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When they say 3db less that means 3db less than the midrange frequency at the corner. They dont draw it like that because it's an asymptotic Bode plot and that is a simplified way of drawing these curves. It's drawn with a corner rather than a curve. The true response is a curve which is at 103db at 10Hz.
Plots like this are often drawn with straight lines when they are really curves, but the straight lines show the major part of the response curve. The 3db down points are drawn as corners. Sometimes these straight line plots are grossly inaccurate and have to be corrected using special curves depending on the damping ratios, but when the response is not too wild straight lines tell the story quite well but without being perfect at the corner frequencies.
In drawing these plots a pole causes a relative slant downward and a zero causes the opposite (relative to what came before it), so after a pole a zero will create a horizontal line again because the zero shifts the slant exactly opposite to what the pole causes so they cancel out and what remains is a horizontal line. Using this simple guideline we can draw these plots by hand knowing only the poles and zeros and WITHOUT a computer or calculator of any kind. (Note: A better way of looking at this is that a pole causes a decrease in slope and a zero causes an increase in slope).
Op amps are DC amplifiers, and DC amplifiers can amplify both AC and DC.
AC amplifiers can only amplify AC.
Decade is a 10 fold increase or decrease in frequency, so 10Hz is a decade lower than 100Hz.
Octave is a 2 fold increase or decrease, so 10Hz is an octave lower than 20Hz.
For a plot that goes -20db per decade, that means that after the corner frequency it decreases an additional 20db for every decade, so a signal that is 0db at 10Hz would be -20db at 100Hz and -40db at 1000Hz.
For a plot that goes -6db per octave if the signal is at 0db at 10Hz it would be -6db at 20Hz and -12db at 40Hz and -18db at 80Hz (an additional -6db every time the frequency doubles).
It is a related problem because it uses logarithmic scale and each x-axis scale value increases by a decade . Please have a look on the attachment. As you can see the x-axis scale starts from "10". Can't it start from "0" because log(1)=0. Do you get me? Please help me with it.
Could you please help me with the queries, Q1 and Q2, included in the attachment?
Q3: For a non-inversion op-amp closed loop gain = 1 +Rf/Ri = 1/B, where the attenuation B = Ri/(Ri+Rf). An inverting op-amp has closed loop gain = -Rf/Ri. In my view there exists no concept of attenuation B in case of inverting op-amp, does there? In the book I use the closed loop critical frequency is: fc_(cl) = fc_(ol) [1+B*A_ol(mid)], where fc_(ol) is open loop critical frequency and A_ol(mid) is midrange open loop gain. But the concept of attenuation B doesn't exist for inverting op-amp then doesn't this mean the given formula for closed loop critical frequency can't be used for an inverting op-amp?
Q1: The ouput load must not be lower than a certain specified minimum value. This limit is set by the output current capabilities and/or the internal outpiut resistance of the opamp. That means: If the load is below this minimum, the ouput voltages cannot reach the theoretical value (due to the gain value that is calculated based on ideal assumptions).
Q2: Because it is impossible in the drawing to discriminate between 600 and 620 ohms I would not spent to much time to think about it. Thats not engineering like.
Q3: The term you call "attenuation" is nothing else than (and should be called so) the "feedback factor FB or beta" with FB=Ri/(Rf+Ri)
*For non-inv. opamp amplifiers and signal input directly at the pos. terminal: Gain G=1/FB (idealized opamp properties).
*For inverting opamp amplifiers and signal input at Ri (connected to the inv. terminal); Gain=-FF/FB with FF=Feedforward factor=Rf/(Ri+Rf)
(the incoming signal is reduced before it enters the inv. opamp input). Thus: G=-Rf/Ri.
Q3: The term you call "attenuation" is nothing else than (and should be called so) the "feedback factor FB or beta" with FB=Ri/(Rf+Ri)
*For non-inv. opamp amplifiers and signal input directly at the pos. terminal: Gain G=1/FB (idealized opamp properties).
*For inverting opamp amplifiers and signal input at Ri (connected to the inv. terminal); Gain=-FF/FB with FF=Feedforward factor=Rf/(Ri+Rf)
(the incoming signal is reduced before it enters the inv. opamp input). Thus: G=-Rf/Ri.
So, this means the concept of attenuation does also work for an inverting op-amp and the critical frequency formula, fc_(cl) = fc_(ol) [1+B*A_ol(mid)], can be used for an inverting op-amp as well.
Could you please also help me with the query included in the attachment? Thank you.
again you are using the term"concept of attenuation" - is this used in your book? Didn't it use the phrase "concept of negative feedback"? Indeed, this would be more suitable.
Another comment to the wording: The table gives values for the "bandwidth" (minimum: 437 kHz; max: 1.5 Mhz.) . It is very unusual to call these values "bandwidth".
In fact it is the gain-bandwidth-product (or the transit frequency). That means: It is the frequency for which the open-loop gain crosses the 0-dB line.
The open-loop bandwidth is equal to the ratio (GBW/open-loop gain Aol) which gives approx. 10 Hz only.
Now the calculation:
Assuming a min value GBW= 500,000 and a closed-loop gain of 50 the closed-loop bandwidth is 10 kHz.
This does not fulfill your requirement (gain of 50 for 20 kHz).
For a GBW=1.5 Mhz the closed-loop BW for a gain of 50 is 30 kHz.
Thus, the 741 opamp can fulfiil your requirement only if the GBW value is approx. at its max. value.
again you are using the term"concept of attenuation" - is this used in your book? Didn't it use the phrase "concept of negative feedback"? Indeed, this would be more suitable.
The open-loop bandwidth is equal to the ratio (GBW/open-loop gain Aol) which gives approx. 10 Hz only.
Now the calculation:
Assuming a min value GBW= 500,000 and a closed-loop gain of 50 the closed-loop bandwidth is 10 kHz.
This does not fulfill your requirement (gain of 50 for 20 kHz).
For a GBW=1.5 Mhz the closed-loop BW for a gain of 50 is 30 kHz.
Thus, the 741 opamp can fulfiil your requirement only if the GBW value is approx. at its max. value.
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