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, equation waveform

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 _________. diagram

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    \$\begingroup\$ Hint: the average current is just the size of the area under the curve trace. \$\endgroup\$ – Bimpelrekkie Dec 14 '16 at 10:42
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    \$\begingroup\$ + averaged over the whole cycle \$\endgroup\$ – RoyC Dec 14 '16 at 10:44
  • \$\begingroup\$ The op is talking about RMS and not average dudes! \$\endgroup\$ – Andy aka Dec 14 '16 at 11:10

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}

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  • \$\begingroup\$ Time period of function is T. So the first equation must be divided with T instead of Tp. \$\endgroup\$ – Vedanshu Dec 15 '16 at 7:32
  • \$\begingroup\$ Division by Tp is the mean part of RMS and I defined Tp as the period. \$\endgroup\$ – RoyC Dec 15 '16 at 8:57
  • \$\begingroup\$ Top equation is for the RMS for the ramp part of the waveform and I do not think it is the same as Andy aka's answer. Use bottom equation to average over a longer time period. \$\endgroup\$ – RoyC Dec 15 '16 at 9:37

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

enter image description here

\$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.

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  • \$\begingroup\$ This method is valid. Thanks a lot. Could you give a hint on how to solve the question? \$\endgroup\$ – Vedanshu Dec 15 '16 at 7:36
  • \$\begingroup\$ Which part of my answer relating to finding the rms current didn't you follow? \$\endgroup\$ – Andy aka Dec 15 '16 at 10:21
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    \$\begingroup\$ RMS = √(RECTANGLE^2+SAWTOOTH^2) is only true for uncorrelated waveforms or a few other special cases \$\endgroup\$ – RoyC Dec 15 '16 at 10:38
  • \$\begingroup\$ @oceanp yes I realized that a few minutes ago and I'm going to fix it or dump it. Darn, I'd hoped to get away without anyone noticing LOL. \$\endgroup\$ – Andy aka Dec 15 '16 at 10:39
  • \$\begingroup\$ @oceanp I think it's fixed now!! \$\endgroup\$ – Andy aka Dec 15 '16 at 10:50

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.

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  • \$\begingroup\$ You are quite correct. Both Andy and I missed the point of the question. Upvoted \$\endgroup\$ – RoyC Dec 16 '16 at 17:05

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