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# Lab 3 - Of the four circuits in Part D(4 circuits a and c...

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ECE 206 Section F10 Experiment 3 Lab Write-up 1) In part B, the circuit in step 1 was a low pass filter, because the output decreased as the frequency increased. Conversely, the circuit in step 5 was a high pass filter, because the output increased as the frequency increased. 2) Tau (measured) = (t2–t1) / ln(V1/V2) Tau (theoretical) = RC Measured Values: t1 = 15 x 10^-6 sec t2 = 112 x 10^-6 sec V1 = 4.219 V V2 = 1.781 V Theoretical Values: R = 10 Kohm C = 0.01 x 10^-6 F Tau (measured) = [(112-15) x 10^-6 sec] / ln(4.219V/1.781V) = 112.5 x 10^-6 sec Tau (theoretical) = (10 kOhm)(0.01 x 10^-6 F) = 100 x 10^-6 sec Percent Error = (|measured – computed|/measured) x 100% = (|112.5-100|/112.5) x 100% = 11.11% The measured value is comparable to the theoretical value, but even without the technical knowledge of circuits, I would guess that the percent error would be higher than an acceptable limit in most other situations. 3)

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Unformatted text preview: Of the four circuits in Part D (4), circuits a and c can be solved using the equation presented in class, because the circuits can be reduced by finding the Thevenin equivalent. Vc(t) = K1 + K2 x e^(t/RC) Vc( 4 ) = K1 = Vf K1 = Vf Vc(0) = Vf + K2 = Vi K2 = Vi-Vf Measured Values R1 = 98.695 kOhm R2 = 98.944 kOhm R3 = 9.914 kOhm C1 = 392.1 pF C2 = 369.1 pF Circuit A Vc(t) = Vf + (Vi-Vf) x e^{-t/[(R1+R3) x C2]} = Vf + (Vi-Vf) x e^(-t/4.01 x 10^-5 sec) Circuit B Vc(t) = Vf + (Vi-Vf) x e^{-t/[(R1+R3||R2) x C2]} = Vf + (Vi-Vf) x e^(-t/1.91 x 10^-5 sec) 4) See attached page for table Circuit C Vout = [R2/(R2+R1+R3)]Vin = [98.944/(98.944+98.695+9.914)] x 5V = 2.38V Circuit D C2 Vout = [R2/(R2+R1+R3)]Vin = [98.944/(98.944+98.695+9.914)] x 5V = 2.38V C1 Vout = [(R1(R2+R1+R3]Vin = [98.695/(98.944+98.695+9.914)] x 5V = 2.38V *References – results from lab notebook on last two pages of report...
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Lab 3 - Of the four circuits in Part D(4 circuits a and c...

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