# Flyback Converter, its working, Waveform, and Flyback converters designs

Table of Contents

**Flyback Con verter:**

A flyback converter is an isolated type of DC-DC converter. It is derived from the boost converter. It is basically dc to dc converter that increase or decrease the voltage compare to the input terminals. The speciality of this converter is that it consists of transformer and there is electric isolation which is provided by the transformer. DC voltage is supplied to the input side of the converter through rectified filter. The reason it is called flyback converter because there is no primary current flowing secondary voltage will be induced and consequently power will be delivered. It consists of an active switch, a transformer, a freewheeling diode and a filter capacitor due to these very fewer amounts of circuitry it is cost effective and efficient. The drain of the MOSFET is connected to one side of the primary side of the transformer and another side is connected to the input capacitor. Here the secondary side of the transformer is connected to anode of the diode and cathode of diode is connected to output capacitor. Unlike other converters fly back converter does not have output inductance filter. So the design and effective cost of the converter decreases. Due to all these qualities, it gives impressive efficiency. This is the basic circuit diagram of the fly back converter.

**Working of the fly back converter:**

We will see the working of this converter, for a better understanding of the converter we will split its operation in 2 different parts i.e. when MOSFET is turned on and when it is turned off.

**Case 1:**

In the fly back converter the winding of the transformer will act as inductor. When the MOSFET is turned on at that time this primary side of the transformer will act like an inductor and this will start storing energy coming from the input. We will triggering the gate pulses. Hence at this time, there will not be any voltage at the output side, this is known as an energy storing phase in the flyback converter. When the current will start flowing in the primary winding it will induce voltage in the secondary winding. Assuming the dot convention when the current is entering the dot convention the voltage induced at the secondary will have opposite polarities. In primary side we have positive sign upward and negative side downward while at the secondary side the negative sign will be upward and positive sign will be downward. Now as the diode will get negative voltage so it will be in reverse bias and it will act as open circuit and there will be no current flowing to the load.

**Case 2:**

Now the transistor is turned off. This time the polarity of the inductor i.e. primary side of the transformer will get reversed and voltage will get induced in the secondary side of the transformer due to electromagnetic induction and diode will get forward biased and magnitude of this induced voltage is given by turns ratio

i.e. Vin/Vo = Np/Ns

The switch will be as open circuit. This is known as the flyback phase of the converter, here primary side is fully off and voltage will be delivered to the output capacitor and it will get charged. It is used for mostly in lower power applications when the power required is less.

In next energy storing phase output capacitor delivers the voltage to the load. These cycles get repeated over a period of time and we get regulated DC output voltage. This is how the flyback converter works.

**Waveform of the flyback converter:**

The waveform associated with it we assume the gate pulses to be have equal width pulse t_{on} and t_{off} .The output voltage will be look like a constant dc voltage but practically we have some ripples which will be increase and decrease due to charging and discharging nature of the capacitor. We can reduce the ripples by using the suitable capacitor. So usually we use large value of the capacitor in order to obtain constant dc voltage.

The primary current will look like first the current will increase when the switch is on and when the switch is turn off the current will start decreasing and will be zero because it will be open circuit and this process will be repeat.

The diode and the secondary current will be quite same initially when the switch is on the diode current will zero because the diode will be in reverse biased and when the switch is off the current will start increasing and this process will repeat with respect to the cycle.

## Flyback converter design:

Now considering some necessary design parameters, Let’s take an example for designing a flyback converter, We will give input voltag i.e. V_{in}=150 Volt DC and Output voltage i.e. Vo which is expected from the converter be 55 Volt DC Load current i.e. (IL) should be 2 Amps Switching frequency i.e. “f_{swt}” will be 30 KHz and it is decided by the designer it can vary in a scale of KHz to MHz as we increase the frequency the size of the converter will decrease, how is that? We will see in the formulae designing:

The Output ripple we will permit in this design as 100 mv peak to peak. The temperature rise in the converter is expected here from 0 to 55℃ and the efficiency of the converter i.e. n= 85 %. Now we will start designing the converter:

1st step is the selection of core and calculation the number of turns. The equation for number of turns is dependent on operating frequency, flux density i.e. delta B and effective area of the core. Initially, we select a core and peak flux density for that core. Then the number of turns and wire size is calculated and it is compared with an effective area of the core and if it is insufficient then the largely sized core is selected and calculations are repeated again.

Till this condition fulfils 1st we need to calculate the output power, which is equal: P_{out}= V_{out} × I_{L×Ƞ}

i.e. equal to 130 watts. If we see this chart, we can select the core of the transformer. For this type of transformer, we can select the E ferrite core. Because it can withstand high currents and number of turns to inductance ratio is approximately linear. While making a transformer we use 2 E cores of the same size and connect them, with this fashion.

There comes a bobbin in between which provides insulation between core and wire. According to this chart, we may select the EE4215 core but to fulfil our design we used EE4012 core which is smaller in size.

We are on the 1st step that is selecting a core and calculating number of turns. We have already selected the core. Now we will calculate number of turns which are to be wound around the core, by calculating primary and secondary number of turns. For calculating primary number of turns, the formula is shown:

N_{mpp}= (V_{p} t_{on})/(∆B A_{cp} )

Where Vp is maximum input voltage which is voltage coming to the primary side of the transformer. For primary side, we have to consider switching losses and primary resistance losses. Let’s consider them as 10% of input voltage. Hence we get Vp as 135 volts.

A_{cp} is effective area of the core which is given by the manufacturer in the cosmo ferrites catalog for E4012 core it is 118mm square. T_{on} is the time in which transistor remains switched on that is also decided by the designer here we will put it as 50% of total time. ∆B is change in flux density, this value varies from 0 to 400 mT and the E core that we have selected has saturation flux density as 360mT as shown in B/H curve.

But one thing that is supposed to be kept in mind that we cannot select this maximum saturation flux density for our core. If we do so the unnecessary problems will arise. Hence we will select the flux density at 220mT to provide good working margin of the core. Hence considering all these values the primary number turns will be 68. We can find out the secondary number of turns by this formula. For the secondary turns, we need to calculate the secondary voltage

V_{S} which is equal to V_{out} + V_{losses}.

These losses will be of rectifier diodes, choke wiring and transformer resistance, let’s consider these losses as 5 V. Hence the secondary turns Ns will be 30. Generally for 2.5 Amps of current 1 mm square cross sectional area of the wire is sufficient. This is not the standard assumption but if we increase this current value per mm square then more heat will dissipate and heating losses will increase. According to our design 2 amps is the load current, the wire having 1mm square area will be sufficient for us. For more details you can see the American wire gauge chart for reference. This wire will have the diameter of 1.12mm. We have calculated the number of turns but there is one more doubt whether these turns will fit into core or not. To find it out we need to compare the required area for the wire to be wound which is:

Cs area = no of turns x cs area of the wire with (effective area of the core – bobbin margin (considering it as 10% of Ae)) if this value is less than the supposed value then we can assume that wire will fit into bobbin. Here if we solve this equation then we will find out that the required wire will fit into this E4012 core and we don’t need bigger core than that if we would have selected E4215 core in that case it would have been wastage of space. After that we have to find out the length of wire to be needed for winding. It is dependent on the bobbin size for one turn x number of turns. After finding all of these values we can wind the wire around core.

Now we will calculate the output capacitor such that it will meet the maximum output ripple voltage specification. The formula for selection of capacitor is shown where Toff is off time of transistor.

C= t_{off}I_{dc}/V_{p-p}

Idc is load current and Vp-p is peak to peak ripple voltage. Hence the output capacitor value is 333uF, select the standard available capacitor value and it should be greater than this. I.e. 390uF 200 volts (operating voltage should be greater than maximum voltage present in the circuit here it is 150 volt). Now we will select the MOSFET suitable for our design. Typically the flyback voltage arise at transistor is twice of the input voltage due to inductors, we will also consider 25% margin for inductive overshoot.

Hence we will select MOSFET with Vdss greater than 340+ voltage and Ids should be also greater than twice of load current. Next we will select the rectifier diode; they are subjected to high large peak and RMS current stress. So the current handling capacity of the diode should be greater. So like that we can select all the required components for our design.

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