Mounting the circuit in Figure 1 in Tinkercad, the OPAMP output should model the following equation Y = 4X – 2 (Adder -
Posted: Fri May 20, 2022 10:45 pm
Mounting the circuit in Figure 1 in Tinkercad, the OPAMP output should model the following equation
Y = 4X – 2 (Adder - Subtractor).
Criteria for Design:
V1= is the voltage of the RL charge.
V2 = is variable.
Polarization of the Operational Amplifier V+ = +15 V and V- = -15V.
V3, R3, R4, R5, RL at the discretion of the designer.
R1, R2, RF1, RF2, RX (if necessary), all these resistors are at the discretion of the designer,
looking for the values that meet the expected output from the equation given.
The entire circuit operates with direct current (DC Voltage).
For the output of the OPAMP they will have a margin of error between + – 0.8.
Example:
Y = 4X – 2, if X = 2 then Y = 6.
Vout = 4V2 – 2, if V2 = 2 then Vout = 6V. However, the result can be between 6.8 and 5.2.
You must fill in the following table of results, for the different values of V2:
V2 0.5 1 1.5 2. 2.5 3 3.5 4 Vout
V2 0.5 1 1.5 2. 2.5 3 3.5 4 Vout
Y = 4X – 2 (Adder - Subtractor).
Criteria for Design:
V1= is the voltage of the RL charge.
V2 = is variable.
Polarization of the Operational Amplifier V+ = +15 V and V- = -15V.
V3, R3, R4, R5, RL at the discretion of the designer.
R1, R2, RF1, RF2, RX (if necessary), all these resistors are at the discretion of the designer,
looking for the values that meet the expected output from the equation given.
The entire circuit operates with direct current (DC Voltage).
For the output of the OPAMP they will have a margin of error between + – 0.8.
Example:
Y = 4X – 2, if X = 2 then Y = 6.
Vout = 4V2 – 2, if V2 = 2 then Vout = 6V. However, the result can be between 6.8 and 5.2.
You must fill in the following table of results, for the different values of V2:
V2 0.5 1 1.5 2. 2.5 3 3.5 4 Vout
V2 0.5 1 1.5 2. 2.5 3 3.5 4 Vout