4f3cc240-bcea-4271-89ec-950caf1cf94d 50 4 5 decimal

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element parameters R.fg 50 : R.in 2580 : R.s R.fg R.in + : R.f 14900 : G.dB1 15.15 : 51 line

There are three voltage gains associated with this circuit. Two ideal gains and one gain obtained via measurement. The textbook ideal gain for the inverting operational amplifier does not account for resistance of the source.

A.cl3 R.f R.in / - : 5.7752 -

Non idealities such as the resistance of the source have an effect on the gain. In fact the measure gain is as low as:

A.v V.dB 20 A.v 10 log * :

A.v A.v V.dB G.dB1 ≡ A.v solve : 5.7214

decibels 20 A.v abs 10 log * 15.15

A more precise ideal gain calculation takes into a ccount the resitance of the function generator. One can see that with this consideration the resultant gain approximates the measured gain differntly.

β R.s R.s R.f + / : 0.15

input damping factor k R.f R.s R.f + / - : 0.85 -

closed loop gain A.cl k β / : 5.6654 -

alternaitve calculation A.cl2 R.f R.s / - : 5.6654 -

With the measured and ideal closed loop gains one can calculate the finite open loop gain.

gain in voltage ratio A.ol A.cl abs 1 + A.cl abs A.v - / 1 + / : 681.3467

gain in decibels 20 A.ol abs 10 log * 56.6674

Found this equation in last years solutions

Alternative calculations again uses some ideal values and some experimental

can't work it in sMath

A k A 1 A β * + / * A.v - ≡ A solve maple : 681.3467 -

the equation on the right is from the Proffessors handouts Section 2 - operational amplifiers - page 12

x f u 11 u + (x / + / - :

u 149002630 / :

I couldn't get this page of attempts to match my CASIO Classpad

G k (A A G β * + / * A.v - ≡ G solve maple : 3866.7667

x f 5.7214 - ≡ x solve maple 680.9803 -

A.ol2 A k A.ol2 1 A.ol2 β * + / * :

A.ol2 A.ol2 A A.v ≡ A.ol2 solve : 3.82 - 3.3491 - 21 mat

A.ol2 A A.v - ≡ A.ol2 solve 0.2299 -

β 0.15

k 0.85 -

k A.ol2 1 A.ol2 β * + / * A.v - ≡ A.ol2 solve 0.2299 -

A.v 5.7214

k A.ol2 1 A.ol2 β * + / * 5.358 - ≡ A.ol2 solve 0.4225 -

test 0.8528 - (A.ol2 1 A.ol2 0.1472 * + / * 5.358 - ≡ A.ol2 solve (: 10.6351

0.8528 - (A 1 A 0.1472 * + / * 5.358 - ≡ A solve Maxima #

0.8528 - (A 1 A 0.1472 * + / * 5.358 - ≡ A solve 10.6351

0.8528 - (test 1 test 0.1472 * + / * 3.5352 -

A 681.3467 -

var 0.85 - (var 1 var 0.15 * + / * 1 ≡ var solve maple : 1 -

G A.cl2 11 A.cl2 + (G / + / A.v - ≡ G solve maple : Error, (in solve) a constant is invalid as a variable, 1300354632043560000/ 3362

A.cl2 11 A.cl2 + (G / + / A.v - ≡ G solve maple A.v 1 A.cl2 + (* A.cl2 A.v + / - ≡

evaluate in blank sheet A.cl2 11 A.cl2 + (G / + / A.v - ≡ G solve maple

G R.f R.s / - 11 R.f R.s / - (G / + / 5.7214 - (≡ G solve maple : (Error, invalid terms in product ans

R.f R.s / - 11 R.f R.s / - (G / + / 5.7214 - (≡ G solve maple 28607 R.s - R.f + (* 5000 R.f * 28607 R.s * - / - ≡

evaluate in blank sheet A.cl2 11 A.cl2 + (G / + / A.v - ≡ G solve maple

alternative calcualtion sourced sfrom:

https://electronics.stackexchange.com/questions/222203/feedback-loop-gainbeta-derivation-for-the-inverting-amplifier