I designed the following amplification circuit:
The amplification of this circuit is set by R6. The higher R6 the higher the amplification. BUT the higher the amplification the HIGHER the cutoff frequency of the lowpass caused by the Gain Bandwidth Product. According to f_c = GBP/A it should be just the other way around.
I got to the correct result by using the feedback calculated by FB = R2||(R6+R9) / (R1 + R2||(R6+R9)) and thus using the amplification of A_FB = A0 / (1 + FB*A0) for the calculation of the cutoff frequency instead of the actual gain of the circuit.
I used the "NI Multisim Component Evaluator" to determine the cutoff frequency. Does anybody have in explanation for this behaviour?
Thanks a lot, Andre
PS, for those of you wondering why not use a different design, this circuit is supposed to be a voltage controlled amplifier. In the end (R6 + R9) is being replaced by a n-channel JFET used as a variable resistor. To reduce the possible sources of errors I substituted the JFET with ideal resistors wich behave equally.
I computed your noise gain if you adjusted the R6 from 1 kohms down to 0 ohms and the gain would be around 1146 V/V up to 3489 V/V. Please see the picture below showing that the opamp's cut off frequency varies with respect to the Gain.
So the parameter that refers to the gain bandwidth product is the noise gain and only in some specific circuits equal to the signal gain. Is that correct?
And the noise gain is calculated as the inverse of the feedback, right?
But what has this to do with seeing it as an summing amp?
thanks a lot for your time! I was stuck with this for quite a while. Problem was, that the noise gain is rarely mentioned anywhere if you do not look for it in particular.
So again, thanks a lot to both of you!