RMS switch current in boost converter

Question

I'm struck in calculating the I(switch,rms) in boost converter. Average current through diode = average o/p current , also average current through the inductor = average current through the source. From this I drew the following waveform. So,

EDIT This was the part of the following question.

A DC-DC boost converter, as shown in the figure below, is used to boost 360V to 400 V, at a power of 4 kW. All devices are ideal. Considering continuous inductor current, the rms current in the solid state switch (S), in ampere, is _________.

Answer 1

From first principles for a waveform repeating over a period Tp being a current rising from Ii to Ii+Id.

\begin{equation} I_{RMS}=\sqrt{\frac{1}{Tp}\int_0^{Tp} (I_i+\frac{I_d.t}{Tp})².dt} \end{equation}

\begin{equation} =\sqrt{\frac{1}{Tp}\int_0^{Tp} (I_i^2+\frac{2.I_i.I_d.t}{Tp}+\frac{I_d².t^2}{Tp²}).dt} \end{equation}

\begin{equation} =\sqrt {\frac{1}{Tp}(I_i^2.t+\frac{I_i.I_d.t^2}{Tp}+\frac{I_d².t^3}{3Tp²})_{t=0}^{t=Tp}} \end{equation}

\begin{equation} =\sqrt{ \frac{1}{Tp}(I_i^2.Tp+\frac{I_i.I_d.Tp^2}{Tp}+\frac{I_d².Tp^3}{3Tp²})} \end{equation}

\begin{equation} I_{RMS}=\sqrt {I_i^2+I_i.I_d+\frac{I_d²}{3}} \end{equation}

If instead you consider this to be a ramp rising from i1 to i2 substitute i1 for Ii and i2-i1 for Id and you get

\begin{equation} I_{RMS}=\sqrt {i1^2+i1.(i2-i1)+\frac{(i2-i1)^2}{3}} \end{equation}

\begin{equation} =\sqrt {i1.i2+\frac{(i2^2-2i1.i2+i1^2)}{3}} \end{equation} \begin{equation} =\sqrt {\frac{(i2^2+i1.i2+i1^2)}{3}} \end{equation}

This is independent of time and i1 and i2 are interchangable so where you have a ramp going from i1 to i2 and then back to i1 in one period this is your result and it is independent of duty cycle.

Now you need to average the result from above out over time.

To average two different RMS currents over a longer period (clue one of these can be zero). Say we have Irms1 for t1 and Irms2 for t2.

\begin{equation} I_{RMS}=\sqrt{\frac{I_{RMS1}^2.t_1+I_{RMS2}^2.t_2}{t1+t2}} \end{equation}

So where you have a ramp going from i1 to i2 for t1 and no current for t2 we get.

\begin{equation} I_{RMS}=\sqrt {\frac{t1}{t1+t2}.\frac{(i2^2+i1.i2+i1^2)}{3}} \end{equation}

Answer 2

I'm struck in calculating the I(switch,rms) in boost converter

\$I_{rms}= \sqrt{d_1\cdot\frac{1}{3}\cdot (I_1^2+I_1\cdot I_2+I_2^2)+ (1-d_1)\cdot I_0}\$

T is assumed to be 1 and hence \$d_1\$ becomes the duty cycle fraction.

For your example, the \$I_0\$ term can be ignored hence the formula reduces to: -

\$I_{rms}= \sqrt{d_1\cdot\frac{1}{3}\cdot (I_1^2+I_1\cdot I_2+I_2^2)}\$

Note - I did get in a pickle with this on previous edits and thanks to @oceanp for provoking me to rethink. Pictures and formula taken from here.

Answer 3

The answer is 3.51 amperes RMS.

The problem doesn't seem to be asking for ramp effects, since there's no mention of allowable output ripple or switching frequency. So, it's safe to assume that the switching frequency is so high that the current through the inductor can be considered constant.

The duty cycle, the part of the time that the switch is on, is 10%. When it's on, the voltage across the inductor is 360. When it's off the voltage is -40 (i.e. 360-400). So it follows that the off time must be 9 times the on time.

From the duty cycle it follows that 90% of the inductor current goes to the load. The load current is 10 amps. (4kW at 400 V). The inductor current is therefore 10/.9 amps.

During the 10% of the time the switch is on the current through it is 10/.9 amps. The RMS current is the square root of the inductor current squared times the duty cycle.