Essential_&_Practical_Circuit_Analysis_Part 1_ DC_Circuits.pptx
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Essential_&_Practical_Circuit_Analysis_Part 1_ DC_Circuits.pptx
- 1. By Solid State Workshop Essential & Practical Circuit Analysis Part 1: DC Circuits
- 2. What is circuit analysis? • Circuit analysis is a toolkit for understanding and designing more complex circuits. • Circuit analysis is a layer of abstraction. It is based off of other abstractions, while other abstractions are built off of it. Physics Electronics Circuit Analysis Systems
- 3. Linear Circuit Elements Ohm’s Law Series & Parallel Circuits Voltage & Current Dividers Kirchhoff’s Current Law (KCL) Nodal Analysis Kirchhoff’s Voltage Law (KVL) Loop Analysis Source Transformation Thévenin & Norton Equivalence Superposition Theorem What will be covered in this video?
- 4. Linear Circuit Elements Resistor Ohm Ω Inductor Henry H Capacitor Farad F Current Source Ampere A Volt V Voltage Source + -
- 5. Nodes, Branches, and Loops • A node is a junction of connecting wires. Every point on a node is at the same potential (same voltage). • A branch just another name for any circuit element between two nodes. • A loop is a closed path that begins and ends at the same node. Loop Loop Loop Branch Branch Branch Branch Branch Node Node Node Node
- 6. Ohm’s Law • 𝑉 = 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 • 𝐼 = 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 • 𝑅 = 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 • 𝑉 = 𝐼𝑅 or 𝐼 = 𝑉 𝑅 or R = 𝑉 𝐼 • Ohm’s Law makes it possible to solve for an unknown current, voltage, or resistance. • In this case, 𝐼 = 30𝑉 75Ω = 0.4𝐴 30V I = ? 75Ω
- 7. Series Circuits • There is only one path for current, so the current is the same everywhere in the circuit. • 𝑅𝑒𝑞 = 𝑅1 + 𝑅2+ . . . + 𝑅𝑛 • 𝑅𝑒𝑞 = 100Ω + 50Ω = 150Ω • Applying Ohm’s Law, 𝐼 = 30𝑉 𝑅𝑒𝑞 = 30𝑉 150Ω = 0.2𝐴 50Ω 100Ω 30V I = ?
- 8. • All resistors share a common voltage, but the currents through each depend on the value of each resistor. • 1 𝑅𝑒𝑞 = 1 𝑅1 + 1 𝑅2 + . . . + 1 𝑅𝑛 • 𝑅𝑒𝑞 = 1 150Ω + 1 100Ω −1 = 60Ω 𝐼 = 30𝑉 𝑅𝑒𝑞 = 30𝑉 60Ω = 0.5𝐴 𝐼 = 30𝑉 150Ω = 0.2𝐴 𝐼 = 30𝑉 100Ω = 0.3𝐴 Parallel Circuits 30V 150Ω 100Ω Ohm’s Law
- 9. Voltage Dividers • A voltage divider creates a voltage which is some fraction of its voltage source. • It is a series circuit where the output voltage is (usually) taken across the second resistor. 𝑅𝑒𝑞 = 50Ω 𝐼 = 10𝑉 50Ω = 0.2𝐴 𝑉𝑜𝑢𝑡 = 𝐼𝑅2 = 0.2𝐴 10Ω = 2.0𝑉 10V 40Ω 10Ω Vout
- 10. Voltage Dividers • We can write a general equation that describes the output of any two-resistor voltage divider. • 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 • 𝐼 = 𝑉𝑠 𝑅𝑒𝑞 • 𝑉𝑜𝑢𝑡 = 𝐼𝑅2 • ∴ 𝑉𝑜𝑢𝑡 = 𝑉𝑠 𝑅𝑒𝑞 𝑅2 = 𝑉 𝑠 𝑅2 𝑅1+𝑅2 • Most generally, 𝑉𝑜𝑢𝑡 = 𝑉 𝑠 𝑅𝑥 𝑅𝑒𝑞 VS R1 R2 Vout
- 11. Voltage Dividers • Attaching a load reduces the divider’s output voltage. • 𝑉𝑜𝑢𝑡 = 𝑉 𝑠 𝑅2 𝑅1+𝑅2 • 𝑉𝑜𝑢𝑡 = 𝑉 𝑠 𝑅2 || 𝑅𝐿 𝑅1+ 𝑅2 || 𝑅𝐿 where 𝑅2 || 𝑅𝐿 < 𝑅2 • If the load resistance is much greater the resistance of R2 then the output voltage only drops a bit. • i.e. The load current should be much less than the divider current. RL VS R1 R2
- 12. Current Dividers • A current divider creates a current which is some fraction of its current source. • It is a parallel circuit where the output current is observed in one of the branches. • 𝑅𝑇 = 1 𝑅1 + 1 𝑅2 −1 • 𝐼𝑜𝑢𝑡 = 𝐼𝑠 𝑅𝑇 𝑅𝑋+𝑅𝑇 IS R1 R2 RX RT Iout
- 13. Kirchhoff's Current Law (KCL) • Kirchhoff’s Current Law states that all currents entering a node equals all currents exiting a node. • The sum of all currents in a node = 0. • 𝐼1 + 𝐼2 + 𝐼3+ . . . + 𝐼𝑛 = 0 • Entering a node = positive (+I) • Exiting a node = negative (-I) 𝑗=1 𝑛 𝐼𝑗 = 0 0 1 2 3 4 5 6 7 Entering Exiting Current (mA) Current Entering & Exiting a Node
- 14. 12mA • Find 𝐼1 and 𝐼2 using KCL. • Ask yourself: Do I need to find one current before I can find the other? • KCL at Node B 𝐼2 + 12𝑚𝐴 − 4𝑚𝐴 = 0 𝐼2 = −8𝑚𝐴 • KCL at Node A −𝐼1 − 𝐼2 + 2𝑚𝐴 = 0 −𝐼1 + 8𝑚𝐴 + 2𝑚𝐴 = 0 𝐼1 = 10𝑚𝐴 Kirchhoff's Current Law (KCL) 4mA 2mA 𝑰𝟏 𝑰𝟐 Node A Node B +8𝑚𝐴 −8𝑚𝐴
- 15. Nodal Analysis with KCL • Nodal analysis is a process that uses KCL to determine node voltages. • A reference node (ground) is used to make life easier. • A current through a resistor is described by Ohm’s Law 𝐼 = 𝑉𝑥 − 𝑉𝑦 𝑅 • To find node voltages, write a KCL equation for each node, except the reference node. (N-1) 𝐼 Vx Vy + - + - Reference Node (ground) R
- 16. 𝑰𝟏 𝑰𝟐 1 2 Vo KCL at 𝐼1−𝐼2 + 𝐼𝑠 = 0 𝑉𝑆−𝑉1 𝑅1 − 𝑉1−𝑉2 𝑅2 + 𝐼𝑠 = 0 𝑉1 − 1 𝑅1 − 1 𝑅2 + 𝑉2 1 𝑅2 = −𝐼𝑠 − 𝑉𝑠 𝑅1 KCL at 𝐼2−𝐼2 = 0 𝑉1−𝑉2 𝑅2 − 𝑉2 𝑅3 = 0 𝑉1 1 𝑅2 + 𝑉2 − 1 𝑅2 − 1 𝑅3 = 0 Nodal Analysis with KCL 1 2 𝑉1 = 14𝑉 𝑉2 = 6𝑉 = 𝑉 𝑜 − 1 𝑅1 − 1 𝑅2 1 𝑅2 1 𝑅2 − 1 𝑅2 − 1 𝑅3 ∙ 𝑉1 𝑉2 = −𝐼𝑠 − 𝑉𝑠 𝑅1 0 −5 4 1 4 1 4 −7 12 ∙ 𝑉1 𝑉2 = −16 0
- 17. Nodal Analysis with KCL KCL at 𝐼1 − 𝐼2 + 𝐼𝑠 = 0 𝑉𝑆−𝑉1 𝑅1 − 𝑉1 𝑅2+𝑅3 + 𝐼𝑠 = 0 𝑉1 𝑅1 + 𝑉1 𝑅2 + 𝑅3 = 𝐼𝑠 + 𝑉𝑆 𝑅1 8 7 𝑉1 = 4 + 12 𝑉1= 14𝑉 𝑉2= 𝑉1 𝑅3 𝑅2+𝑅3 = 14 3 4+3 𝑉2 = 6𝑉 = 𝑉 𝑜 1 𝑰𝟏 𝑰𝟐 1 2 Vo
- 18. Kirchhoff's Voltage Law (KVL) • Kirchhoff’s Voltage Law states that all voltage “drops” must equal all voltage “rises” in a closed loop. • That is, the sum of all voltages in a closed loop = 0. • 𝑉1 + 𝑉2 + 𝑉3+ . . . + 𝑉 𝑛 = 0 𝑗=1 𝑛 𝑉 𝑗 = 0 V1 V2 V3 V4
- 19. Kirchhoff's Voltage Law (KVL) V3 V2 V4 V1 𝑰 𝑰 Voltage Drop (+V) 𝑰 Voltage Drop (+V) 𝑰 Voltage Rise (-V)
- 20. Kirchhoff's Voltage Law (KVL) 𝑰 A B 15V 6V 3Ω 2Ω 3Ω 1Ω • Find 𝑉𝐴𝐵 using KVL. • First, we need to find the current around the loop. • KVL starting at B 15 + 3𝐼 + 1𝐼 − 6 + 3𝐼 + 2𝐼 = 0 9 = −9𝐼 𝐼 = −1𝐴 𝑉𝐴𝐵 = 𝐼(2Ω) 𝑉𝐴𝐵 = −1𝐴 2Ω = −2𝑉
- 21. Loop Analysis with KVL • Loop analysis is a process that uses KVL to determine loop currents. • Loop currents are assigned to all independent loops. • To solve for all loop currents, write a KVL equation for all independent loops and then solve. 𝑰𝟏 𝑰𝟐 … + 𝑅 𝐼1 − 𝐼2 … … + 𝑅 𝐼2 − 𝐼1 …
- 22. Loop Analysis with KVL KVL for −15 + 2𝐾 𝐼1 − 𝐼2 + 1𝐾 𝐼1 = 0 −15 1𝐾 + 2𝐾 1𝐾 𝐼1 − 𝐼2 + 1𝐾 1𝐾 𝐼1 = 0 1𝐾 −15𝑚𝐴 + 2 𝐼1 − 𝐼2 + 𝐼1 = 0 3𝐼1 − 2𝐼2 = 15𝑚𝐴 KVL for 2𝐾 𝐼2 + 2𝐾 𝐼2 − 𝐼1 + 4𝐾 𝐼2 − 𝐼3 = 0 2𝐾 2𝐾 𝐼2 + 2𝐾 2𝐾 𝐼2 − 𝐼1 + 4𝐾 2𝐾 𝐼2 − 𝐼3 = 0 2𝐾 𝐼2+ 𝐼2 − 𝐼1 + 2 𝐼2 + 5𝑚𝐴 = 0 4𝐼2−𝐼1 = −10𝑚𝐴 15V 1K 2K 4K 𝑰𝟑 𝑰𝟐 𝑰𝟏 2K 5mA 𝑰𝟏 2 3𝐼1 − 2𝐼2 = 15𝑚𝐴 → 6𝐼1 − 4𝐼2 = 30𝑚𝐴 −𝐼1 + 4𝐼2 = −10𝑚𝐴 → −𝐼1 + 4𝐼2 = −10𝑚𝐴 5𝐼1 = 20𝑚𝐴 𝐼1 = 4𝑚𝐴
- 23. Source Transformation • Thévenin circuit: Voltage source in series with a resistor • Norton circuit: Current source in parallel with a resistor • Source transformation allows simple conversion between these circuits. 𝐼𝑁𝑜 = 𝑉𝑇ℎ 𝑅 𝑉𝑇ℎ = 𝐼𝑁𝑜 ∙ 𝑅 𝑅𝑇ℎ = 𝑅𝑁𝑜 = 𝑅
- 24. Thévenin’s & Norton’s Theorems Black Box Circuit Thévenin Equivalent Circuit VTh RTh Norton Equivalent Circuit RNo INo
- 25. Thévenin’s & Norton’s Theorems • To find 𝑅: • 1.) Detach the load resistor (if present). • 2.) Set all sources equal to 0. Voltage source Short-circuit Current source Open-circuit • 3.) Find the equivalent resistance “looking in” from the two output terminals.
- 26. Thévenin Equivalent Circuits • To find 𝑉𝑇ℎ : • 1.) Detach the load resistor (if present). • 2.) Find the open-circuit voltage 𝑉𝑂𝐶 across the circuit’s output terminals. 𝑉𝑂𝐶 = 𝑉𝑇ℎ Use any form of analysis (nodal, loop, voltage divider, source transformation, etc.) VOC Known Circuit
- 27. Thévenin Equivalent Circuits • Find 𝑉𝐴𝐵 using Thévenin's theorem. 𝑅𝑇ℎ = 2𝐾 + 1𝐾 = 3𝐾Ω 2𝑚𝐴 − 𝑉1 3𝐾 = 0 𝑉1 3𝐾 = 2𝑚𝐴 𝑉1 = 6𝑉 𝑉1 −6𝑉 − 3𝑉 + 𝑉𝑂𝐶 = 0 𝑉𝑂𝐶 = 9𝑉 = 𝑉𝑇ℎ KCL KVL 𝑉𝐴𝐵 = 9𝑉 6𝐾 6𝐾+3𝐾 𝑉𝐴𝐵 = 6𝑉
- 28. Norton Equivalent Circuits • To find 𝐼𝑁𝑜 : • 1.) Detach the load resistor (if present). • 2.) Short circuit the output terminals • 3.) Find the short-circuit current 𝐼𝑆𝐶 through the shorted output. 𝐼𝑆𝐶 = 𝐼𝑁𝑜 Use any form of analysis (nodal, loop, voltage divider, source transformation, etc.) ISC Known Circuit
- 29. Norton Equivalent Circuits • Find 𝑉𝐴𝐵 using Norton's theorem. 𝑅𝑁𝑜 = 2𝐾 + 1𝐾 = 3𝐾Ω −3𝑉 − 6𝑉 + 3𝐾𝐼𝑆𝐶 = 0 3𝐾𝐼𝑆𝐶 = 9𝑉 𝐼𝑆𝐶 = 3𝑚𝐴 = 𝐼𝑁𝑜 𝑅𝑒𝑞 = 1 3𝐾 + 1 6𝐾 −1 = 2𝐾Ω 𝑉𝐴𝐵 = 2𝐾 ∙ 3𝑚𝐴 = 6𝑉
- 30. Superposition Theorem • Superposition says that a circuit with multiple sources can be solved by this process: 1.) Set all sources = 0, except one. 2.) Solve necessary currents and voltages, using only that source. 3.) Repeat Step 2 for each source. 4.) “Superimpose” the solutions onto each other. • Use the symbol prime ( ′ ) to differentiate variables with the same name. 𝑰𝟏 = 𝑰′𝟏 + 𝑰′′𝟏 𝑰𝟐 = 𝑰′𝟐 + 𝑰′′𝟐
- 31. Superposition Theorem • Find 𝑉𝑂 using superposition. + − + − 𝐼1 = 6𝑚𝐴 ∙ 2 3 = 4𝑚𝐴 𝑉′𝑂 = −4𝑚𝐴 ∙ 6𝐾 = −24𝑉 𝑉′′𝑂 = 12𝑉 6𝐾 6𝐾+6𝐾+6𝐾 𝑉′′𝑂 = 12𝑉 ∙ 1 3 = 4𝑉 𝑉𝑂 = 𝑉′𝑂 + 𝑉′′𝑂 𝑉𝑂 = −24𝑉 + 4𝑉 𝑉𝑂 = −20𝑉 𝐼1 = 6𝑚𝐴 6𝐾+6𝐾 6𝐾+6𝐾+6𝐾 6mA + − 𝑰𝟏
- 32. Ending Remarks • The key to solving circuit problems quickly and correctly is practice! • Each technique is not terribly difficult on its own. The challenging part is identifying which technique (or combination of techniques) is appropriate for a given circuit. • Experiment with circuits on your own! The components used in this video are inexpensive and easily obtained through distributors such as Mouser and DigiKey. Use simulation software such as LTSpice (free) to verify your findings and to learn more about circuit behavior.
- 33. Thanks for watching! I’d love to hear your feedback! Subscribe
Editor's Notes
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- Changed wires to freeform lines