# Lab 1 - Vd = Vin – Vout I3 =(12 – Vout R3 I12 =...

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ECE 206 Section F10 Experiment 1 Lab Write-up 1) Percent Error = (|R-measured – R-computed| / R-measured) x 100% A. Percent Error for R1||R2: R1||R2 measured = 4.964 kOhm R1||R2 computed = 5 kOhm Percent Error = (|(4.964-5) kOhm| / 4.964 kOhm) x 100% = 0.725% B. Percent Error for R1||R2 + R3: R1||R2 + R3 measured = 14.904 kOhm R1||R2 + R3 computed = 15 kOhm Percent Error = (|(14.904-15) kOhm| / 14.904 kOhm) x 100% = 0.241% C. Percent Error for I: I measured = 0.829 mA I computed = 0.8 mA Percent Error = (|(0.829-0.8) mA| / 0.829 mA) x 100% = 3.49% 2) A. By comparing each terminal in the circuits, the schematic of the circuit shown is the exact same as the circuit in part B, step 20. B. Results are placed in a table – see attached sheet The Equations used are:

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Unformatted text preview: Vd = Vin – Vout I3 = (12 – Vout) / R3 I12 = Vout/(R1||R2) Id = I12 – I3 P = Id x Vd 3) By comparing my calculated values for Vd with the recorded values from the multimeter in step 16, I concluded that the value recorded from the meter was the turn on voltage for the diode. The attached table shows the voltage and current for the diode at various Vin and Vout voltages. Using KVL, I found that the voltage through the diode was related to Vin and Vout, through the equation: Vd = Vin – Vout. After Vd gets to (approximately) the recorded voltage from the meter, the current through the diode begins to increase. In step 25 we determined that Vout started to increase when Vin got to approximately 4.5 V....
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## This note was uploaded on 01/18/2010 for the course ECE 205 taught by Professor Zhang during the Spring '09 term at University of Illinois at Urbana–Champaign.

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Lab 1 - Vd = Vin – Vout I3 =(12 – Vout R3 I12 =...

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