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These are the Solved Assignment of Electronics which includes Simple Inverting Amplifier, Input Resistance, Current, Load Resistor, Input Voltage, Given Circuit, Voltage Equations, Node, Confirmed etc. Key important points are: Non Inverting Amplifier, Voltage Gain, Approximately, Simple Model, Non Inverting Amplifier, Positive Input Terminal, Input Resistance, Resistor Ratio, Branched, Negative
Typology: Exercises
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Assignment #
1-9. Design a non-inverting amplifier with a voltage gain of approximately 12 with an input resistance of 12 kΩ. Use SPICE to confirm the result using a simple model of the OpAmp.
Solution: The non-inverting amplifier has an input resistance of infinity. Therfore, one alternative is to place a 12k resistor from the positive input terminal of the OpAmp to ground
To yield a gain of 12, choose resistor ratio in the branched from the negative terminal of the
OpAmp to have a ratio of 11:1 since the non-inverting gain is: where Rf is the feedback resistor connecting the output to the negative OpAmp input terminal and RG is the resistor connecting the negative OpAmp input terminal to ground. Using standard value resistors, I chose RG = 2k and Rf = 22k.
An alternate scheme could be to include a series resistance connected to vs. In order to meet the specified input resistance the sum of that series resistance and a resistor from the positive OpAmp input terminal must equal to Rin. Since this resistor network at the source is a voltage divider, the values of Rf and Rg must be altered to meet the voltage gain of 12. I chose the following: Rs1 = 6.04k, Rs2 = 6.04k, Rg = 1k, and Rf = 23.2k. This yields a gain of approximately 12.
1-12. Design an operational amplifier circuit to meet the following specifications:
Solution: The basic configuration of the difference amp is used:
R 4
R 3
R 1 R 2
−
v (^) o v (^) i
v (^) i
Use superposition to solve for the two signal inputs. The input resistance of the inverting path is R 1 = Ri1 and the input resistance of the non-inverting path is (R 3 + R 4 ) = Ri2 = 25k.
In the inverting amplifier path: Ri1 = R 1 =15k so R 2 = -Av R 1 where Av = -5.0, and R 1 = 15k yielding R 2 = 75k.
For the non-inverting case: Given the values for R 1 and R 2 the non-inverting gain at the positive OpAmp input terminal is 6.0. However, since vo = 3.0 vi 2 - 5.0 vi 1 , the non-
inverting gain must be 3.0 which is half of 6.0. Therefore, let R 3 = R 4 = 12.5k ≈ 12.4k. The resulting circuit is shown below.
R (^4)
R (^3)
R 1 R (^2)
−
v (^) o v (^) i
v (^) i
12.4k
12.4k
15k 75k
1-14. What is the expression of the output voltage for the circuit shown for
Using superposition and Thevenin equivalent analysis for input resistances, we know that Rina = RA and Rinb = RC + RD. So let Rina = RA ≥ 22k ⇒ RA = 33k, Rinb = RC + RD ≥ 22k So RB = 67(33k) ≈ 2.2M ; let RC = 33k and RD = 2.2M ⇒ Rinb = 2.233M > 22k (b) ADMvDM > 90ACMvCM
so vDM >90vCMACM /ADM = 20 10 20
CMRR
CM CM
CMRR
CM v A
v = (^) ⇒ vDM > 0.0032V