percentage gain for op-amps

Discussion in 'Homework Help' started by lemon, Mar 14, 2010.

Jan 28, 2010
125
2
Hi:
I am trying to calculate the percentage error for an inverting and non-inverting op-amp.
For the inverting op-amp the error is 0.7V. This is the difference between the theoretical gain -Rf/Rin=100k/10k=10, and measured gain Vout/Vin=16.6/0.2=9.3

error is 0.7V.
Percentage error = 18.6/2 =9.3
0.7/18.6x100=3.8%
0.7/2x100=35%
3.8%+35%=38.8%
0.338x9.3=3.1V
For the non-inverting op-amp the error is 0.1V.
Percentage error = 3.80/2 =1.9
0.1/3.8x100=0.03%
0.1/2x100=0.05%
0.03%+0.05%=0.08%
0.0008x1.9=0.0015V

Have I done this right?

thank you

2. Bychon Member

Mar 12, 2010
469
41
16.6/0.2 = 83 not 9.3
I think you meant 16.6/2.0 = 8.3
The calculated voltage output is 20
The real voltage result is 16.6
16.6/20 x 100 = 83%
100-83 = 17% error

for 18.6/2 = 9.3 demonstrated gain
10 is calculated gain
9.3/10 x 100 = 93%
The error is 7%

That's the end. The error is 7%

Then you divided the voltage error by the real output voltage
Then you divided the voltage error by the input voltage
Then you multiplied that sum by the wrong answer from the first paragraph
and you did it all without labels!

Then you declared .1 volt error for the non-inverting amp
Then you divided 3.80 somethings by the 2 volts you amplified
Then you divided the declared .1 volt error by the 3.8 somethings and rounded off 2.63 to 3 and wrote it down as .03%
Then you divided the declared voltage error by the 2 volts you were amplifying and wrote 5 as .05%
then you multiplied a percentage by the answer to 3.8 somethings divided by the 2 volts you were amplifying
and rounded off .00152 to .0015 volts

Not even close.