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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!
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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)
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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)
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