Electronic – What actually is the output current of a boost conveter

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}

This is a broad-brush explanation how to get to the inductance required in a discontinuous boost converter.

Try and think of things in terms of power (as per my answer to your linked question). Your output power is 180 volts x 30 mA = 5.4 watts so, if you transfer energy 100,000 times per second then the energy transfer in one cycle is 54 uJ.

Knowing that you need to store energy in the first half of the switching cycle and release it in the second half of the cycle you can use the inductor energy formula: -

W (energy) = \$\dfrac{LI^2}{2}\$ therefore I = \$\sqrt{\dfrac{2\times 54\times 10^{-6}}{L}}\$.

Also knowing that V = \$L\dfrac{di}{dt}\$ we can put numbers of di and dt.

• di is the change in current needed to charge energy into the coil (as per I from the energy equation above)
• dt can be half a switching cycle (5 us)
• V is the 12 volts input supply

This boils down to doing a bit of algebra to find L: -

L = \$\dfrac{(12 \times 5\times 10^{-6})^2}{2\times 54\times 10^{-6}}\$ = 33 uH.

I found the inductance to be around 400uH

You have to use the correct approach.

If you work out the current charged into and discharged from the inductor using my approximate approach the peak current in the inductor is 1.818 amps and this is also the peak to peak ripple current because of discontinuous operation.

Ripple current is what the inductor sees - it doesn't actually flow into the load because most of it is soaked-up in the output capacitor. The load will draw what current it needs from the 180 volts but the trick is keeping the output voltage stable because: -

A booster is a power regulator - it regulates power not voltage

To regulate voltage you have to have a control loop around the basic power regulator to keep the mark-space ratio correct so that voltage is regulated by controlling power.

Is it typical for ripple current to be larger (for an average load of 30mA) in a boost converter operating in CCM or DCM?

My simplified example above is for DCM and this will have a peak-to-peak ripple current that bears little relationship with load current.

In CCM, the inductor is always conducting and this means the peak-to-peak ripple current can be much smaller; the energy in the inductor isn't depleted to zero therefore the energy transfer per cycle is based around: -

W (energy) = \$\dfrac{L.I^2_{max}}{2}-\dfrac{L.I^2_{min}}{2}\$

In other words, if Imax is high then Imin need only be a little bit smaller to get the same energy per cycle (compared to DCM).

I decided to do a quick simulation to see how things panned out against my formulas: -

For 100 kHz switching at 50:50 duty I got a stable peak voltage of 186 volts with a peak inductor current of 1.9 amps with a load of 6 kohm. The output capacitor is only 330 nF just so that the output would charge up quicker in the sim. Inductor is as calculated - 33 uH.

Remember - this is a fixed load scenario - to make a booster with a regulated output you need an overall control system that tweaks the duty cycle as output load and input voltage varies. There is no such thing as a working simplified boost circuit with good voltage regulation.