2007
1,068
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The Electrician,

Right you are. I am not going to use the excuse of it being late at night either. I should know better. I guess I had the right approach, but a simple mistake in current direction makes all the difference. One other thing. I originally said that the resistors had to be balanced or the OP amp could not maintain zero volts across its input terminals. That isn't so. The currents in the upper and lower branches simply change to compensate for whatever differences there are in the two R2's. So like you said, the current through RL will be stable and subject to the ability of the OP amp to supply the required output voltage. Thanks again for the correction.

Ratch

17. ### steveb Senior Member

Jul 3, 2008
2,433
469
Now that you got it, I'd like to explain what I was trying to say originally. The method I was recommending is as follows.

Note that both $R_2$ currents are equal and flow through $R_1$ in the upper side and in $R_1||R_L$ on the lower side. Those currents are equal because the resistances are equal and both terminals are attached to the same voltage potentials; i.e. $V_a$ and $V_{pos}=V_{neg}$. Also, you are assuming opamp inputs currents are zero. (i.e. ideal opamp)

Hence: ${{(V_{pos}-V{ref})}\over{R_1}}={{V_{pos}\;(R_1+R_L)}\over{R_1\;R_L}}$ (note the use of the product over sum formula for the parallel resistors.)

From this it's easy to rearrange to: $V_{pos}={{-V_{ref}\;R_L}\over{R_1}}$

Then just divide by $R_L$ to find $i_o$: $i_{o}={{-V_{ref}}\over{R_1}}$

I just show this to point out that there are always different methods, and sometimes if you try to visualize what is happening in the circuit, you can see an easy way. In this case, the method I recommended removed $R_2$ from the expressions right away.

Last edited: Mar 14, 2009
18. ### Ratch New Member

Mar 20, 2007
1,068
3
steveb,

Looks like the same method I was using minus the mistake in current directions. You implemented it by the parallel resistance expression directly, whereas I divided each resistance into the node voltage. Same thing.

Ratch

19. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Here is a more general analysis from a slightly different approach. It also shows the true differential capability of the circuit. Note that RL never appears in the equations.

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20. ### steveb Senior Member

Jul 3, 2008
2,433
469
Well, maybe. I still haven't looked at your approach in detail since it was obviously wrong. However, if you look at the second post in this thread, I explained the method first. Then I hoped that the OP, and not you, would post an attempt at the solution.