Can we solve this circuit

WBahn

Joined Mar 31, 2012
29,976
Hi, I don't know why but I'm unable to start new thread. So I decided to post my question here.

I have this simple op amp based circuit



And I try to find Vo voltage.

But I want to ask you is there any easier way to find solution for this circuit?
Maybe you know some trick that we can use here to speed up finding a solution?
Maybe I'm missing something in the exchanges here, but this doesn't seem like an overly tricky circuit or one that should be requiring a computer or matrices to solve efficiently. I timed how long it took to do it completely by hand, including all the arithmetic to get a numerical answer, in just under three minutes.

Perhaps some people, such as The Electrician, use matrices to solve so many things that it is sufficiently second nature to them that they could use that approach to solve the problem, from the very first glance at it, in well under three minutes. I know it would take me about that long just to even fire up MatLab, assuming I was even near a computer that had it on it, and at least that long again to come up with the correct matrix to enter into it. But I don't do that kind of thing very often and so it is definitely not second nature to me.

Here's the approach I used:

1) We know that the voltage at the two inputs of the opamp have to be equal (assuming ideal opamp here). That means that the voltage gain from Vn, the inverting input, due to the input voltage source has to be dropped across Rin. Thus the current out of the Vin source and into Rin is simply

Iin = Vin/Rn

This current has no place to go but through R1. Thus the voltage at Vp, the non-inverting input, (and hence Vn) is

Vp = Vn = Iin*R1

Given the voltage at Vn, we know the current going downward in R2 is

I2 = Vp/R2

Summing the current at junction between RL and R2 gives the current flowing downward in RL to be

IL = Iin + I2

and the voltage at the output is thus

Vout = Vp + IL*RL

Plugging back into things we get

Vout = Iin*R1 + (Iin + I2)RL

Vout = Iin*R1 + (Iin + Vp/R2)RL

Vout = Iin*R1 + (Iin + Iin*R1/R2)RL

Vout = Iin*(R1 + RL + R1*RL/R2)

Vout = (Vin/Rin)*(R1 + RL + R1*RL/R2)

Av = Vout/Vin = (R1 + RL + R1*RL/R2)/Rin

Av = Vout/Vin = (1kΩ + 100Ω + (1kΩ/250Ω)*100Ω)/1kΩ

Av = Vout/Vin = (1kΩ + 100Ω + 4*100Ω)/1kΩ

Av = Vout/Vin = 1.5kΩ/1kΩ

Av = Vout/Vin = 1.5

So let's check that.

Let Vin = 10V which means Vout should be 15V

The voltage at Vn and Vp will then be 10V with 10mA running through the source, Rin and R1. With Vout being 15V and Vn being 10V, that means that 5V appears across RL whichp puts 50mA through it. 10mA is diverted over to Vin with the remaining 40mA flowing down through R2 which agrees with the voltage at Vn being 10V.
 

LvW

Joined Jun 13, 2013
1,752
The reason I was critical of your method is because is didn't seem to be easier and speedier, which is what Jony130 was asking for; perhaps you overlooked this request of his in post #11.
Hi Electrician - why this discussion?
Jony130 was searching for a another method - and I have described another alternative to solve the shown circuit.
For your opinion, this method is not easier and speedier. OK - but as mentioned earlier - everybody has its own preferences.
Nevertheless, for my opinion, it is not a bad idea to try several methods and then to decide which methods seems to be most appropriate. I think, it is not easy from the beginning - that means: without trying several approaches - to see which method is easier and speedier.
Besides this argument, it is a good exercise - in particular for other interested forum members - to learn about different ways to solve such a problem.
In this context, I also consider WBahn`s contribution as helpful.
 
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Thread Starter

Jony130

Joined Feb 17, 2009
5,487
Thank you guys for your input. I really appreciate your help.
The original circuit was a voltage to current converter/current loop (XTR115).



And when I first so this circuit i say no way this circuit can work.
And to simplify the analysis i remove BJT's. And after some time I manage to confirm that this circuit indeed work as a voltage to current converter. Very clever little circuit.
 

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The Electrician

Joined Oct 9, 2007
2,971
I think, the wanted "trick" could be the following:

The closed-loop gain of an idealized opamp (Aol infinite) with feedback can be expressed as Acl=Hf/Hr
with Hf=(Vp-Vn)/Vin for Vout=0
and Hr=(Vp-Vn)/Vout for Vin=0.
There seems to be a problem with this method as described here. I'm going to start a new thread to discuss it.
 
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