# Find output voltage as a function of time

1. Apr 11, 2014

### zealeth

1. The problem statement, all variables and given/known data

The op amp in the circuit seen in the figure is ideal. Find the steady-state expression for vo(t). vg = 1cos(106t) V. Express your answer in terms of t, where t is in microseconds. Enter the phase angle in radians.

Hints: (1) Let the node above the capacitor be Node a, and the node at the inverting (i.e., minus) input be Node b. (2) Write KCL equations at Node a and at Node b. Hints: This gives equations in terms of Va and Vo. Also note that the 40 K-Ohms load resistor is not part of the equations

2. Relevant equations

Current = 0 into input terminals of op amp
ZC = 1/(jωC), where j=$\sqrt{-1}$
KCL
V = IZ

3. The attempt at a solution

Vg is given in the format: v(t) = Vmaxcos(ωt+ø), where ω is the frequency and ø is the phase angle.

Vb = 5V

For KCL, ∑ientering = ∑ileaving

Node a:
i100kΩ + i20kΩ = i5kΩ + i100pF

Using V=IZ,

$\frac{V_o - V_a}{100k}$+$\frac{V_b - V_a}{20k}$ = $\frac{V_a - V_g}{5k}$ + Va*(jωC)

$\frac{V_o - V_a}{100k}$ + $\frac{5 - V_a}{20k}$ = $\frac{V_a-1}{5k}$ + Va*j*10^6*100*10^(-12)

Node b:
i10pF = i20k

Vb*(jωC) = $\frac{V_b-V_a}{20k}$
Vb*(jωC) = $\frac{5-V_a}{20k}$

Solving, I get Va = 5-j and Vo = 95+j24

Converting Vo to polar, Vo = 98∠14.2° V
Converting to radians, Vo = 98∠0.25 V
Putting this in terms of t,

Vo(t) = 98cos(10^6t+0.25) V

Converting to microseconds,

Vo(t) = 98cos(10t+0.25) V which is incorrect. Not exactly sure what I've done wrong here.

Last edited: Apr 11, 2014
2. Apr 11, 2014

### Staff: Mentor

Why do you say Vb = 5V?

And when you say this "expression for vo(t) = vg = 1cos(106t) V", you really mean "expression for vo(t). vg = 1cos(106t) V", right?

3. Apr 11, 2014

### zealeth

Because it is directly connected to the 5V terminal of the op amp.

Yes, that was a typo. I fixed it now.

4. Apr 11, 2014

### Staff: Mentor

I thought b is the inverting input of the opamp...

5. Apr 11, 2014

### zealeth

I believe I set it up the way my professor intended, as this was the method we used in class. He also specifies the node "at" the inverting input, rather than something along the lines of "node b is the inverting input."

6. Apr 11, 2014

### Staff: Mentor

The problem says it is an ideal opamp. What is the difference in voltage between the - and + input terminals for an opamp that is in a negative feedback configuration? What is the voltage on the + terminal?

7. Apr 11, 2014

### zealeth

I'm not sure what you're getting at here. The voltage on the + terminal is -5V which is given, so the difference between the terminals is -10V. However, I don't see what the positive terminal has to do with the problem.

8. Apr 11, 2014

### Staff: Mentor

No. I think I see your confusion now. The +/-5V shown are the power supply pins for the opamp. They are not connected to the +/- inputs of the opamp. The drawing should have been a bit clearer about that, since you are just learning about opamps.

So, what is the + input connected to (what does that down-arrow represent)? And that means the - input is at what voltage? Then see if you get a better answer for Vout...

9. Apr 11, 2014

### zealeth

That might be it, I had never seen an opamp shown with the power supply pins yet. So doing this with V+ = V- = 0, I have,

Node a:

$\frac{V_o - V_a}{100k}$+$\frac{V_b - V_a}{20k}$ = $\frac{V_a - V_g}{5k}$ + Va*(jωC)
$\frac{V_o - V_a}{100k}$+$\frac{0 - V_a}{20k}$ = $\frac{V_a - 1}{5k}$ + Va*(j*106*100*10-12)

and for node b:

Vb = V- = 0

(Vo-Vb)*(jωC) = $\frac{V_b-V_a}{20k}$
(Vo)*(j*106*10*10-12) = $\frac{-V_a}{20k}$

Solving, I came up with 3.77∠1.38 which, when plugged into the equation, is vo(t) = 3.77cos(10t+1.38) V (already converted to microseconds/radians). However, this still isn't the correct answer. Any ideas?

EDIT: Just realized I did the unit conversion to microseconds wrong and that was what was causing me the issue here. Thanks for the help!

Last edited: Apr 11, 2014
10. Apr 11, 2014

### Staff: Mentor

Ah, thanks for the Edit! I was just about to spend a lot of time going through the equations. Glad you figured it out!