class: title-slide count: false .logo-title[] ## ELECTENG 311 # Electronics Systems Design ### Flyback Converter Analysis & Design .TitleAuthor[Duleepa J Thrimawithana] --- layout: true name: template_slide .logo-slide[] .footer[[Duleepa J Thrimawithana](https://www.linkedin.com/in/duleepajt), Department of Electrical, Computer and Software Engineering (2021)] --- name: S1 # Learning Objectives - Energy storage elements - What is a flyback converter? - Converter topology and how it works - A simplified model of a flyback converter - Steady-state analysis of an ideal flyback converter - What are the expected voltage and current waveforms - Deriving relation between circuit parameters and operating conditions - Example design - Determining the circuit parameters to meet a set of design specifications - Determining the operating conditions - Practical considerations - Understanding the impact of non-ideal transformer - Practical waveforms --- class: title-slide layout: false count: false .logo-title[] # What is a Flyback Converter? ### An Introduction --- layout: true name: template_slide .logo-slide[] .footer[[Duleepa J Thrimawithana](https://www.linkedin.com/in/duleepajt), Department of Electrical, Computer and Software Engineering (2021)] --- name: S2 # Inductors .left-column[ - An inductor is an energy storage element - Energy is stored in the magnetic field that is generated when there is a current flowing through the windings of an inductor, \\[ E\_L = \frac{1}{2} L I\_L^2 \\] - To store energy we can connect an inductor across a DC voltage source and let the current ramp-up to a desired value since, \\[ I\_L = \frac{1}{L} \int\_{0}^{T\_{dur}} V\_{in} \, \mathrm{d}t = \frac{V\_{in}}{L} T\_{dur} \quad \because V\_{in} \textrm{ is constant} \\] - Assume initial current in L is 0A and S_p turned-off at T_dur - **I_L cannot change abruptly** so D circulates I_L when S_p is off ] .right-column[ .center[<img src="img/Inductor.png" height="420px">] ] --- name: S3 # Capacitors .left-column[ - A capacitor is an energy storage element - Energy is stored in the electric field that is generated when there is a voltage across the terminals of a capacitor, \\[ E\_C = \frac{1}{2} C V\_C^2 \\] - To store energy we can connect a capacitor across a DC current source and let the voltage ramp-up to a desired value since, \\[ V\_C = \frac{1}{C} \int\_{0}^{T\_{dur}} I\_{in} \, \mathrm{d}t = \frac{I\_{in}}{C} T\_{dur} \quad \because I\_{in} \textrm{ is constant} \\] - Assume initial voltage of C is 0V and S_p turned-on at T_dur - **V_C cannot change abruptly** so D isolates V_C when S_p is off ] .right-column[ .center[<img src="img/Capacitor.png" height="420px">] ] --- name: S4 # Converter Topology .center[<img src="img/Flyback_Main.png" height="180px">] - A flyback converter consists of a *flyback transformer (Tx)*, *switch (S_p)*, *diode (D_s)* and a *capacitor (C_o)* - The switch is operated at a certain duty-cycle, D_p, to control the output voltage, V_out - When S_p is on during D_pT_s, the primary winding, L_p, of the transformer is connected across the input source, V_in, causing I_in to build-up thus storing energy in L_p - When S_p is off during (1-D_p)T_s, the energy stored in L_p is released to the output, charging C_o - D_s makes sure C_o is not discharged through L_s - As we learnt in [PWM control](https://uoa-ee311.github.io/presentations/intro/presentation.html#46), S_p is operated at a fixed frequency f_s and therefore T_s = 1/f_s --- name: S5 # Flyback Operation - An Analogy (PI) .center[<img src="img/Analogy.png" height="220px">] - We can compare a flyback converter to an analogy where a smaller water bucket is used to fill a bigger water bucket that has a hole - Here the hole represents the load resistor (i.e., smaller R_L bigger hole) and the water in smaller bucket represents energy stored in L_p and transferred to output - The larger bucket represents the output capacitor that holds V_out relatively constant - However, there is a voltage ripple at V_out since the water level dips and rises between each fill --- name: S6 # Flyback Operation - An Analogy (PII) .left-column[ - If very small or no hole (i.e., large R_L or open-circuit ) water will overflow (i.e., large voltage **damaging** system) - V_out can build up to a very large voltage - **electric shock** hazard and **flying debris** from exploding components - Before turning-on a flyback converter make sure **protection circuitry** works and an appropriate **load is attached** - We can reduce the ripple at V_out by - Increasing the size of the larger bucket (i.e., larger C_out) - Filling more frequently (i.e., faster f_s) - To transfer more power (i.e., higher V_out at smaller R_L) we can - Increasing the size of the smaller bucket (i.e., larger E_L) - Filling more frequently (i.e., faster f_s) ] .right-column[ .center[<img src="img/Analogy_2.png" height="420px">] ] --- class: title-slide layout: false count: false .logo-title[] # Analyzing a Flyback Converter ### A Steady-State Model --- layout: true name: template_slide .logo-slide[] .footer[[Duleepa J Thrimawithana](https://www.linkedin.com/in/duleepajt), Department of Electrical, Computer and Software Engineering (2021)] --- name: S7 # Transformer Model - We know how a flyback converter works in general terms, so lets mathematically model its operations - The mathematical model will help us determine the parameters of components we should use - To simplify our analysis, we could replace the flyback transformer with it equivalent cantilever model - In our transformer model, L_p and L_s are the self-inductances of primary and secondary windings - k, the coupling factor between the two windings is typically greater than 0.98 and also, \\[ \acute{V}\_s = n V\_s \quad \text{and} \quad \acute{I}\_s = \frac{1}{n} I\_s \quad \text{and} \quad \acute{R}\_s = n^2 R\_s \\] \\[ \text{where for T-model} \quad n = \sqrt{ \frac{L\_p}{L\_s} } = \frac{N\_p}{N\_s} \quad \text{and for cantilever model} \quad n = k\sqrt{ \frac{L\_p}{L\_s} } \approx \sqrt{ \frac{L\_p}{L\_s} } \approx \frac{N\_p}{N\_s}\\] .zoom175[ .center[<img src="img/Tx_Models.png" height="120px">] ] --- name: S8 # A Model of the Flyback .center[<img src="img/Flyback_Model.png" height="190px">] - Lets replace the flyback transformer with its equivalent cantilever model to help our mathematical analysis - For now we are going to **assume k is 1** (later we will investigate the impact of a k less than 1) - In an SMPS, such as this flyback converter, the switch, S_p, can only be either on or off - We can draw two equivalent circuits, one for when S_p is on and other for when S_p is off - We will analyse the circuit assuming it has reached a steady-state - The average change in current through L_p and voltage across C_o over one switching period is 0 --- name: S9 # Equivalent Circuits - S_p is On .center[<img src="img/Flyback_On.png" height="210px">] - With PWM control, at the beginning of each switching cycle, S_p will be turned-on for a duration [DT_s](https://uoa-ee311.github.io/presentations/intro/presentation.html#47) - Turning on S_p shorts L_p across V_in causing I_in to ramp from 0 to a higher value storing energy in L_p - Since V_in is directly applied across the primary side of the ideal transformer V_s is -V_in/n - This is because the positive end of transformer secondary is reversed as indicated by the dots - The anode end of D_s is negative while the cathode end is positive, thus D_s is reverse biased (i.e. blocking) - We will assume that C_o is quite large and holds V_out constant (though it is discharging a bit) --- name: S10 # Equivalent Circuits - S_p is Off (PI) .center[<img src="img/Flyback_Off.png" height="210px">] - After DT_s, S_p will be turned-off until the beginning of the next switching cycle - Turning off S_p disconnects L_p from V_in and thus I_in is 0A - Since current in L_p cannot change abruptly, it finds a path to continue through the transformer and D_s - This is called **flyback time** and energy that was stored in L_p during DT_s now replenishes C_o - As L_p losses energy, I_D as well as I^'_D ramps down to 0A before the end of the switching cycle - We call this **discountinuous mode** operation since current in L_p is 0 during part of T_s --- name: S11 # Equivalent Circuits - S_p is Off (PII) .center[<img src="img/Flyback_Disc.png" height="210px">] - After I_D reach 0A, apart from C_o, the other components do not contribute to the functionality until the beginning of the next switching cycle - We will call this time the **discountinuous time** - Having a discountinuous time ensures, all the energy we stored in L_p is delivered to the output - One of our design goals will be to select components that ensure discountinuous mode operation - This is the preferred operation mode for a flyback converter --- name: S12 # Steady-State Waveforms .zoom15[ .center[<img src="img/Waveforms.png" height="440px">] ] --- name: S13 # Energy Stored in L_p .left-column[ - During DT_s (i.e., when S_p is on), current in L_p, which is the same as I_in, ramps-up from 0 and reaches I_Lp(pk) when S_p is turning off, \\[ I\_{Lp(pk)} = \frac{1}{L\_p} \int\_{0}^{DT\_s} V\_{in} \, \mathrm{d}t = \frac{V\_{in}}{L\_p} DT\_{s} \quad \because V\_{p} = V\_{in} \equiv \textrm{ constant} \\] - Using I_Lp(pk) we can find peak energy stored in L_p, \\[ E\_{Lp(pk)} = \frac{1}{2} L I\_{Lp(pk)}^2 = \frac{V^2\_{in}}{2L\_p} D^2T^2\_{s} \\] - Thus, we can use D to control the peak energy we store in L_p - L_p, V_in and T_s are fixed in a given design - PWM controller sets D to regulate V_out by controlling E_Lp(pk) ] .right-column[ .center[<img src="img/Lp_Energy.png" height="420px">] ] --- name: S14 # Output Power .left-column[ - All of the energy stored in L_p (i.e., E_Lp(pk)), is delivered to the output between DT_s and T_s - All this energy is consumed by R_Load because the average voltage across C_o is not changing - In steady-state C_o is effectively not absorbing/loosing energy when averaged over one time period - However, inside each time period C_o charge and discharge by the same amount leading to a small ripple in V_out - So we can relate output power to E_Lp(pk) assuming all components are lossless, \\[ P\_{out} = E\_{Lp(pk)} f\_s = \frac{V^2\_{in}}{2f\_s L\_p} D^2 \quad \because T\_{s} = 1/f\_{s} \\] ] .right-column[ .center[<img src="img/Lp_Energy.png" height="420px">] ] --- name: S15 # Calculating Required L_p - Since P_out is proportional to D², maximum P_out is achieved at maximum D we choose to operate our flyback converter - Therefore, we can determine the value of L_p that should be used in a flyback converter designed to deliver up to P_out(max) \\[ L\_{p(req)} \leqslant \frac{V^2\_{in}}{2f\_s P\_{out(max)}} D^2\_{max} \\] - We know P_out(max), V_in and f_s from design specifications - D_max can be selected to be 0.5 as it simplifies the design - Since we assumed a lossless flyback converter for this analysis, when designing your converter pick a L_p value that is slightly smaller than what you calculate - If V_in is not constant, use the smallest V_in you expect to operate at (in your design V_in is constant) --- name: S16 # Calculating Required Turns Ratio (PI) .left-column[ - n determines whether we operate in the discountinuous mode - Note that if in discountinuous mode `\( \lgroup D + D' \rgroup \leqslant 1 \)` - Since the average value of I_Lp doesn't change in steady-state, \\[ \int\_{0}^{T\_s} V\_p \, \mathrm{d}t = V\_{in}DT\_s - nV\_{out}D'T\_s = 0\\] - Substituting `\( D' = V_{in}D \, / \, nV_{out} \)` in `\( \lgroup D + D' \rgroup \leqslant 1 \)` we get, \\[ n \geq \frac {V\_{in}D} {V\_{out}(1-D)} \\] - Thus, to guarantee discountinuous operation, select an n that is greater than the max value of `\( V_{in}D \, / \, V_{out} (1-D)\)` ] .right-column[ .center[<img src="img/Voltage_Stress.png" height="420px">] ] --- name: S17 # Calculating Required Turns Ratio (PII) .left-column[ - n also directly influences maximum voltage stress experienced by S_p and D_s \\[ V\_{Sp(max)} = V\_{in} + nV\_{out} \quad \text{and} \quad V\_{Ds(max)} = V\_{in}/n + V\_{out} \\] - When selecting devices to implement S_p and D_s we need to make sure they have voltage ratings greater than the maximum voltage stress they experience - Higher voltage rated devices are more expensive - Performance tends to be also poor (e.g., higher losses) - Smaller n leads to lower voltage stress across S_p at the expense of the voltage stress across D_s - We need to find a balance between these contradicting needs ] .right-column[ .center[<img src="img/Voltage_Stress.png" height="420px">] ] --- name: S18 # Average Currents .left-column[ - Since ripple voltage at V_out is negligible, I_out is a constant DC value give by, \\[ I\_{out} = \frac{V\_{out}}{R\_{Load}} \\] - The average value of I_in is, \\[ I\_{in(av)} = \frac{1}{T\_{s}} \int\_{0}^{T\_s} I\_{in} \, \mathrm{d}t = \frac{1}{2} DI\_{L(pk)} = \frac{V\_{in}}{2L\_p} D^2T\_{s} \\] - The average value of I_D should be the same as I_out and is, \\[ I\_{D(av)} = \frac{1}{T\_{s}} \int\_{0}^{T\_s} I\_{D} \, \mathrm{d}t = \frac{1}{2} nD'I\_{L(pk)} = I\_{out} \\] ] .right-column[ .center[<img src="img/Average_Currents.png" height="400px">] ] --- name: S19 # RMS Currents .left-column[ - RMS currents dictate conduction loss in transformer and S_p - To calculate RMS of I_in, which is a linear function, it can be written as `\( I_{Lp(pk)} t /DT_s \)` for `\( 0 \leq t \leq DT_s\)` - RMS of I_in is therefore give by, \\[ I\_{in(RMS)} = \sqrt { \frac{1}{T\_s} \int\_{0}^{T\_s} I^2\_{in} \, \mathrm{d}t } = \sqrt { \frac{1}{T\_s} \int\_{0}^{DT\_s} \left( \frac {I\_{Lp(pk)}} {DT\_s} t \right)^2 \, \mathrm{d}t } \\] \\[ I\_{in(RMS)} = \sqrt { \frac{1}{T\_s} \left( \frac {I\_{Lp(pk)}} {DT\_s} \right)^2 \left[ \frac{t^3} {3} \right]^{DT\_s}\_0 } = I\_{Lp(pk)} \sqrt {\frac {D} {3}} \\] - Similarly we can show that `\( I_{D(RMS)} = nI_{Lp(pk)} \sqrt{D' \, / \, 3 } \)` ] .right-column[ .center[<img src="img/Average_Currents.png" height="400px">] ] --- name: S20 # Output Voltage .left-column[ - In a flyback converter, like in any switched-mode converter, we use D to control the output voltage across a load R_load - Since we already know what [P_out](#S14) is, we can derive the relation between V_out and D, \\[ P\_{out} = \frac {V^2\_{out}} {R\_{Load}} = \frac{V^2\_{in}}{2f\_s L\_p} D^2 \quad \Rightarrow \quad V\_{out} = \sqrt{ \frac{R\_{Load}}{2f\_s L\_p} } \, V\_{in}D \\] - As we expected, V_out is proportional to D - However, V_out also depends on R_load unlike in some other switched-mode converters - An open circuit at output (i.e., R_load → ∞) causes V_out to reach ∞ as predicted by the [water bucket](#S6) analogy ] .right-column[ .center[<img src="img/OutputWaveforms.png" height="400px">] ] --- name: S21 # Output Voltage Ripple .left-column[ - In practice C_o is a finite value and the AC current, I_Co, flowing through it cause a very small ripple voltage, ΔV_out, at output - From KCL we know that I_Co = I_D - I_out - Assume ΔV_out is negligible so V_out and I_out are DC signals - Using trigonometry we can determine duration C_o is charging, \\[ \frac {nI\_{Lp(pk)} - I\_{out}} {nI\_{Lp(pk)}} = \frac {T\_{Chrg}} {D'T\_s} \quad \Rightarrow \quad T\_{Chrg} = \frac { (nI\_{Lp(pk)} - I\_{out}) D'T\_s} {nI\_{Lp(pk)}} \\] - Using this information we can evaluate ΔV_out since, \\[ \Delta V\_{out} = \frac {1} {C\_o} \int\_{0}^{T\_{Chrg}} I\_{Co} \, \mathrm{d}t = \frac {1} {2C\_o} \left( nI\_{Lp(pk)} - I\_{out} \right)T\_{Chrg} \\] ] .right-column[ .center[<img src="img/RippleVout.png" height="425px">] ] --- name: S22 # Demo: A Conceptual Flyback Converter .questions[ When f_s=100kHz, D=0.5 and R_load=100Ω, calculate I_Lp(pk), P_out, V_out, D', V_Sp(max), V_Ds(max), I_in(av), I_D(av), I_in(RMS), I_D(RMS) and ΔV_out. Show answers match simulation. Explore behavior when D and R_load change. ] .center[ <iframe src="example2.html" width="900" height="350" frameborder="0"></iframe> ] --- class: title-slide layout: false count: false .logo-title[] # How to Design a Flyback Converter ### An Example --- layout: true name: template_slide .logo-slide[] .footer[[Duleepa J Thrimawithana](https://www.linkedin.com/in/duleepajt), Department of Electrical, Computer and Software Engineering (2021)] --- name: S23 # Design Specifications .left-column[ - Using what we learnt so far lets design a flyback converter that meets the specifications listed in the table - In this design both the input and output voltages are fixed - In many cases you will need to cater for a range of input/output voltages - We are going to take as D_max = 0.5, η = 100% and design for discountinuous mode operation - To 'Design' this converter first we have to determine suitable values for L_p, n, L_s and C_o - Then we analyse the voltage stress across S_p and D_s, average and RMS current stresses as well as operating D - We will do this analysis at P_out = 50W and P_out = 100W ] .right-column[ .center[<img src="img/FlybackDesign.png" height="170px">] <table class="tg" style="undefined;table-layout: fixed; width: 300px; margin-left:auto; margin-right:auto;"> <colgroup> <col style="width: 150px"> <col style="width: 150px"> </colgroup> <thead> <tr> <th class="tg-dzaw"><span style="color:white">Parameter</span></th> <th class="tg-dzaw"><span style="color:white">Specification</span></th> </tr> </thead> <tbody> <tr> <td class="tg-jayl">V_in</td> <td class="tg-jayl"> 20V </td> </tr> <tr> <td class="tg-sabo">V_out</td> <td class="tg-sabo"> 200V </td> </tr> <tr> <td class="tg-ig71"> P_out</td> <td class="tg-ig71"> 0W to 100W </td> </tr> <tr> <td class="tg-ig71">f_s</td> <td class="tg-ig71">100kHz</td> </tr> <tr> <td class="tg-sabo">ΔV_out</td> <td class="tg-sabo">2V</td> </tr> </tbody> </table> ] --- name: S24 # Parameters of the Transformer - Since we know the maximum power the converter is expected to deliver from the specifications (i.e., 100W) we can calculate the [required L_p](#S15), \\[ L\_{p(req)} \leqslant \frac{V^2\_{in}}{2f\_s P\_{out(max)}} D^2\_{max} = 5 \mu H \\] - Since we are assuming our converter has no losses, to be on the safe side lets **choose L_p = 4.5µH** - We can also determine the [minimum turns-ratio](#S16) of the transformer, \\[ n \geq \left [ \frac {V\_{in}D} {V\_{out}(1-D)} \right]\_{(max)} \quad \Rightarrow \quad n \geq 0.1 \\] - Noting that a smaller *n* results in [larger V_Ds(max) and a larger *n* in larger V_Sp(max)](#S17) lets **choose *n* = 0.5** \\[ V\_{Sp(max)} = V\_{in} + nV\_{out} = 120V \quad \text{and} \quad V\_{Ds(max)} = V\_{in}/n + V\_{out} = 240V \\] --- name: S25 # Operating Conditions (PI) - Since we have now determined the parameters of the transformer (i.e., L_p = 4.5µH & *n* = 0.5) next will be determine [operating D and D'](#S14) at each output power level \\[ D = \sqrt{ \frac {2 P\_{out} f\_s L\_p} {V^2\_{in}} } \quad \Rightarrow \quad D\_{@100W} = 0.47 \quad \& \quad D\_{@50W} = 0.34 \\] - Now we can determine both [I_Lp(pk)](#S13) (i.e., same as I_in(pk)) and I_Ls(pk) (i.e., same as I_D(pk)), \\[ I\_{Lp(pk)} = \frac{V\_{in}}{L\_p} DT\_{s} \quad \Rightarrow \quad I\_{Lp(pk@100W)} = 21.08A \quad \& \quad I\_{Lp(pk@50W)} = 14.91A \\] \\[ I\_{Ls(pk)} = nI\_{Lp(pk)} \quad \Rightarrow \quad I\_{Ls(pk@100W)} = 10.54A \quad \& \quad I\_{Ls(pk@50W)} = 7.45A \\] --- name: S26 # Operating Conditions (PII) - [Average currents](#S18) can be calculated as, \\[ I\_{in(av)} = \frac{1}{2} DI\_{L(pk)} \quad \Rightarrow \quad I\_{in(av@100W)} = 5A \quad \& \quad I\_{in(av@50W)} = 2.5A \\] \\[ I\_{D(av)} = I\_{out} = \frac{V\_{out}}{R\_{Load}} \quad \Rightarrow \quad I\_{D(av@100W)} = 0.50A \quad \& \quad I\_{D(av@50W)} = 0.25A\\] - [RMS currents](#S19) can be calculated as, \\[ I\_{in(RMS)} = I\_{Lp(pk)} \sqrt {\frac {D} {3}} \quad \Rightarrow \quad I\_{in(RMS@100W)} = 8.38A \quad \& \quad I\_{in(RMS@50W)} = 4.38A \\] \\[ I\_{D(RMS)} = nI\_{Lp(pk)} \sqrt {\frac {D'} {3}} \quad \Rightarrow \quad I\_{D(RMS@100W)} = 1.87A \quad \& \quad I\_{D(RMS@50W)} = 1.11A \\] --- name: S27 # Capacitor Size and Ripple Voltage - We can now calculate the [C_out required](#S21) to ensure a voltage ripple less than what is specified (i.e., 2V), \\[ C\_o \geqslant \frac {1} {2 \Delta V\_{out} } \frac { \left( nI\_{Lp(pk)} - I\_{out} \right)^2 } {nI\_{Lp(pk)}} D'T\_s \quad \Rightarrow \quad C\_o \geqslant 2.26 \mu F \\] - As the size and cost targets are not specified we can use a larger capacitor and in this example lets **choose C_out = 10µF** - Lets calculate ripple voltage at each power level to make sure it is acceptable \\[ \Delta V\_{out} = \frac {1} {2 C\_o} \frac { \left( nI\_{Lp(pk)} - I\_{out} \right)^2 } {nI\_{Lp(pk)}} D'T\_s \quad \Rightarrow \quad \Delta V\_{out(@100W)} = 0.45V \quad \& \quad \Delta V\_{out(@50W)} = 0.23V\\] - As a percentage our voltage ripple is less than 0.23% of V_out (i.e., our assumption of V_out being a DC voltage is accurate) --- class: title-slide layout: false count: false .logo-title[] # What to Expect from a Practical Flyback Converter ### Impact of Transformer Leakage Inductance --- layout: true name: template_slide .logo-slide[] .footer[[Duleepa J Thrimawithana](https://www.linkedin.com/in/duleepajt), Department of Electrical, Computer and Software Engineering (2021)] --- name: S28 # Leakage Inductance (PI) .center[<img src="img/Real_Flyback.png" height="220px">] - It is impossible to build a transformer that has perfect coupling (i.e., k = 1) as we assumed earlier - k of a flyback transformer is quite high and is typically over 0.998 - This means the analysis we conducted assuming k = 1 is still very accurate - Having a k of less than 1 means that there is going to be leakage inductance (i.e., `\( L_{lk} = (1-k^2)L_p \)`) - We can use the Cantilever model to explore impact of leakage inductance on the operation of the flyback converter --- name: S29 # Leakage Inductance (PII) .center[<img src="img/Real_FlybackW.png" height="170px">] - From Cantilever model we can see that the I_in flows through the leakage inductance (i.e., `\( L_{lk} = (1-k^2)L_p \)`) when S_p is turned-on storing energy in it - When S_p is turning-off I_in rapidly fall from I_Lp(pk) to 0A over a few 10s of nano-seconds because a real switch cannot turn-off instantaneously - V_lk can be very large since the time S_p takes to turn-off, t_off, is very small, \\[ V\_{lk} = L\_{lk} \, \frac {\mathrm{d}I\_{in}} {\mathrm{d}t} = -(1-k^2)L\_p \frac {I\_{Lp(pk)}} {t\_{off}} \\] --- name: S30 # Parasitic Capacitance of Switch (PI) .center[<img src="img/Real_FlybackW2.png" height="170px">] - A practical switch also has a small inbuilt capacitance across it due to the way it is manufactured - Size of C_Sp depends on the switch and you will learn more about this later - When S_p turns-off, this parasitic capacitance, C_Sp, absorbs the energy stored in the leakage inductance - As a result V_Sp rises up to a voltage greater than V_in + nV_out - The large voltage across V_Sp when S_p is off causes this parasitic energy to circulate between C_Sp and the leakage inductance at their resonant frequency - The circulating energy is dissipated as losses and the resonance eventually disappears --- name: S31 # Parasitic Capacitance of Switch (PII) .center[<img src="img/Real_FlybackW3.png" height="170px">] - After the energy stored in k²L_p is delivered to output, the flyback enters [discontinuous time](#S11) - Note that during the [flyback time](#S10) C_Sp is charged to V_in + nV_out - In the ideal converter I_in and therefore V_p was 0 during [discontinuous time](#S11) - This causes C_Sp that is charged to V_in + nV_out to discharge in to V_in - However L_p is in this discharge path thus causing this parasitic energy to circulate between C_Sp and L_p at their resonant frequency - As this circulating energy is dissipated in losses the resonance starts decaying --- name: S32 # Practical Steady State Waveforms .zoom15[ .center[<img src="img/Real_FlybackW_Summary.png" height="453px">] ] --- name: S33 # Demo: A Practical Flyback Converter .questions[ Calculate the resonant frequency of the parasitic circulating energy during the flyback time and the discontinuous time. Show answers match simulation. Explore behavior when D and R_load change. ] .center[ <iframe src="example2.html" width="900" height="350" frameborder="0"></iframe> ] --- class: title-slide layout: false count: false .logo-title[] # Questions?