The circuit configuration for flyback converter is shown in Fig. 1. It consists of a power MOSFET $M1$, transformer for isolation purposes, diode D capacitor C and load. An uncontrolled rectifier converts a.c to d.c output which is fed to flyback SMPS as shown in Fig.1.

When power MOSFET is turned on, supply voltage $V_{s}$ is applied to the transformer primary, i.e. $v_{1}=V_{s}$ . A corresponding voltage $v_{2}$ , with the polarity as shown in Fig.2(a), is induced in transformer secondary, i.e. $v_{2}=\frac{V_{s}}{N_{1}} N_{2} .$ As $v_{2}$ reverse biases diode D equivalent circuit of Fig.2(a) is obtained. Filter capacitance C is assumed large enough so that capacitor voltage $v_{c}(t)=$ load or output voltage $V_{0}$ is taken as almost constant. When $M 1$ is turned off, a voltage of opposite polarity is induced in primary and secondary winding as shown in Fig.2(b) Voltage across transformer secondary is $v_{2}=-V_0=-\frac{V_{s}}{N_{1}} N_{2}$ . Diode D is forward biased and starts conducting a current $i_{D} .$ As a result, energy stored in the transformer core is delivered partly to load and partly to charge the capacitor C .

Waveforms for $v_{1}, v_{2},$ transformer magnetizing current $i_m$ and diode current $i_{D}$ are shown in Fig.3. During the time $M 1$ is on, $v_{1}=V_{s}, v_{2}=\frac{V_{s}}{N_{1}}, N_{2}$ . For magnetizing current, it is assumed that transformer core is not demagnetized completely at the end of periodic time $T=T_{o n}+T_{o f f}$. In other words, it means that transformer magnetizing current at $t=0$ is not zero but has some positive value $I_{m o} .$ Therefore, during $T_{o n}$ , magnetizing current rises linearly from its initial value $I_{m o}$ to $I_{m 1}$ at $t=T_{o n} .$ With the rise of $i_{m}$ during $T_{o n},$ magnetic energy gets stored in the transformer core. The variation of $i_{m}$ as shown in Fig.3 can be expressed as under:

$$i_{m}(t)=I_{m o}+\frac{V_{s}}{L} t \ldots \ldots 0 \lt t \lt T_{o n}-----(1)$$

where $L=$ transformer magnetizing inductance, H

At $$t=T_{o n}, \quad i_{m}\left(T_{o n}\right)=I_{m 1}=I_{m o}+\frac{V_{s}}{L} \cdot T_{o n}-----(2)$$

When $\mathrm{M} 1$ is turned off, the emfs induced in primary and secondary windings are reversed as shown in Fig.2(b). Diode D is now forward biased. A current in transformer secondary winding begins to flow through D . As this current $i_{D}$ or magnetizing current $i_{m}$ reduces from $I_{m 1}$ to $I_{m o}$ at $t=T,$ transformer core energy is delivered to load.

During $T_{\text { off }} M 1$ is off and $v_{2}=-V_{0}$ . This voltage when referred to primary is $v_{1}=-\frac{V_{0}}{N_{2}} N_{1}$. The fall of current $i_{m}$ during $T_{o f f}$ can be expressed as under.

$$i_{m}(t)=I_{m 1}-\frac{V_{0}}{N_{2}} N_{1} \cdot \frac{1}{L}\left(t-T_{o n}\right) \quad \ldots . T_{o n}\lt t \lt T-----(3)$$

At $t=T$

$$i_{m}(T)=I_{m 1}-\frac{V_{0}}{N_{2}} \cdot N_{1} \cdot \frac{1}{L}\left(T-T_{o n}\right)$$

Substituting the value of $I_{m 1}$ from Eq.2 in the above expression, we get

$$i_{m}(T)=I_{m o}+\frac{V_{s}}{L} \cdot T_{o n}-\frac{V_{0}}{N_{2}} \cdot N_{1} \cdot \frac{1}{L}\left(T-T_{o n}\right)-----(4)$$

Since the net energy stored in core over periodic time $T$ is zero, $$i_{m}(O)=i_{m}(T)$$

$$I_{m o}=I_{m o}+\frac{V_{s}}{L} \cdot T_{o n}-\frac{V_{0}}{N_{2}} \cdot N_{1} \cdot \frac{1}{L}\left(T-T_{o n}\right)$$

$$V_{s} \cdot T_{o n}=\frac{V_{0}}{a}\left(T-T_{o n}\right)$$

$\therefore$ Load voltage,

$$\quad V_{0}=\frac{a \cdot V_{s} \cdot T_{o n}}{T-T_{o n}}=\frac{a \cdot V_{s} \cdot k}{1-k}-----(5)$$

where $a=\frac{N_{2}}{N_{1}}$ , transformer turns ratio from secondary to primary

and $\quad k=\frac{T_{o n}}{T},$ duty cycle of flyback converter.

It is seen from Fig.2(b) that open circuit voltage across $M 1$ is $$V_{oc}=v_{1}+V_{s}=\frac{V_{0}}{N_{2}} \cdot N_{1}+V_{s}=\frac{V_{0}}{a}+V_{s}$$

From Eq.5 $$\quad V_{oc}=\frac{V s \cdot k}{1-k}+V_{s}=\frac{V_{s}}{1-k}-----(6)$$

Eq.3 gives current on primary side of the transformer. This current, when referred to secondary side, is equal to diode current $i_{D}$ .

$$\begin{aligned} i_{D}(t) &=i_{m}(t) \cdot \frac{N_{1}}{N_{2}} \\ &=\frac{N_{1}}{N_{2}}\left[I_{m 1}-\frac{V_{0}}{N_{2}} \cdot N_{1} \cdot \frac{1}{L}\left(t-T_{o n}\right)\right] \\ &=\frac{I_{m 1}}{a}-\frac{V_{0}}{a^{2} \cdot L}\left(t-T_{o n}\right)-----(7) \end{aligned}$$

Flyback converter offers simple SMPS and is useful for applications below about 500 $\mathrm{W}$