反激变压器Flyback-transformer

Step 1：The turns ratio of primary and secondary

When Q is on, the dropped voltage across the primary is VLVI1. From the ohms-law

VLVI1Ldi(t) dtVI1ti(0) LSo, the i(t) can be described as i(t)IF the flyback is operated at discontinuous mode, i(0)0. So,

VI(min)1V1i(t)It. It leads to IPTon

LLP

When Q is off state, the diode at secondary side is turn on and the mirror voltage will reflect to the primary side

NNVPPVS . The voltage VSVo1. That is VPP(Vo1).

NSNS

1

So, the transistor is sustained the voltage stress Vms

VmsVIVPVI(max)NP(Vo1) NSThe turn ratio of

Step 2：

NPVmsVI(max) NSVo1 The core must be guaranteed not to saturated. The voltage-time product of

“on-time” must be equal to “off-time”. That is

NPVi(minTV1Tr )onoNSand the circuit must be remained in discontinuous mode.

TonTrTtdT TonTr0.8T NP(0.8TTon)NS

NPNP(VI(min)1(Vo1))TonVo10.8TNSNS(VI(min)1)Ton(Vo1)NP0.8TNSso, Ton

NPVI(min)1(Vo1)NS(Vo1)step 3: Determine the inductance LP of primary

When the transistor is “on”, the energy storage in the primary is equal to W1W122LPIP joules. So, the input power PinLPIP 2T2Tnow, the efficiency is assumed to 80%, we have

(VI(min)1)22VI(min)11112Pin1.25PoLPIPLPTonLPTon 22T2TLP2TLP2That is LP2(VI(min)1)2Ton2.5PoT

Step 4： The primary wire turns VPVI(max)1NPAeVI(max)1Ton8B810  NP10 TonAeBStep 5: The secondary wire turns

2

VI(max)1Vo1NP NSStep 6：The primary rms current and wire size must be calculated. From the rms formula of the primary is IprmsTonI12i(t)dtP and the T3TIPTTonNp12rms formula of the secondary is Isrms i(t)dtTTNs3And wire size is specified by 300~500 circular miles per rms ampere

P300IPTon

3TS300IP0.8TTonNP

TNS32837. So, when the primary and secondary wire miles are larger fstep 5: considered the skin effect. skin depth Sthan skin depth, more wire numbers will be better a single larger wire size The parallel of primary wire numbers are nsfix(s) and the secondary wire s1numbers are ns

fix(s)。 s13

Flyback transformer design（CCM Mode） Illustration：

Step 1：The turns ration of primary and secondary

When Q is off state, the diode at secondary side is turn on and the mirror voltage will reflect to the primary side

NNVPPVS . The voltage VSVo1. That is VPP(Vo1).

NSNSSo, the transistor is sustained the voltage stress Vms

VmsVIVPVI(max)NP(Vo1) NSStep 2：

The core must be guaranteed not to saturated. The voltage-time product of “on-time” must be equal to “off-time”. That is

NPVi(minTV1Toff )onoNSand the circuit must be remained in continuous mode. That is, TonToffT So,

Vi(minT)onVo1NP(TTon). NSNPTNsIt leads to Ton

NPVI(max)(Vo1)NS(Vo1)step 3: Determine the inductance LP of primary

TIcprTonIcsrtoffFrom above fig. the output power is given

ToffTPoVoIcsrVoIcsr(1on) That is IcsrTTThe input power is given

4

PoTVo1onT

TPin1.25Po(VI(min)1)Icpron That is IcprT1.25PoT(VI(min)1)onT

Considered the boundary condition between the DCM mode and CCM mode, the following condition is satisfied

dIpIcpr

2The slop of the ramp dIP is given as

dIPVI(min)1TonLP

So, the inductance of the primary is

LP2(VI(min)1)2Ton2.5PinT

Step 4：The primary rms current and wire size must be calculated.

TFrom the rms formula of primary IprmsIcpronT1.25PoTon

TonT(VI(min)1)TAnd wire size is specified by 300~500 circular miles per rms ampere

P300Irmsp

From the rms formula of primary IrmssIcsrToffTPoToffTTVo1onT

S300Irmss

step 5: considered the skin effect. skin depth S2837. So, when the primary and secondary wire miles are larger fthan skin depth, more wire numbers will be better a single larger wire size The parallel of primary wire numbers are nsfix(s) and the secondary wire s1 5

s)。 numbers are nsfix(s1Step 6： The primary wire turns VPVI(max)(VImax1)Ton108B81NPAe10  Np

Tonand NVsNpo1V

I(max)1

BAe6