Hello, I need help trying to solve this problem in example 2.26 of my Microelectronics Book

" An internally compensated omp amp is specified to have an open loop dc gain of 106dB and a unity gain bandwidth of 3MHz.Find Fb."

Oringally I thought that I could find fb since

ft=wt/(2*Pi)
fb=wb/(2*Pi)

and since wt=open loop dc gain*wb

I thought I could find fb by doing

ft/open loop dc gain=fb

however it turns out the answer is 15Hz

 

Hello, I need help trying to solve this problem in example 2.26 of my Microelectronics Book

" An internally compensated omp amp is specified to have an open loop dc gain of 106dB and a unity gain bandwidth of 3MHz.Find Fb."

Oringally I thought that I could find fb since

ft=wt/(2*Pi)
fb=wb/(2*Pi)

and since wt=open loop dc gain*wb

I thought I could find fb by doing

ft/open loop dc gain=fb

however it turns out the answer is 15Hz

\(20log(A_{0}) = 106 dB\)

\(A_{0} = 199526.23\)

\(f_{t} = A_{0} f_{b}\)

\(f_{b} = \frac{f_{t}}{A_{0}} = 15.04Hz\)

 

\(20log(A_{0}) = 106 dB\)

\(A_{0} = 199526.23\)

\(f_{t} = A_{0} f_{b}\)

\(f_{b} = \frac{f_{t}}{A_{0}} = 15.04Hz\)

OOO! I was afraid to take 20log(A0) because I did not know whether to multiply the log by 20 or 10 (I read somewhere that the gain of a power amplifer is 10log(power gain)

Thank you so much!

 

You are given the unity-gain frequency in MHz, not Radians/second, so you don't need to worry about conversion. Dividing the unity-gain frequency by the dc loop gain does give about the figure you expect, so are you sure you got the right gain value?

In the usual definition of voltage gain in dB, we calculate 20*log(ratio). (Purists may object to this except where input and output impedances are equal, but that's life.) Assuming this is so, divide the gain in dB by 20 and raise 10 to the answer to get the gain as a number.

Edit: I see others have beaten me to it. Ah well...

 

OOO! I was afraid to take 20log(A0) because I did not know whether to multiply the log by 20 or 10 (I read somewhere that the gain of a power amplifer is 10log(power gain)

Thank you so much!

That is the power gain.

\(\text{Gain} = 10log\left( \frac{P_{out}}{P_{in}} \right) \quad dB\)

For the voltage gain, make the substitution \(P = \frac{V^{2}}{R},\)

\(\text{Gain} = 10log\left( \frac{\frac{V_{out}^{2}}{R_{out}}}{\frac{V_{in}^{2}}{R_{in}}} \right) \quad dB\)

Under the assumption that the input and output impedances are equal(This is usually the case),

\(\text{Gain} = 10log\left( \frac{V_{out}}{V_{in}} \right)^{2} dB\)

\(\Rightarrow \text{Gain} = 20log\left( \frac{V_{out}}{V_{in}} \right) \quad dB \)