Let me do this for you.

- First, lets define the duty cycle d = T(Q1-Q4-D1)/(T(Q1-Q4-D1) + T(Q1-Q2-Q3-Q4)

- Note that T(Q1-Q4-D1) = T(Q2-Q3-D2)

- Now, the relationship holds:

v + vloss1 =( Vg - vloss2) * duty /n

(vloss1, vloss2 being the loss voltages in the transistors, coils, and diodes)

-> you should set n so that v can be created from Vg with reasonable duty cycle. There may be limits imposed on the duty cycle by the driver circuit. I would try to aim for a duty cycle in the range of 50 % to 70 %.

- R is the load. That should be a given, depending on the purpose of the circuit. This can widely vary depending on the application.

- The inductor L determines the magnitude of the ripple current from the source Vg and into the load (R,C). The value of the ripple current is basically a compromise between:

(1). What the source Vg can deliver. (no information here.)

(2). The effect of the cost and losses in the bridge transistors and the transfomer. Higher ripple mean higher cost and losses in the bridge and the transformer.

(higher cost because transistors and transformer need to be able to handle higher current.)

(3). The cost of the inductor L. Lower ripple means larger inductor value, so more costly inductor and more losses in the inductor.

(4). The size and cost of the capacitor C. Large ripple means you need a larger capacitor to smooth it out.

Typically, the ripple can be anywhere between 20 % and 200 % of the average current. Its given by the equation:

Peak-peak ripple = Vg/(L * T(Q1-Q2-Q3-Q4))

-> define the ripple you want, then calculate L.

- The main factor in determining the capacitor is that it needs to sustain the load current while Q1-Q2-Q3-Q4 are conducting. So, in approximation we have:

peak-peak Vripple = T(Q1-Q2-Q3-Q4)*v/R /C

-> define the acceptable ripple voltage, then do some math, and you will get the capacitor value.

Note: You forgot to ask how to set the switching frequency. This is basically a trade-off between:

(1) Switching losses in the transistors, diodes and inductors

(2) Size of the inductors and capacitors .

(3) EMI considerations

- Higher switching frequency means higher losses, but smaller inductors and capacitors, and more EMI interference.

- This is rather difficult to give advice.  You need to look to datasheets of transistors, IGBT's, transformers and inductors to see what can be done. Typically, the transistors are determining factor, you have to look in datasheets what is acceptable these days. Its quite high when using mosfets.

Regards,

Henri.

You can refer to the papers (both available on the web) below for some guidelines:

1) "Design procedure of a push pull current-fed dc-dc converter”, D. Maiti, N. Mondal, S. K. Biswas, Proceedings of National Power Electronics Conference (NPEC-2010), Roorkee, India, June 2010, paper no.DCC003.

2) "Optimization of a bi-directional hybrid current-fed-voltage-fed converter link”, D. Maiti, N. Mondal and S. K. Biswas, Proceedings of the IEEE PEDES 2010, New Delhi, India, Dec 2010, Paper ID 297.

Thank you Mr. Biswas

Let me do this for you.

- First, lets define the duty cycle d = T(Q1-Q4-D1)/(T(Q1-Q4-D1) + T(Q1-Q2-Q3-Q4)

- Note that T(Q1-Q4-D1) = T(Q2-Q3-D2)

- Now, the relationship holds:

v + vloss1 =( Vg - vloss2) * duty /n

(vloss1, vloss2 being the loss voltages in the transistors, coils, and diodes)

-> you should set n so that v can be created from Vg with reasonable duty cycle. There may be limits imposed on the duty cycle by the driver circuit. I would try to aim for a duty cycle in the range of 50 % to 70 %.

- R is the load. That should be a given, depending on the purpose of the circuit. This can widely vary depending on the application.

- The inductor L determines the magnitude of the ripple current from the source Vg and into the load (R,C). The value of the ripple current is basically a compromise between:

(1). What the source Vg can deliver. (no information here.)

(2). The effect of the cost and losses in the bridge transistors and the transfomer. Higher ripple mean higher cost and losses in the bridge and the transformer.

(higher cost because transistors and transformer need to be able to handle higher current.)

(3). The cost of the inductor L. Lower ripple means larger inductor value, so more costly inductor and more losses in the inductor.

(4). The size and cost of the capacitor C. Large ripple means you need a larger capacitor to smooth it out.

Typically, the ripple can be anywhere between 20 % and 200 % of the average current. Its given by the equation:

Peak-peak ripple = Vg/(L * T(Q1-Q2-Q3-Q4))

-> define the ripple you want, then calculate L.

- The main factor in determining the capacitor is that it needs to sustain the load current while Q1-Q2-Q3-Q4 are conducting. So, in approximation we have:

peak-peak Vripple = T(Q1-Q2-Q3-Q4)*v/R /C

-> define the acceptable ripple voltage, then do some math, and you will get the capacitor value.

Note: You forgot to ask how to set the switching frequency. This is basically a trade-off between:

(1) Switching losses in the transistors, diodes and inductors

(2) Size of the inductors and capacitors .

(3) EMI considerations

- Higher switching frequency means higher losses, but smaller inductors and capacitors, and more EMI interference.

- This is rather difficult to give advice.  You need to look to datasheets of transistors, IGBT's, transformers and inductors to see what can be done. Typically, the transistors are determining factor, you have to look in datasheets what is acceptable these days. Its quite high when using mosfets.

Regards,

Henri.

I want to simulate with these parameters:

including Losses, vg within [18, 40]V, v=55V, Pload within [10, 150]W, fs=100kHz, Delta(i)=2%, Delta(V)=3%

Hello Omid,

Could you try to do the math, then I will correct it ?.

Good question.. Please share me the best answer might you trust...

Regards…