Comments on: Open-loop, Closed-loop and Feedback Questions and Answers

By: Adrian S. Nastase

Adrian S. Nastase — Mon, 20 May 2019 00:09:10 +0000

In reply to Aarohi. Glad to hear that.

By: Aarohi

Aarohi — Wed, 01 May 2019 08:53:19 +0000

This was the most helpful content for the loop gain and stuff. Love your site. <3

By: nisha

nisha — Wed, 17 May 2017 11:25:47 +0000

Thanks for the informative blog

By: Adrian S. Nastase

Adrian S. Nastase — Thu, 07 Jan 2016 23:10:30 +0000

In reply to Yinglai. In Fig 5 the graph shows and amplifier with the two resistors 100kohm and 1kohm but, indeed, they can be 10kohm and 100ohm.

By: Yinglai

Yinglai — Wed, 23 Dec 2015 06:31:59 +0000

Hi Eric,
That is really helpful. I wonder in Fig. 5 the 2 resistors should be 100ohm and 10kohm right?

By: Adrian S. Nastase

Adrian S. Nastase — Thu, 30 Apr 2015 17:59:16 +0000

In reply to vishwas. The configuration does not matter. The maximum current capability of the op amp is what limits the capacitor current charge.

By: Eric Lamarque

Eric Lamarque — Wed, 08 Apr 2015 05:15:35 +0000

In reply to Adrian S. Nastase. Thanks for the explanation.

By: Adrian S. Nastase

Adrian S. Nastase — Tue, 07 Apr 2015 18:30:47 +0000

In reply to Eric Lamarque.

Eric, this sounds like homework, so all I can do for you is to tell you how to approach this problem, not to solve it for you. You need to solve it yourself.

First of all, you should verify the drop rate between each frequency. Simply calculate the slope. I calculated that between 0.7 MHz and 3 MHz you have 20 dB/decade, and between 3 MHz and 18 MHz there is 40 dB/decade. You need to prove that before going further. Therefore, the transfer function is a multiplication of 3 first-order one pole transfer functions like this:

TF(f) = Gain * (1/(1+i f/f0)) * (1/(1+i f/f1)) * (1/(1+i f/f0))

where Gain = 10^4 (or 80 dB), f0 = 0.7 MHz, f1 = 3 MHz, f2 = 18 MHz

From here you have three methods to find the cross-over frequency:
1. Graphical solution: You can use a program like Mathcad or Matlab to draw the graph of |TF(f)|and then check at what frequency it crosses over.

2. Equation solving: Make |TF(f)| = 1 and again use Mathcad to solve this equation in f.

3. Asymptote approximation: Use the asymptotes to determine at what fraction of the decade between 18 MHz and 180 MHz is the cross-over frequency. Using the graph linearization with asymptotes, will give you an approximate frequency. Here is how:

After 18 MHz the drop is 60 dB/decade. Doing a quick calculation I got that the cross over frequency falls at 0.6 of the decade between 18 MHz and 180 MHz. This value is calculated with the 60dB/decade slope and the 3dB correction at 18 MHz. In other words, you consider 36 dB at 18 MHz not 33 dB. To calculate the frequency you apply to 0.6 the inverse of logarithm in 10, multiplied by 18 MHz. I calculated 71.65 MHz, very close to the book result.

By: Eric Lamarque

Eric Lamarque — Sat, 04 Apr 2015 15:06:59 +0000

Dear Adrian,
my question is not on feedback but on cross over frequency.I am stuck with a problem that is really getting on my nerves because i can’t solve it.It is really frustrating and i have no professional around me to turn to.I tried hard though going thru phase angles and polar equations but to no avail.The data for the response curve is given below.
Aol gain=8odB,-3dB gain=0.7Mhz,64dB=3Mhz,33dB=18Mhz.
The question is to calculate the crossover frequency and the book’s answer is 72.3Mhz..I didn’t come even close to that answerAnyway how is the gain 64 and 33dB at 3Mhz and 18 Mhz respectively? Please can you help ? you will make me a happy apprentice again.thanks.

By: Adrian S. Nastase

Adrian S. Nastase — Thu, 05 Feb 2015 18:03:46 +0000

In reply to tanmoy. This sounds like the first question in this article. My answer shows the relation between open-loop and closed-loop.