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EECS 16B Designing Information Systems and Devices II UC Berkeley Fall 2023 Note 3: Inductors and RL Circuits 1 Inductors 1.1 Introduction to Inductors Here, we introduce a new passive component, the inductor. This new component will help us design more interesting circuits and introduce oscillations within our circuits. Definition 1 (Inductor) An inductor is denoted as in Figure 1 . I L ( t ) L + − V L ( t ) Figure 1: Example Inductor Circuit The voltage across the inductor is related to its current as follows: V L ( t ) = L d I L ( t ) d t (1) where L is the inductance of the inductor. The SI unit of inductance is the Henry (H). The following are important facts about inductors: 1. The current through an inductor cannot change instantaneously. 2. Immediately after a current is passed through the inductor, the inductor acts as an open circuit, but as t → ª§¦ª , the inductor acts like a short. (a) Why is this the case? Well, note that steady-state (i.e. when t → ª§¦ª ) is when the circuit achieves an equilibrium state. In other words, the current across the inductor will be con- stant. Given the inductor definition equation V L ( t ) = L d I L ( t ) d t = ( L )( 0 ) , we can confirm that an inductor is indeed a short in steady-state. Notice that the voltage-current relationship written in eq. ( 1 ) is similar to that of a capacitor, but with voltage and current swapped. The short term and long term behavior of inductors and capacitors are also opposites of each other. 1

EECS 16B Note 3: Inductors and RL Circuits 2023-09-08 14:21:57-07:00 Theorem 2 (Series Equivalence) Consider the two inductors in series configuration in Figure 2 , and suppose we wish to find the series equivalent as in Figure 3 . I test ( t ) L 1 + − V 1 ( t ) L 2 + − V 2 ( t ) I test Figure 2: Series Inductor Circuit I test ( t ) L eq + − V eq ( t ) I test Figure 3: Equivalent Series Inductor Circuit The equivalent series inductance is L eq = L 1 + L 2 . Proof. We use the test current source, I test ( t ) , depicted in Figure 2 and Figure 3 to find the equivalent voltage across both inductors, i.e., V eq ( t ) . Using KVL, we have V 1 ( t ) + V 2 ( t ) = V eq ( t ) (2) L 1 d I L ( t ) d t + L 2 d I L ( t ) d t = V eq ( t ) (3) ( L 1 + L 2 ) | {z } L eq d I L ( t ) d t = V eq ( t ) (4) as desired. © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 2

EECS 16B Note 3: Inductors and RL Circuits 2023-09-08 14:21:57-07:00 Theorem 3 (Parallel Equivalence) Consider the two inductors in parallel configuration in Figure 4 , and suppose we wish to find the parallel equivalent as in Figure 5 . I eq ( t ) I 1 ( t ) L 1 + − V 1 ( t ) I 2 ( t ) L 2 + − V 2 ( t ) − + V test Figure 4: Parallel Inductor Circuit I eq ( t ) L eq + − V test ( t ) − + V test Figure 5: Equivalent Parallel Inductor Circuit The equivalent inductance is given by L eq = 1 L 1 + 1 L 2 − 1 . Proof. We can apply the test voltage V test as depicted in Figure 4 and Figure 5 to find the equivalent current through both inductors, i.e., I eq ( t ) . By NVA, we have that V 1 ( t ) = V 2 ( t ) = V test ( t ) (5) L 1 d I 1 d t = L 2 d I 2 d t = L eq d I eq d t (6) and from KCL we have I eq ( t ) = I 1 ( t ) + I 2 ( t ) (7) d I eq d t = d I 1 d t + d I 2 d t (8) d I eq d t = L eq L 1 d I eq d t + L eq L 2 d I eq d t (9) 1 L eq = 1 L 1 + 1 L 2 (10) © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 3

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EECS 16B Note 3: Inductors and RL Circuits 2023-09-08 14:21:57-07:00 L eq = 1 L 1 + 1 L 2 − 1 (11) as desired. Theorem 4 (Stored Energy) The stored energy in an inductor can be written as E = 1 2 Li 2 L (12) where i is the current through the inductor. Proof. The formula for power can be manipulated as follows: P L = v L i L (13) P L = L d i L d t i L (14) P L d t = Li L d i L (15) Integrating both sides to find stored energy, we have Z P L d t = Z Li L d i L (16) E = 1 2 Li 2 L (17) Definition 5 (Mutual Inductance) The mutual inductance between two inductors L 1 and L 2 is given by M = N 2 Φ 21 i 1 = N 1 Φ 12 i 2 (18) where N 1 and N 2 are the number of windings in the coils for inductors L 1 and L 2 respectively, and i 1 and i 2 are the current through the respective inductors. Φ 12 is the flux passing through coil 1 from the magnetic field induced by coil 2, and Φ 21 is the flux passing through coil 2 from the magnetic field induced by coil 1. Theorem 6 (Induced Voltage from Mutual Inductance) Consider the circuit below, with two inductors L 1 and L 2 , with mutual inductance M . i 1 + − v 1 L 1 i 2 + − v 2 L 2 M The dots in the circuit indicate the orientation of the inductors. For the given orientation, the following © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 4

EECS 16B Note 3: Inductors and RL Circuits 2023-09-08 14:21:57-07:00 equations hold: v 1 = L 1 d i 1 d t + M d i 2 d t (19) v 2 = M d i 1 d t + L 2 d i 2 d t (20) If the orientation of L 2 is flipped, as shown in the circuit below i 1 + − v 1 L 1 i 2 + − v 2 L 2 M then the following equations hold: v 1 = L 1 d i 1 d t − M d i 2 d t (21) v 2 = − M d i 1 d t + L 2 d i 2 d t (22) Proof. Understanding of this proof is optional , but knowing/understanding mutual inductance is still in scope. Here, we will only prove the first part of the theorem since the second part follows by a symmetry argument, with a negated value of EMF to account for the flipped orientation. First, we can find the induced EMF in L 2 due to mutual inductance only. We can apply Faraday’s law, E 2,mutual = − N 2 d Φ 21 d t (23) = − N 2 d d t Mi 1 N 2 (24) = − M d i 1 d t (25) where in eq. ( 24 ) we apply Definition 5 . Now, notice that there is also current flowing through the second inductor, so we have an induced EMF from that. We can compute that, using Definition 1 , as follows E 2,current = L 2 d i 2 d t (26) Combining these two EMFs using superposition and taking care to note the orientation of L 2 , we obtain v 2 = −E 2,mutual + E 2,current = M d i 1 d t + L 2 d i 2 d t (27) We can apply the exact same argument symmetrically to L 1 to obtain v 1 = L 1 d i 1 d t + M d i 2 d t (28) © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 5

EECS 16B Note 3: Inductors and RL Circuits 2023-09-08 14:21:57-07:00 1.2 Physics behind Inductors Inductors store energy in a magnetic field. In the same way that a capacitor separates charge ( Q ) and this leads to an electric field ( ⃗ E ), anytime current flows down a conductor, it creates a magnetic field ( ⃗ B ), and this magnetic field can store energy. Inductors’ behavior can be described using Faraday’s Law of Induction . The magnitude of magnetic field created by a straight wire is pretty small, so we usually use other geometries to create useful inductances. A solenoid is a good example, where we wind a wire around a conductor like a copper rod: A I S L N turns ℓ L = N 2 µ A ℓ [ H ] Figure 6: The Inductance of a Solenoid: a wire coiled around something. Note that the inductance ( L ) depends on the geometry and a material property called magnetic perme- ability ( µ ) of the solenoid core material. In the case of the solenoid in fig. 6 , the inductance depends on the number of turns ( N ), the length of the solenoid ( l ) and the area ( A ) of the loops. Inductors are useful in many applications such as wireless communications, chargers, DC-DC converters, key card locks, trans- formers in the power grid, etc. But in many high speed applications, their presence might be undesirable as they create delays in the time response of the circuit (analogous to capacitors). Contributors: • Anish Muthali. • Neelesh Ramachandran. • Kristofer Pister. • Utkarsh Singhal. • Aditya Arun. • Kyle Tanghe. • Anant Sahai. • Kareem Ahmad. • Nikhil Shinde. • Chancharik Mitra. • Nikhil Jain. © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 6

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