Question

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 application, a bandwidth of 32 kHz is required. What is the highest gain available under these conditions?

Answer 1

To find the highest available gain for an op-amp circuit with a required bandwidth, we use the Gain-Bandwidth Product (GBP), which is a constant. For a GBP of 768 kHz and a bandwidth of 32 kHz, the maximum available gain is 24 V/V.

Step-by-step explanation:

The question concerns determining the highest available gain for an op-amp circuit given a required bandwidth.

The Gain-Bandwidth Product (GBP) is a constant for an op-amp and is found by multiplying the current gain with its corresponding 3-dB frequency.

With an initial gain of 96 V/V and a 3-dB frequency of 8 kHz, the GBP can be calculated as 96 V/V * 8 kHz = 768 kHz.

To meet the requirement of a 32 kHz bandwidth with the same GBP (because GBP is constant), we can calculate the maximum gain as follows:

GBP = Gain * Bandwidth, which gives us

Gain = GBP / Bandwidth.

Plugging in the numbers, we get

Gain = 768 kHz / 32 kHz, resulting in a maximum gain of 24 V/V under the conditions of a 32 kHz bandwidth.

Answer 2

`Av_2 =24\ V/V`

Step-by-step 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= (96* 8)/(32)`

`Av_2 =24\ V/V`

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