Magnetics in Switched-Mode Power Supplies Agenda • Block Diagram of a Typical AC-DC Power Supply • Key Magnetic Elements in a Power Supply • Review of Magnetic Concepts • Magnetic Materials • Inductors and Transformers 2 Block Diagram of an AC-DC Power Supply Input AC Rectifier PFC Input Filter Power Trans- Output DC Outputs Stage former Circuits (to loads) 3 Functional Block Diagram Input Filter Rectifier PFC L + Bus G PFC Control + Bus N Return Power StageXfmr Output Circuits + 12 V, 3 A - + Bus + 5 V, 10 A - PWM Control + 3.3 V, 5 A + Bus Return - Mag Amp Reset 4 Transformer Xfmr CR2 L3a + C5 12 V, 3 A CR3 - CR4 L3b + Bus + C6 5 V, 10 A CR5 Q2 - + Bus Return • In forward converters, as in most topologies, the transformer simply transmits energy from primary to secondary, with no intent of energy storage. • Core area must support the flux, and window area must accommodate the current. => Area product. 4 3 ⎛ PO ⎞ 4 AP = Aw Ae = ⎜ ⎟ cm ⎝ K ⋅ΔB ⋅ f ⎠ 5 Output Circuits • Popular configuration for these CR2 L3a voltages---two secondaries, with + From 12 V 12 V, 3 A a lower voltage output derived secondary CR3 C5 - from the 5 V output using a mag CR4 L3b + amp postregulator. From 5 V 5 V, 10 A secondary CR5 C6 - CR6 L4 SR1 + 3.3 V, 5 A CR8 CR7 C7 - Mag Amp Reset • Feedback to primary PWM is usually from the 5 V output, leaving the +12 V output quasi-regulated. 6 Transformer (cont’d) • Note the polarity dots. Xfmr CR2 L3a – Outputs conduct while Q2 is on. + C5 12 V, 3 A – Secondary Vpeaks = +Bus • Ns/Np CR3 - • Note the coupled output choke, L3. CR4 L3b + Bus + – Windings must have same turns ratios C6 5 V, 10 A CR5 as transformer, which is the same as Q2 - output voltages plus diode drops of CR3 and CR5. + Bus Return • With output chokes in continuous conduction, each output voltage is the average of its secondary voltage (neglecting diode drops). • Therefore, each output voltage is its secondary peak voltage times the duty ratio of the primary bus voltage, +Bus, (neglecting diode drops and Q2’s ON voltage). 7 Review of Some Magnetic Concepts • Units used in the design of magnetic components • Current and magnetic flux • Characteristics of magnetic materials • Faraday’s Law (the “transformer equation”) 8 Units and Their Symbols Symbol Description SI Units H field strength A-t/m B flux density tesla (T) μ permeability T-m/A-t2 F magnetomotive force A-t φ flux weber/t (Wb/t) R reluctance A-t2/Wb P permeance henry/t2 I current ampere (A) L inductance henry (H) N winding turns turn (t) • Units named for famous people are not capitalized (ampere, henry, volt), but their symbols are (A, H, V). •Always separate the value from the unit symbol (10 uH, not 10uH)---it’s not an option. 9 Right-Hand Rule Flux Direction as a Result of Current Flow • Wrap one’s right hand around a conductor with thumb pointing in the direction of current flow. Fingers point in the direction of flux lines. 10 Material Characteristics B flux density in tesla (1 tesla = 10,000 gauss) Slope = B/H = μ = permeability Air μ(relative) = 1 H magnetic field strength in ampere turns / meter • Bulk property of the material • H = NI/le = ampere ⋅ turns per meter – Classic definition is amperes per meter (assumes only one turn) –le = magnetic path length -7 • µ = permeability, usually relative to air (µair = 4⋅π ⋅10 H/m) 11 Core Characteristics φ = B Ae flux in webers (1 weber = 1 tesla square meter) Slope = φ /F = P = permeance "Inductance Factor" in H / t2 F = H le magnetomotive force in ampere turns • Core with no winding. • Material characteristics, with Ae and le added. –Ae = core area, le = effective magnetic path length – Common unit for the slope is “Inductance Factor,” usually given in nH / t2 12 Wound Coil Characteristics Ν φ flux turns in weber turns = volt seconds Slope = L = inductance (henries) I current in amperes • Using volt-seconds and amperes, the wound component can be analyzed easily by circuit engineers using time-domain analysis. 13 The “Transformer Equation” (Faraday’s Law) E = 4B ⋅ A ⋅ f N e • B in tesla, Ae in m2, f in Hz – Modern SI units • The saturation flux density, Bmax, determines the maximum volts per turn that can be applied to a given transformer or inductor winding at a given frequency. 14 Watch closely, now: Transformer Equation from Faraday’s Law ΔΦ E = N Δt Φ = B ⋅ Ae ΔΦ = ΔB ⋅ Ae = 2B ⋅ Ae B 1 1 1 ΔB Δt = T = ⋅ 2 2 f E ΔΦ 1 = = 2B ⋅ A ⋅ = 4B ⋅ A ⋅ f N Δt e 1 e 2 f • Note: This applies to square waves (where Δt = half of the period). 15 An Extremely Important Fact E ΔΦ = = 4B ⋅ A ⋅ f N Δt e • Unless the flux is changing, there will be no voltage. • If the flux swings back and forth, so will the voltage. • In order for there to be a net dc voltage, the flux must be continually increasing. • Therefore, our chances of inventing a magnetic rectifier are ZERO. • The average voltage (dc) across a winding (neglecting winding resistance) is ALWAYS ZERO. This is one of the most useful facts in our bag of tools. 16 Popular Materials Loss @ 0.1 T, Permeability Bsat Material 100 kHz Usage (relative) (tesla) (mW/cm3) Power Transformers Ferrite 2500 0.5 80 Filter Inductors (gapped) (Mag. Inc. P) PFC Inductors (gapped) Ferrite EMI Filters 10,000 0.42 250 (Mag. Inc. W) (common-mode only) Molypermalloy Filter Inductors 60 0.75 340 (Mag. Inc. MPP) PFC Inductors Sendust Filter Inductors 60 1 850 (Mag. Inc. Kool-Mu) PFC Inductors Powdered iron Filter Inductors 75 1.4 3200 (Micrometals 52) PFC Inductors 80% Cobalt tape 100,000 0.55 90 Mag. Amps (Honeywell 2714A) • Note the wide range of permeability and power loss. 17 Inductors and Transformers • Inductor operation (example: buck regulator) • Conduction modes – Continuous mode – Critical conduction mode – Discontinuous mode • The boost regulator • Transformer operation – Flyback converter – Forward converter 18 Buck Regulator (Continuous Conduction) i • Inductor current is continuous. 3 i 1 Vout = 5 V – Vout is the average of the voltage at v 1 its input (V ). 1 Vin = Load i – Output voltage is the input voltage 15 V 2 (R) times the duty ratio (D) of the switch. – When switch in on, inductor current flows from the battery. – When switch is off, it flows through v 1 0 the diode. – Neglecting losses in the switches and inductor, D is independent of i 1 0 load current. • A characteristic of buck regulators i 2 0 and its derivatives: – Input current is discontinuous i3 (chopped), and output current is 0 continuous (smooth). time 19 Buck Regulator (Critical Conduction) i3 i 1 Vout = 5 V v 1 • Inductor current is still Vin = Load i continuous, but just 15 V 2 (R) “touches” zero as the switch turns on again. – This is called “critical conduction.” v 0 • Output voltage is still equal 1 to the input voltage times i D. 1 i 2 i 3 0 time 20 Buck Regulator (Discontinuous Conduction) i3 • In this case, the current in the i 1 i 2 Vout = 5 V inductor is zero during part of v 1 i 1 each period. Vin = Load i • Output voltage is still (as 15 V 2 (R) always) the average of v1. • Output voltage is NOT the input voltage times the duty ratio (D) v 1 of the switch. 0 • While the load current is below i the critical value, D varies with 1 load current (while Vout i 2 remains constant). i3 time 21 Boost Regulator i 1 i3 • Output voltage is always greater v Vout = 15 V than (or equal to) the input 1 Vin = Load i voltage. 5 V 2 (R) • Input current is continuous, and output current is discontinuous Vout = 15 V (the opposite of a buck regulator). Vin = 5 V • Relationship of the output v 1 0 voltage to the duty ratio, D, is not i as simple as in the buck 1 regulator. In the continuous- 0 conduction case, it is: i 2 ⎛ 1 ⎞ 0 Vo = Vin⎜ ⎟ ⎝1− D ⎠ i3 • In this example, Vin = 5, 0 Vout = 15, and D = 2/3. time 22 Transformer (No Energy Storage) IN ΔΦ OUT • Ampere-turns of all windings sum to zero. – Right-hand rule applies to the applied current and the resulting flux. The opposite occurs on the output winding. 23 Transformer (Energy Storage) • This is a conventional flyback transformer. • Energy is delivered to the magnetic core during the pulse applied to the primary. • Energy is transferred from the core to the load during the remaining portion of the cycle. • Ampere-turns of all windings do not sum to zero over each cycle when in continuous- 1 2 conduction mode. This is consistent with energy storage ( /2 L I ). 24 Transformer Operation Including Effect of Primary Inductance turns ratio: i 1 : 2 i sec. v0 i i v sec. Load (R) time 1 2 i0 v sec. 0 i1 0 i2 0 i sec. 0 • This is an example of a “step-up” transformer (secondary voltage is higher than the primary voltage). • Transformer is shown as an ideal transformer, with its primary (magnetizing) inductance as an inductor in parallel with the primary. 25 Flyback Transformer Really a Multi-Winding Inductor turns ratio: i sec. i pri. 1 : 2 Vout Vin v pri. v sec. Load (R) v pri.