Inverting

**Amplifiers** or any operational amplifier for that matter and these are.

- No Current Flows into the Input Terminals
- The Differential Input Voltage is Zero as V1 = V2 = 0 (Virtual Earth)

Then by using these two rules we can derive the equation for calculating the closed-loop gain of an inverting amplifier, using first principles.

Current ( i ) flows through the resistor network as shown.

Then, the **Closed-Loop Voltage Gain** of an Inverting Amplifier is given as.

and this can be transposed to give Vout as:

Non-inverting

In the previous Inverting Amplifier tutorial, we said that for an ideal op-amp “No current flows into the input terminal” of the amplifier and that “V1 always equals V2”. This was because the junction of the input and feedback signal ( V1 ) are at the same potential.

In other words the junction is a “virtual earth” summing point. Because of this virtual earth node the resistors, Rƒ and R2 form a simple potential divider network across the non-inverting amplifier with the voltage gain of the circuit being determined by the ratios of R2 and Rƒ as shown below.

### Equivalent Potential Divider Network

Then using the formula to calculate the output voltage of a potential divider network, we can calculate the closed-loop voltage gain ( A_{V} ) of the **Non-inverting Amplifier** as follows:

Then the closed loop voltage gain of a **Non-inverting Operational Amplifier** will be given as:

Summing

In this simple summing amplifier circuit, the output voltage, ( Vout ) now becomes proportional to the sum of the input voltages, V_{1}, V_{2}, V_{3}, etc. Then we can modify the original equation for the inverting amplifier to take account of these new inputs thus:

However, if all the input impedances, ( R_{IN} ) are equal in value, we can simplify the above equation to give an output voltage of:

### Summing Amplifier Equation

Differential Amplifier

By connecting each input in turn to 0v ground we can use superposition to solve for the output voltage Vout. Then the transfer function for a **Differential Amplifier** circuit is given as:

When resistors, R1 = R2 and R3 = R4 the above transfer function for the differential amplifier can be simplified to the following expression:

### Differential Amplifier Equation

Op-amp Integrator Circuit

To simplify the math’s a little, this can also be re-written as:

notes:

if V_{in} is constant voltage so R_{in} voltage drop always equal V_{in} so it is feeding C with constant current source for example if we have capacitor with 1H and constant current source 1A every seconds the capacitor it will increase 1 volt constantly V_{in}/(R_{in}.C)

Op-amp Differentiator Circuit