i. Circuit Schematic:
```
+Vin -----R1-----+--------- Output
|
-Vin ---R2-----| |
|
Rf
|
-Vcc
```
To design a non-inverting op-amp amplifier with a voltage gain of 25 dB, we can use the following formula:
Voltage Gain (Av) = 1 + (Rf/R1)
Since the desired voltage gain is 25 dB, we can convert it to a linear gain as follows:
Av (linear) = 10^(25/20) = 18.7
Now, let's assume a value for R1. We can choose a common value, such as R1 = 10 kΩ.
Substituting the values into the voltage gain formula, we can solve for Rf:
18.7 = 1 + (Rf/10kΩ)
Rf = 18.7 * 10kΩ - 10kΩ = 187kΩ - 10kΩ = 177kΩ
Therefore, we can use R1 = 10 kΩ and Rf = 177 kΩ for the desired voltage gain of 25 dB.
For the input signal with a frequency of 100 kHz, a peak-to-peak voltage of 200 mV, and a DC offset of 1 V, the op-amp needs to have a gain-bandwidth product (GBW) and slew rate (SR) that can handle the signal faithfully without distortion.
The minimum gain-bandwidth product can be calculated using the formula:
GBW = Av * f
Where Av is the voltage gain and f is the frequency.
GBW = 18.7 * 100 kHz = 1.87 MHz
So, the minimum gain-bandwidth product required is 1.87 MHz.
The minimum slew rate (SR) can be calculated using the formula:
SR = 2π * f * Vpp
Where f is the frequency and Vpp is the peak-to-peak voltage.
SR = 2π * 100 kHz * 200 mV = 125.66 V/μs
Therefore, the minimum slew rate required is 125.66 V/μs.
To design a non-inverting op-amp amplifier with a voltage gain of 25 dB, we can use resistor values of R1 = 10 kΩ and Rf = 177 kΩ. The op-amp should have a minimum gain-bandwidth product of 1.87 MHz and a minimum slew rate of 125.66 V/μs to handle an input signal with a frequency of 100 kHz, a peak-to-peak voltage of 200 mV, and a DC offset of 1 V.