Circuit design of electronic devices based on operational amplifiers Totalizer Differentiator Integrator

Circuit design of electronic devices based on operational amplifiers

Deep Negative Operational Amplifiers
For the construction of various decisive amplifiers, op amps with deep negative feedback are used.
Basic switching circuit of operational amplifiers
• Non-inverting amplifier;
• Inverting amplifier;
• Amplifier with differential input 1


Fig. 1. Diagrams of switching on the operational amplifier: inverting (a); non-inverting (b); differential (c).

Non-inverting amplifier


Inverting amplifier
Kuoc = - R2 / R1

With differential switching on the OU, input voltages are supplied to both inputs through resistors R1 and R3. In this case, the equality of the potentials at the inputs is preserved.



Therefore



With equality, R1 = R3; R2 = R4 should:


where in brackets the voltage difference taking into account the sign

Adder.

Figure 2 shows an amplifier circuit based on an op amp that serves as an adder.


Figure 2. The OA-based adder with frequency independent elements.

Based on the 1st Kirchhoff Law for node 1
or

That is, the voltage at the output of the amplifier circuit is equal to the voltages at the input elements with weighting factors equal to the gain for the input in question:

Differentiator.
The use of reactive elements makes it possible to realize analog operations of differentiation and integration. If in the circuit the input element is a capacitor with capacitance C, and the feedback element is a resistor with resistance R, then from condition I1 = I2 it follows that

from where

those. the output voltage is an input voltage differential with a weighting factor RC.

Integrator.
In the case when a resistor is connected to the inverting input and a capacitor is installed in the negative feedback circuit, the circuit with the operational amplifier acts as an integrator:



those. the output voltage is an integral of the input voltage.

Example.
If the input resistor voltage is changed abruptly from 0 to U0, then the output of the OU voltage will change according to the law i.e. grow in proportion to time. This corresponds to the case of charging the capacitor With a constant current Iin = Uin / R