# Calculating the peak current of my boost converter

I am stuck while calculating the peak current in the boost converter through the inductor.

Can you please check if I am right?

Ipeak=(Max Voltage/ R2) = 18/1000=0.018. So, my peak current through the inductor is just 18mA? Am I doing it correct. If this is correct, how will I account for the Vce 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 transistor outside the IC instead of using the darlington switch in the IC for thermal purposes.

Can you please help me with the calculation of Vswce and Ipeak. The calculation of Vswce is not mentioned in the datasheet. So, please help. And the calculation of Peak current is mentioned in the datasheet. But is my approach of calculating the peak current through the inductor right? 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

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

Since a range of input voltage is given in your schematics, you will at least have to check for 8V and 18V 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.

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

• Thank you very much. Now, I clearly understand this calculation and working. Can you please help with Vswce? How do I calculate Vswce? It is not given in datasheet. – Freshman Jan 29 '19 at 4:39
• Why would you want to know Vswce? – Huisman Jan 29 '19 at 10:45
• In the datasheet of NCV3063, Fig.15 we need to calculate Vswce to find Ton/Toff. – Freshman Jan 29 '19 at 10:55
• It is given in the datasheet, page 5, especially Figure 9, but also in Figures 7,8 and 10. – Huisman Jan 29 '19 at 11:20
• Yes. But I am not able to understand. Please explain how to calculate Vswce like how you clearly explained Ipeak calculation. Please. Sorry for the trouble – Freshman Jan 29 '19 at 11:23