# What is the output voltage function of this opamps [closed]

What is the output voltage function of this opamps:

• Integrator (with extra resistor):
• Differentiator (with extra resistor):
• Is this where kids come to get their homework answers?
– Jon
Feb 11, 2016 at 19:06
• It is not, We're working on a project, and we get a lot of noise on our signal so we want to know if we can solve it my using this extra resistors but we've to know the output functions Feb 11, 2016 at 19:07
• -1 For lack of online search nor any work shown. Feb 11, 2016 at 19:08
• I did, but I couldn't find anything usefull Feb 11, 2016 at 19:11
• I suggest you describe the problem you are having, and your proposed solution and ask a question about that. Simulation would also help you to evaluate your solutions. Feb 11, 2016 at 19:11

We're not here to give you the solution for homework, we're here to help you solve problems, or more importantly clarify confusing steps in work put into solving problem. So instead of answering a question of two problems, I'll show you an approach for these kinds of problems.

## Circuit Analysis of Op-Amp using KCL

We're analyzing here for small signal analysis. We're going to treat these op-amps as ideal op amps to neglect extra parameters. An Ideal Op Amp has infinite input impedance, meaning it consumes no current via the input node, and entails any current coming from input source goes straight to the feedback line.

Because we have a capacitor, this circuit becomes a first-order problem, meaning that KCL or KVL would let you find a differential equation to mathematically describe the circuit.

We can avoid solving a differential equation problem by converting it to the S-Domain, using a Laplace transform.

For capacitors we convert $$C_! -> 1/jwC_1$$

Next, we want to remove as many components as we can by either adding components in series, or finding the result of components in parallel. So we convert all components to the resulting impedance

Finally we use KCL to analyze the circuit. So KCL at node where the voltage is 0 (yes, the negative and positive polarity inputs of the Op-Amp are virtually connected, and in this case it's obvious they're grounded.)

The sum of currents going in = sum of currents going out. However in this picture I drew the picture of both currents going in. So $$I_1+I_2 = 0$$

or

$$I_2 = - I_1$$

Also $$I_1 = V_{out}/Z_1$$ $$I_2 = V_{in}/Z_2$$

Sorry, but your question (you should ask the question not just in the title) could be answered with a basic google search http://www.electronics-tutorials.ws/opamp/opamp_7.html. If you ever have a feedback path you can reduce the impedances by solving the circuit with laplacian elements. Then you end up with two impedance resistors that can be solved just like the inverting amplifier equation you are used to. Vout/Vin = -Rf/Rin changes to Vout/Vin = -Zf/Zin where Z is the simplified circuit impedance. This is why we learn math and remember it! In the first case it would more difficult to solve the feedback path with other circuit solving methodologies. You end up getting

Vout/Vin = -(R/(CR+Cs))+L*S)/R2

In the second case (which you are interested in) you get

Vout/Vin = -Zf/Zin

Zf = R10

Zin = R+1/C*s

so

Vout/Vin = -(R10)/(R+1/Cs), you can subsitute s = jw and w = 2*pi*f if you want to know what it does in the frequency and phase domain

simulate this circuit – Schematic created using CircuitLab