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ipn [44]
3 years ago
9

A noninverting op-amp circuit with a gain of 96 V/V is found to have a 3-dB frequency of 8 kHz. For a particular system applicat

ion, a bandwidth of 32 kHz is required. What is the highest gain available under these conditions?
Physics
1 answer:
umka21 [38]3 years ago
6 0

Answer:

Av_2 =24\ V/V

Explanation:

given,

op-amp circuit with a gain of = (Av₁) = 96 V/V

Band width  = (Bw₁) = 8 kHz

Required bandwidth(Bw₂) = 32 kHz

Highest gain available =(Av₂) = ?

For the given system Bandwidth product is constant

                           Av₁ Bw₁ = Av₂ Bw₂

                           96 x 8 = Av₂ x 32

                           Av_2= \dfrac{96\times 8}{32}

                           Av_2 =24\ V/V

the highest gain available under these conditions Av_2 =24\ V/V

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