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反激变压器设计(DCM模式)Flybacktransformer.docx

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    • 反激变压器设计〔DCM 模式〕Flyback transformer design〔DCM Mode〕Illustration:Step 1:The turns ratio of primary and secondaryWhen Q is on, the dropped voltage across the primary is V = V - 1 . From theL Iohms-lawV = VL I- 1 = L di( t )dtV -1So, the i( t ) can be described as i( t ) = ILt + i( 0 )IF the flyback is operated at discontinuous mode, i( 0 ) = 0 . So,V -1i( t ) = I t . It leads to IV - 1= I (m i n ) TL P L onP1When Q is off state, the diode at secondary side is turn on and the mirror voltage will reflect to the primary sideN N2(V = P V P N SS. The voltage VS= V + 1 . That is V =o PP V + 1) .NoSSo, the transistor is sustained the voltage stress VmsN(V = Vms I+V = VP+I (m ax)NP V + 1)oSThe turn ratio ofN VNP = ms-VI ( m ax)V + 1S oStep 2:The core must be guaranteed not to saturated. The voltage-time product of“on-time” must be equal to “off-time”. That is( )NV T = V + 1 P Ti(m i n ) on o N rSand the circuit must be remained in discontinuous mode.T + T + Ton r td= T Þ T + Ton r)(( .))N= 0.8TÞ ( VI (m i n )-1 Ton= V +1oP 0 8T - TN onS( VI (m i n )-1+ V(oN+1) NPST = (V)on o+1)NPNS0.8T()V + 1oNNP 0.8Tso, T =on VS- 1 + ( VN+ 1) NPI (m i n ) oSstep 3: Determine the inductance LP of primaryWhen the transistor is “on”, the energy storage in the primary is equal to1 W 1W = L I 2 joules . So, the input power P = = L I 2 2 P P in T 2T P Pnow, the efficiency is assumed to 80%, we have1 1 æ V-1 ö2 1 ( V- 1) 2P = 1.25P= L I 2 = L çI (m i n )T ÷ = LI (m i n ) T 2èin o2T P P2T P LPon ø2T P L2 on PL = (VI (min) - 1)2 T 2That is Pono2.5P TStep 4: The primary wire turns( )V = V- 1 = NA DB 10-8VN = I (m ax)-1 Ton 108P I (m ax)P e TonP A DBeStep 5: The secondary wire turnsVI ( m ax)- 1 N=PV + 1 No S1 i 2 (t)dt TIP3TonTStep 6:The primary rms current and wire size must be calculated.From the rms formula of the primary is I=prms= and therms formula of the secondary is I= =1 i2 (t)dt TIP3T - TNon pTNssrmsAnd wire size is specified by 300~500 circular miles per rms ampereIP3TonTF = 300PIP30.8T - T NonPT NSF = 300Sstep 5: considered the skin effect.2837fskin depth S =. So, when the primary and secondary wire miles are largerthan skin depth, more wire numbers will be better a single larger wire sizefix( F s )s1FThe parallel of primary wire numbers are n =sand the secondary wirefix( F s )s1Fnumbers are n = 。

      sFlyback transformer design〔CCM Mode〕 Illustration:Step 1:The turns ration of primary and secondaryWhen Q is off state, the diode at secondary side is turn on and the mirror voltage will reflect to the primary sideN NV = P V . The voltage V = V + 1 . That is V = P ( V + 1) .P N S SS o P N oSSo, the transistor is sustained the voltage stress VmsNV = Vms I+V = VP+I (m ax)P ( VN oS+ 1)Step 2:The core must be guaranteed not to saturated. The voltage-time product of“on-time” must be equal to “off-time”. That is( ) NV T = V + 1 P Ti(m i n ) on o NSoffand the circuit must be remained in continuous mode. That is, Ton+ T = ToffSo, V T = (V+ 1)NP ( T - T ) .i(m i n ) ono NS( V + 1)oonNP TNIt leads to Ton=VI (m ax)+ ( Vos+ 1) NPNSstep 3: Determine the inductance L of primaryPTIcpr IcsrTontoffFrom above fig. the output power is givenP = Vo oIcsrToffT= V IocsrT( 1 - on ) That is IT=csrVPoæ T öç1- on ÷The input power is giveno è T øP = 1.25Pin o= ( VI (m i n )- 1) IcprTon That is IT=cpr1.25Poæ T ö( V -1) ç on ÷I (m i n )è T øConsidered the boundary condition between the DCM mode and CCM mode, the following condition is satisfieddII = pcpr 2PThe slop of the ramp dI is given as(dI = VI (m i n )-1)TonP LPSo, the inductance of the primary is(VL = I (mi n )-1) 2 T 2onP 2.5P TinStep 4:The primary rms current and wire size must be calculated.From the rms formula of primary Iprms= Icpron =TT1.25P To onæ T ö T(V - 1)ç on ÷I (min)And wire size is specified by 300~500 circular miles per rms ampereF = 300I。

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