An OpAMP has a GBWP of 100MHz and a slew rate of 10V/us and AVoc = 160dB at DC (

Question

An OpAMP has a GBWP of 100MHz and a slew rate of 10V/us and AVoc = 160dB at DC (?=0). Assume VCC = +/- 15 Volts. a. If the opamp is configured as a non inverting amplifier with voltage gain of 50, what is the maximum frequency the opamp will operate linearly at? b. With the same configuration (gain of 50), what is the maximum input for linear operation (limited by VCC)? c. What is the maximum frequency that this output (from b) will allow without distortion (limited by SR)? *NOTE: Please write out full, complete work.

Explanation / Answer

voltage gain=1+Rf/R1 so Rf=49R1....... given GBWP=gain*BW so10^8=10^8*BW BW=1Hz............ so the max FREquency is=1Hz.................. (B) Max input voltage=Max Vo/Gain=15/50=0.3v.... (C)slew rate is =-dVo/dt=max rate of change of output with respect to the time slew rate =2*pi*f*peak amplitude of voltage... so f=0.11Mhz

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.