Calculating the peak current of my boost converter

Question

I am stuck with calculating the peak current through the inductor in a boost converter. Can you please check if I am right?

I_peak = (V_max / R2) = 18 / 1000 = 0.018 A. So, my peak current through the inductor is just 18 mA? Am I doing it correctly?

If this is correct, how will I account for the V_ce drop inside the pin1 and pin2 of the IC?

The Darlington transistor inside the IC is just like a normal transistor. I have added two transistors outside the IC instead of using the Darlington switch in the IC for thermal purposes.

Can you please help me with the calculation of V_swce and I_peak? The calculation of V_swce is not mentioned in the datasheet.

The calculation of I_peak is mentioned in the datasheet, but is my approach of calculating the peak current through the inductor right?

Answer 1

From Figure 4. Block Diagram in the datasheet, you can see pin 6 is compared with pin 7 of the IC.
Note that the + input of the comparator lies 200mV lower than Vcc.
When pin 7 becomes lower than (Vcc-200mV) the peak detection will be activated (and so, the Darlington transistor will be disabled).

From your posted schematic, the (peak) current through the inductor causes a voltage drop across R1.
This voltage drop is divided by resistors R2 and R3.
So, the voltage on pin 7 will be:
\begin{equation}(V_{cc} - I * R_1 ) * R_3 / (R_2+R_3)\end{equation}

The peak detection will be activated when \begin{equation} V_{cc} * R_3/(R_2+R_3) - I * R_1 *R_3/(R_2+R_3) < V_{cc}-200mV \end{equation} Note that this depends on the current through the inductor as well as the applied voltage Vcc.

R3 is unclear to me, but 0M18 = 180K makes sense, and pick Vcc = 15V \begin{eqnarray*} 15V * 180/181 - I * 0.082E * 180/181 &<& 14.8V\\ 14.917V - I * 0.082E * 180/181 &<& 14.8V\\ 117mV &<& I * 0.082E * 180/181\\ 118mV &<& I * 0.082E\\ 118mV / 0.082E &<& I\\ I &>& 1.44A \\ \end{eqnarray*} So, max peak cuurent through inductor is 1.44 A

When e.g. Vcc = 10V, the same circuit limits the current to 1.78 A

Answer 2

In a boost converter the inductor peak current (relevant for saturation) is inductor average current plus maximum ripple. Your modulation index is

\$d = 1 - \frac{V_{IN}}{V_{OUT}}\$

The average inductor current, depending on your specific load \$I_{OUT}\$, is

\$I_{L,avg} = \frac{I_{OUT}}{1-d}\$

The ripple current is (depending on your switching frequency \$f\$)

\$\Delta I_{L,peak-peak} = \frac{V_{IN}}{L} \cdot \frac{d}{f}\$

Finally we get for continuous conduction mode

\$I_{L,peak} = I_{L,avg} + \frac{1}{2} \Delta I_{L,peak-peak}\$

while for discontinuous conduction mode the maximum current is simply

\$I_{L,peak} = \Delta I_{L,peak-peak}\$

Since a range of input voltage is given in your schematics, you will at least have to check for 8 V and 18 V which gives the higher inductor peak current.

Your maximum input current is specified as 0.6 A, so the maximum inductor peak current will be higher than this.