ADA4528 Rout

Question

Hi, 
 I am working with the operational amplifier ADA4528 and I am facing some instability problems due to the capacitive load. 
 After some research, I found the article “Practical Techniques to Avoid Instability Due to Capacitive Loading" . I decided that circuit of the figure 3 called in-the loop compensation circui t in page 1 of this article it is what I want to use for my application. 
 My question is, what it is the Rout of the ADA4528 . How I can find this value? 
 On the other hand, I wonder if you know how they came-out with the equation in page 2 (up-right corner): f-3dB=1/2pi(Cf*Rf) . I don’t understand how the bandwidth will be determined only by Cf and Rf. Why the other external elements such as Rx and CL do not affect the bandwidth? 
 Thank you very much for your assistance and your time. 
 Sincerely, 
 Jorge Reynaga

EZAdministrator · Accepted Answer

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ADIApproved · Answer

Hi Jorge, 
 Closed-loop output impedance is a function of gain at a particular frequency and the open-loop output impedance. 
 R_closed-loop = R_open-loop / (1+Aol*beta) 
 Figure 28 and 31 (data sheet Rev 0) show typical graphs of closed-loop output impedance vs frequency at 3 different gain configurations, Av = 1, 10 and 100. 
 You can read open-loop output impedance off the graph when the op amp runs out of loop gain (Aol*beta=0). 
 Open-loop output impedance is about 100ohms.