=> AF =- 7.78kΩ/1kΩ= -7.78. 5. The band width of the peaking amplifier is expressed as
A. BW = (fp× XL)/ (RF+Rp)
B. BW =[ fp×(RF+Rp)× XL ] / (RF×Rp)
C. BW =[ fp×(RF+Rp)] / (RF×Rp)
D. BW = [fp×(RF+Rp) ]/ XL
Answer:A
=> AF =- 7.78kΩ/1kΩ= -7.78. 5. The band width of the peaking amplifier is expressed as
A. BW = (fp× XL)/ (RF+Rp) B. BW =[ fp×(RF+Rp)× XL ] / (RF×Rp) C. BW =[ fp×(RF+Rp)] / (RF×Rp) D. BW = [fp×(RF+Rp) ]/ XL
Explanation: The bandwidth of the peaking amplifier,