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Electrical Analog Circuit Analysis, Cheat Sheet of Electronics

National Institute of EducationElectronics

A detailed analysis of an electrical analog circuit, including calculations of voltage, current, and power dissipation across various components. It covers topics such as thevenin's theorem, diode characteristics, and parallel resistance calculations. A comprehensive understanding of the circuit's behavior and can be useful for students studying electrical engineering, electronics, or circuit analysis. The analysis covers four main sections, each addressing specific aspects of the circuit, such as voltage across a resistor, diode voltage drop, load current and voltage, and voltage and current across a resistor. The document demonstrates the application of fundamental electrical engineering principles and provides a valuable resource for students to enhance their understanding of analog circuit design and analysis.

Typology: Cheat Sheet

2022/2023

Uploaded on 01/13/2023

keopheakdey
keopheakdey 🇰🇭

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Electrical Analog
I3-GEE-B1
KEANG Sakda (ID e20200856)
KEO Jiankor (ID e20200856)
KENH Sothavy (ID e20200856)
KHAI Cheypiseth (ID e20200856)
KEO Pheakdey (ID e20200856)
December 1, 2022
1. (a) Calculate the voltage across the loos R3.
We have Vth −Rth −0.7−R3I= 0
But Rth =R1×R2
R1+R2(R1= 2kΩ, R2= 1kΩ, Vs= 24v)
=2×106
3×103= 0.66 ×103Ω
Vth =R2
R1+R2×Vs=24
3= 8V
⇒VR3=R3I= 103×4.39 ×10−3= 4.39V
Thus VR3= 4.39V
(b) Calculate the voltage across the R2
VR2−VD−VR3= 0
⇒VR2=VD+VR3
⇒VR2= 0.7+4.39 = 5.09v
Thus VR3= 4.39V
(c) Calculate the current i.
i=Vth−0.7
0.66×103+1 = 4.39mA
Thus i= 4,39mA
1
pf3
pf4

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Electrical Analog

I3-GEE-B

KEANG Sakda (ID e20200856)

KEO Jiankor (ID e20200856)

KENH Sothavy (ID e20200856)

KHAI Cheypiseth (ID e20200856)

KEO Pheakdey (ID e20200856)

December 1, 2022

  1. (a) Calculate the voltage across the loos R 3. We have Vth − Rth − 0. 7 − R 3 I = 0 But Rth = R R^11 ×+RR^22 (R 1 = 2kΩ, R 2 = 1kΩ, Vs = 24v) = 23 ××^101063 = 0. 66 × 103 Ω Vth = (^) R 1 R+^2 R 2 × Vs = 243 = 8V ⇒ VR 3 = R 3 I = 10^3 × 4. 39 × 10 −^3 = 4. 39 V Thus VR 3 = 4. 39 V (b) Calculate the voltage across the R 2 VR 2 − VD − VR 3 = 0 ⇒ VR 2 = VD + VR 3 ⇒ VR 2 = 0.7 + 4.39 = 5. 09 v Thus VR 3 = 4. 39 V (c) Calculate the current i. i = (^0). 66 Vth×− 100. (^37) +1 = 4. 39 mA Thus i = 4, 39 mA

(d) Diode will destroyed if the power rating is 3w. PD = VDI = 0. 7 × 4. 39 × 10 −^3 = 13. 17 × 10 −^3 w Explain : According to PD < 3 W so this diode will not be destroy.

  1. (a) Calculate the load current and voltage of this circuit. I = Vs− RV D= 510 −−^0. 37 = 4. 3 mA V = RI = 4. 3 × 1 = 4. 3 V (b) Draw the forward current of is circuit.

(c) Suppose tat we check the voltage of the load and it is display 0.If we check voltage on the load and it display 0V. There are many reason that it display 0V. First for all thediode can be burn, destroy or maybe you put diode in the wrong direction. Secondly the resistance can be burn or broken. Finally maybe there is some problem with the powersource.

  1. (a) Calculate the current and voltage across the resistance R 2 Ohm’s Law:I = PP^ ER = (^) (1+1+1)^20 −^0.^7 −×^010.^73 = (^3) ×^1810.^6 − 3 = 6. 2 mA ⇒ IR 2 = I = 6. 2 mA ⇒ VR 2 = R 2 IR 2 = 10^3 × 6. 2 × 10 −^3 = 6. 2 v Thus VR 2 = 6. 2 v