#### Junus2012

##### Advanced Member level 5

Basically the gain error and as also explained by the TEXAS Instrument handbook, can be calculated as :

Gain Error = [(ideal expected closed loop gain- actual closed loop gain)/ideal expected closed loop gain]*100%.

where actual closed loop gain= AOL/(1+Beta. Aol), and that 1+Beta.Aol represent the loop gain illustrated by the graph.

Also from the graph, it can be clearly seen that loop gain is the difference between the Aol and Acl, and the author abbreviated it L

From the Texas Instrument procedure, the loop gain is substituted for the value at zero or DC frequency, hence a loop gain of 40 dB give a gain error of 1% if the AOL is 80 dB and so on as shown in the table below:

My question is, the table above should only be true for the DC or near DC level signals because as the signal frequency increases the loop gain decreases and the gain error increases. Hence it is not enough to preserve the required loop gain for a certain gain error for the entire interesting range of frequency. Therefore the GBW of the amplifier will play the rule for insuring this loop gain.

Practically speaking and by considering the discussion above, suppose I have an input signal of 100 kHz and the required closed loop gain is 100 (40 dB) with gain error of 1%, what should be the Aol and the GBW of the op-amp ?

please solve it again if the accepted error is about 10 %

Thank you in advance for your help

Regards