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Да се пресметне изходното напрежение на инвертиращ суматор, акоRF=150K, R1=10K, R2=8K, u1=20mV, u2=10mV.

Решение:

i1=U1/R1

i1= (20/10) x 10-6 = 2 ?A

i2= U2/R2

i2= (10/8) x 10-6 = 1.25 ?A

-iF = i1+i2 = 2 + 1.25 = 3.25 ?A

iF=-3.25 ?A

Uo = −iF X RF = (3.25x 10-6) X (150x1000)

Uo = 187.5 x (1/1000)

Uo = 187.5 mV

 
Last modified 23 May 2011 10:56 PM by Cvetelina X.  
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