Given that the slew rate of an op-amp is 20v/us. If the expected amplifier output voltage is 10Vp, Determine the maximum operating frequency of the amplifier.
a) 159khz
b) 318khz
c) 500khz
d) 250khz
The maximum operating frequency can be found using the formula:
f = SR/(2*pi*Vp)
where SR is the slew rate, Vp is the peak voltage, and f is the maximum frequency.
Substituting the given values, we get:
f = 20/(2*pi*10) = 0.318 kHz = 318 kHz
Therefore, the maximum operating frequency of the amplifier is option (b) 318 kHz.
To determine the maximum operating frequency of the amplifier, we need to consider the slew rate and the expected amplifier output voltage. The maximum operating frequency is limited by the slew rate, which is the maximum rate of change of the output voltage.
The slew rate is given as 20 V/μs.
The expected amplifier output voltage is 10 Vp, which means a peak-to-peak voltage of 10 volts.
The maximum rate of change of the output voltage can be determined using the formula:
Slew Rate = 2πfAmax
where f is the maximum frequency and Amax is the expected amplifier output voltage.
Substituting the given values, we have:
20 V/μs = 2πf(10 V)
Simplifying, we get:
10 V/μs = πf
f = 10 V/μs / π
f ≈ 3.183 kHz
Therefore, the maximum operating frequency is approximately 3.183 kHz.
However, the answer choices provided are in kHz. To convert the frequency to kHz, we divide it by 1000:
f (kHz) = 3.183 kHz / 1000
f (kHz) ≈ 0.003183 kHz
Therefore, the maximum operating frequency of the amplifier is 0.003183 kHz, which is closest to 318 kHz.
So the correct answer is:
b) 318 kHz