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.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?
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.