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Hi, I have a question about DC-Link Capacitor for PMSM, How to calculate the volume of the DC-Link Capacitor for PMSM? For example the DC-Link Voltage is 85V， the MAX current of PMSM is 300A, the delta Voltage is 2.5%*85V，the switching frequency is 10KHZ.

Solved! Go to Solution.

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Hi, @wanglipeng00000 ,

具体情况具体分析，这边给您举个例子，您可以参考一下：

直流310V电源供电，系统峰值电流20A，开关频率20KHz，电压纹波要求小于10%。功率器件开通时间最大30us（60%占空比）。

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 * 10% *U = 5% U

In extreme conditions, the capacitor provides all energy:

Q = W => 2*C*U*ΔU = U*I*Ton

=> Cmax = (Ipeak*Ton)/(2*ΔU) = (Ipeak*Ton)/(10% *U) = （20 *30）/（10% * 310）uF ≈ 20uF

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*(Ipeak*Ton)/(10% *U) ≈ 10uF

Hope this can help you.

BR,

Owen

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Hi,

To calculate the volume of the DC-Link capacitor for a PMSM, we need to consider several parameters such as the DC-Link voltage, maximum current of the PMSM, delta Voltage, and switching frequency. The DC-Link Capacitor is usually used to smooth the rectified output of the main bridge and reduce the ripple current flowing through the inverter.

And you are calculating the **volume(容量)** of capacitor, right?

BR,

Owen

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Yes, I am calculating the volume of capacitor, I don't know how to calculate,

For example If the DC-Link Voltage is 85V， the MAX current of PMSM is 300A, the delta Voltage is 2.5%*85V，the switching frequency is 10KHZ. How to calculate the volume of capacitor?

Thanks for your help.

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Hi, @wanglipeng00000 ,

具体情况具体分析，这边给您举个例子，您可以参考一下：

直流310V电源供电，系统峰值电流20A，开关频率20KHz，电压纹波要求小于10%。功率器件开通时间最大30us（60%占空比）。

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 * 10% *U = 5% U

In extreme conditions, the capacitor provides all energy:

Q = W => 2*C*U*ΔU = U*I*Ton

=> Cmax = (Ipeak*Ton)/(2*ΔU) = (Ipeak*Ton)/(10% *U) = （20 *30）/（10% * 310）uF ≈ 20uF

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*(Ipeak*Ton)/(10% *U) ≈ 10uF

Hope this can help you.

BR,

Owen

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Thanks very much, @Owen_Su , I exactly know how to calculate, but I still have a question,

功率器件开通时间最大30us（60%占空比）

I don't know whether the duty cycle is max 60% in control the PMSM, or can be bigger than 60%?

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Hi，

It's possible to have a duty cycle higher than 60% in control the PMSM, but this might lead to issues such as increased motor heating and reduced efficiency. It is recommended to consult the motor and control system specifications to determine the appropriate maximum duty cycle for your specific application. Hope this can help you.

BR,

Owen

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Thanks very much, @Owen_Su , I now know the reason of duty cycle below 60%, but I still don't know whether this means the input power is only output 60%？

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Hi,

If the duty cycle is below 60%, it usually means that the output is active or on for less than 60% of the total time of the signal. In other words, the output is off for more than 40% of the time. This could be due to various reasons, such as the design of the system, the load being powered, or other factors.

However, the duty cycle alone does not provide information about the efficiency of the system. The efficiency of a system refers to the ratio of the output power to the input power, and it is often expressed as a percentage. Therefore, a duty cycle below 60% does not necessarily mean that the input power is only output at 60%.

I think that you can go through this application note and you will learn more about the motor control. Hope this can help you.

BR,

Owen