Electronic Circuit I Lab report, Cheat Sheet of Electrical Circuit Analysis

Download Electronic Circuit I Lab report and more Cheat Sheet Electrical Circuit Analysis in PDF only on Docsity! Lab Report Course Name: Electronic Circuit-I Course Code: EEE102 Lab Section: 2 Group: 8 Experiment No: 01 I-V Characteristics and Modeling of Forward Conduction of A Diode Submitted To: Dr. M. Ryyan Khan Submitted By: Niaz Morshed Razon Student ID: 2022-2-80-008 Dept. of Electrical and Electronic Engineering East west University Page 1 of 12 OBJECTIVE: In this experiment, we will measure the I-V characteristics and determine the large and small signal models of forward conduction of a p-n junction diode. Background: An ideal diode acts like a one-way valve for electric current, acting as a short circuit for current flow in one direction (forward bias connection) while behaving as an open circuit for current flow in the opposite direction (reverse bias connection). The characteristics of practical diodes are however somewhat different from those of ideal ones. The p-n junction diodes are one of the most popular types of diodes used in the industry. The forward bias current-voltage characteristic of a p-n junction diode will be measured in this experiment. The reverse bias I-V characteristic requires the diode to operate in the break down region. Circuit Diagram: Fig. 1: Circuit set up to measure forward bias I-V characteristics of a diode Equipment: 1. Diode (1 pc). 2. Resistor (1K). 3. Digital multimeter. Forward Bias I-V Measurement: 1) To measure the I-V characteristic of diode in forward bias mode, follow the following steps exactly. Page 2 of 12 As we can see from the graph tangent line touch the X-axis at the point 0.52. therefore, the build in voltage is VDO = 0.52 Volt. Answer to the question No: 3 Let, 𝐼𝐷1= 2 mA is point 1 And 𝐼𝐷2 = 2.5 𝑚𝐴 is point 2 Given, ID1 = 2 mA ID2 = 2.5 mA VD1 = 0.537 v VD2 = 0.546 v n = ? Is = ? We know that, Page 5 of 12 Now, (1) ÷ (2)   n =  n= 0.546-0.537 0.026 ln 2.5 2 ∴ n = 1.55 From (1) => Is= Id1 e Vd1 nVt Is= 2 e 0.537 1.55× 0.026 Is = 3.267× 10−6 mA Answer to the question no: 4 Here, 𝐼𝐷1 = 2.0 mA, 𝐼𝐷2 = 2.5 mA 𝑉𝐷1= 0.537 V, 𝑉𝐷2 = 0.546 mA So, 𝑉𝐷1 = 𝑉𝐷𝑜 + 𝐼𝐷1 × 𝑟𝐷  0.537 = 𝑉𝐷𝑜 + 2 × 𝑟𝐷……………(01) And, 𝑉𝐷2 = 𝑉𝐷𝑜 + 𝐼𝐷2 × 𝑟𝐷  0.546 = 𝑉𝐷𝑜 + 2.5 × 𝑟𝐷……………(02) Now, (2) – (1) => Page 6 of 12 0.546 – 0.537 = 2.5𝑟𝐷 − 2𝑟𝐷 => 0.5𝑟𝐷 = 0.009 ∴ 𝑟𝐷 = 0.018 KΩ (1) => 𝑉𝐷𝑜 = 0.537 - 2𝑟𝐷 =>0.537 – 2 × 0.018 ∴ 𝑉𝐷𝑜 = 0.501 Volt Answer to the question no: 5 The graph shows the identification points of corresponding to ID=3mA. Here, for corresponding to 𝐼𝐷=3mA; 𝑉𝐷3=0.55V We Know, Exponential model, V D3=n×V T×ln Id Is V D3 = 1.55 ×0.026 × ln 3 3.267×10−6 V D3 = 0.553 V Page 7 of 12 Signed Datasheet Page 10 of 12 EE bA02- (2 Experiment Ho. O1. Group 07, Nitg- Menished TD! 9029-2- 90-00 Ahmed Sedct Reger TD* 2022-1 -20- 052 } R= 990 A 9% No Wott) My (vole) Vp. (volt) 0 ier oO o-3 o-ae0 | 9-099 04 0+ 440 0030 oF 0: 47d. 0°350 1-0 0° 495 0-560 44a | Osos. 0940 ak 01 598 4:20 wo 0630 |) 529 peaes 25 0537 1981 22 ory 2°317 30 O54 D 495 34 0°55) 2°887 3°6 0° 554 3:08 32 0's56 3-309 40 oO SSsF 2457 as O'S? B20 pe Oa: G Qe oe organ | On 6:0 Oss “54593 GS oO: 59? | £996 ae O-°s7! 6S 90 Eto) Page 11 of 12 Page 12 of 12