How to design a forward switching power supply?

How to design a forward switching power supply, as we said before, the research and development first puts forward the requirements, and then designs according to the specific situation. Then we will use a simple 5.5V/20A as an example to illustrate the forward switch. Power supply design process and calculation of various parameters.

A simple drawing is shown below:

First of all, we need to clarify some technical indicators of the power supply: input voltage: Vin=220V; voltage variation range: 106~235V; input frequency: f=50Hz; output voltage: Uo=5.5V; full load current: Io=20A; output power Po =110W.

We can calculate the power supply period T=1/f=5uS from the working frequency; for forward switching power supply, our general duty cycle range is between 40% and 45%, then we choose 45% to calculate , the maximum on-time Ton(max)=5*0.45=2.25uS can be obtained.

Next, we need to calculate the secondary winding output voltage and turns ratio of the transformer. From the simple diagram above, we can see that when the switch tube is turned on, the secondary winding voltage U2 of the transformer is equal to the output voltage Uo plus the inductor consumption voltage UL. Adding the voltage UF consumed by diode VD1, we can simply write the minimum output voltage of the transformer's secondary winding as: U2(min)=(Uo+UL+UF)*T/Ton=(5.5+0.3+0.5)* 5/2.25=14V.

Since the forward transformer only plays the role of transmitting energy, we can think that the ratio of the primary winding winding N1 and the secondary winding N2 is equal to the ratio of the primary winding voltage U1 and the secondary winding U2 of the transformer, which can be written as N=U1/U2 ;According to our above output voltage range between 106V~235V, we can get U1=200~350V; so we can take U1(min)=200V, so N=200/14=14.3.

From N=U1(min)/U2(min) and U2(min)=(Uo+UL+UF)*T/Ton we can get N=U1(min)*D(max)/(Uo+UL+UF ); And because there is the following relationship between the number of primary winding turns N1 of the transformer and the magnetic flux density: N1>=(U1(min)*Ton(max)*10000)/(Bm*S); According to the output power and magnetic core Regarding the relationship between sizes, we choose EI-28 magnetic core; its effective cross-sectional area S=85 square millimeters; magnetic flux density Bm=3000 Gauss at 100°C, but because the power supply we designed is a forward type, it is It is a one-way excitation, so a margin for magnetic reset must be reserved, so we take Bm=2000 Gauss.

Substituting the data into the above formula, we can get N1=26.5. We take 27 turns and get N2=2 turns according to N2=N1/N.

Based on the calculated number of turns, we can get the accurate duty cycle D(max)=(Uo+UL+UF)*N/U1(min)=(5.5+0.5+0.3)*13.5/200=42.5%; That is to say, the primary winding of the transformer N1=27, the secondary winding N2=2; the maximum duty cycle of the power supply D(max)=0.425; the maximum conduction time of the switch tube Ton(max)=2.1uS; the secondary winding outputs the lowest voltage U2(min)=14.8V.

Next, we need to calculate the output inductor L1. The inductance L=[[U2(min)-(UF+Uo)]*Ton(max)]/IL; because the output current Io=20A; generally we choose the current flowing through the inductor IL is 10%~30% of the output current Io; we take the middle value 20%, then the current flowing through the inductor IL=0.2*20=4A; substituting 4A into the above formula, we can calculate the inductance L=(14.8-0.5- 5.5)*2.1/4=4.6uH.