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Level 2
Level 2
Distributor - WPG(GC)
` `

Could you know l me how much load is the Cdclink 1000uF for on the BTN 9970 Spec?

Design for output load Rating is continuous: 7A, peak(over current): 25A. How many capacitors should be used to meet such application conditions?Can you provide the calculation method? Thanks!

1 Solution

Moderator
Moderator

Hi @Brianwu

You can use the following formula for simple calculation:

C = I * Ton/ (U*5%) = 25*20/(12*5%) uF = 833.3uF

C:Cdc_link ; I = 25A; Ton: Mosfet on time = 20us; U = Vbat=12V; Voltage ripple：5%

(I shall be taken according to actual requirements)

Detail：

Vbat = 12V, MOSFET switch frequency = 20KHz, Max Ton = 20us(duty = 40%),Voltage ripple：5%,I= 25A;

We can get one cycle of MOSFET switch the bus provide power:

W = U*I*Ton

The energy released by the bus capacitor in one switching cycle is:

Q = C*((U+ΔU)^2-(U-ΔU)^2)/2 = 2*C*U*ΔU

ΔU = 1/2 Vripple = 1/2 * 5% *U = 2.5% U

In extreme conditions, the capacitor provides all energy:

Q = W => 2*C*U*ΔU = U*I*Ton

=> C = (I*Ton)/(2*ΔU) = (I*Ton)/(5% *U)

Generally, we think that capacitors provide half of the conduction energy, This gives the minimum capacitance：

Q = 1/2*W => 2*C*U*ΔU = 1/2*U*I*Ton

Cmin = 1/2*(I*Ton)/(5% *U)

3 Replies

Moderator
Moderator

Hi @Brianwu

You can use the following formula for simple calculation:

C = I * Ton/ (U*5%) = 25*20/(12*5%) uF = 833.3uF

C:Cdc_link ; I = 25A; Ton: Mosfet on time = 20us; U = Vbat=12V; Voltage ripple：5%

(I shall be taken according to actual requirements)

Detail：

Vbat = 12V, MOSFET switch frequency = 20KHz, Max Ton = 20us(duty = 40%),Voltage ripple：5%,I= 25A;

We can get one cycle of MOSFET switch the bus provide power:

W = U*I*Ton

The energy released by the bus capacitor in one switching cycle is:

Q = C*((U+ΔU)^2-(U-ΔU)^2)/2 = 2*C*U*ΔU

ΔU = 1/2 Vripple = 1/2 * 5% *U = 2.5% U

In extreme conditions, the capacitor provides all energy:

Q = W => 2*C*U*ΔU = U*I*Ton

=> C = (I*Ton)/(2*ΔU) = (I*Ton)/(5% *U)

Generally, we think that capacitors provide half of the conduction energy, This gives the minimum capacitance：

Q = 1/2*W => 2*C*U*ΔU = 1/2*U*I*Ton

Cmin = 1/2*(I*Ton)/(5% *U)

Level 2
Level 2
Distributor - WPG(GC)

Thanks @LinGuohui, Could i find it  from infineon website ?