# Mistake in calculation of RMS current flyback primary switch node

I'm trying to calculate the RMS current of a flyback converters primary switch node. The Waveform is shown below:

I'm okay with assuming it is a triangle wave with a duty cycle of 50%. The peak value is 1.25A and the Period time is 4u seconds.

I can calculate it using the following formula I found on the internet:

And I will find a RMS current of 0.51A what is right.

But my problem is when I try to calculate it with another method i believe is right I get a different answer. I hope someone can point out me where I'm wrong. See my calculation below:

$$I_{RMS} = \sqrt{\frac{1}{T}\cdot\int_{0}^{T}I^2(t)dt}$$

I then simplify the integral to an easy area calculation

$$\int_{0}^{T}I^2(t)dt = (0.5 \cdot 0.5T \cdot I_{PK})^2$$

I work this all out to the following:

$$I_{RMS} = \sqrt{\frac{1}{T}(0.5 \cdot 0.5T \cdot I_{PK})^2}$$

When i fill everything in:

$$I_{RMS} = \sqrt{\frac{1}{4\mu}(0.5 \cdot 0.5 \cdot 2\mu \cdot 1.25)^2}$$ $$I_{RMS} = 3.125\cdot10^{-4} A$$

I end up wrong. I think my problem is with the simplification. But it doesn't make sense to me why it should be wrong.

$$(A) \equiv \sqrt{(s^{-1}) \left[ (A) (s) \right]^2} = \sqrt{ A^2 s} = A \sqrt{ s}$$
The curve you need to integrate is a quadratic, from 0 to DT, and the remainder (from DT to T) is zero. The antiderivative of the square is 1/3 of the cube, which lies is under the radical, hence where the $$\\sqrt{3}\$$ fraction comes from.