Dealing with the start-up current

The RRC3570 is a high-power battery. The total internal impedance is in the range of 100 mΩ only. With a voltage of 29.4 V, the battery can supply up to approx. 300 A as an inrush current.
Make sure your application can deal with that current.

 

Do I really need a start-up current limiter?

Generally, on start-up, all capacitors are empty. The inrush current is so high because it needs to fill these capacitors. The current is highest at the very beginning, when the capacitors represent a "short-circuit".
 

Easy start-up current limiting circuit

As an easy current-limiting circuit, you can use a resistor in series to your circuit. Put a switch (preferably a FET transistor) in parallel to that resistor, as shown in the picture below:

_{Figure 1 – Easy current limiting circuit using a switch an a resistor.}
 

Upon start-up, the FET shall be open, and all the current passes through the resistor. The resistor slows down the inrush current. After a specific time, or after reaching a defined voltage level, either an MCU or an analog circuit (slowed down by an RC filter) can close the FET, thus permitting full load current.

You can use an NFET in the GND line or a PFET in the (+) path. If you prefer using the GND path, pay special attention to the SGND line: In any case, the SGND connection should be established only after the startup phase has finished! You must ensure that the startup current won’t pass through the 10 Ω protection resistor in the SGND line (compare to Figure 1)  as it would not be able to survive the stress.
 

^{Figure 2 – Both SGND and PGND are internally connected to CGND. But because the supply current mainly passes through PGND, variations in supply current do not induce a voltage noise on the GND line. For lowest noise induction, keep SGND separate to PGND in your application as well and interconnect them in a kind of star connection, as well.}

When choosing the resistor, pay attention to its impulse load. Let's make an example:

Example on choosing a suitable resistor

In the LTSpice simulation below, we see a simple model of the battery on the left-hand side, the application on the right-hand side, and a switch in-between. The switch S1 and the resistor R1 serve only for simulation purposes – they are needed so that the capacitor C1 is empty at the beginning of the simulation.

The example application incorporates a big 2.2 mF input capacitor. We want to limit the inrush current with a series resistor of 100 Ω. Earlier, we spoke about a FET parallel to that resistor – in the simulation, we omit that FET for clarity.

 

_{Figure 3 - Inrush current simulation.}
 

As you can see in the simulation result, the capacitor needs approximately 500 ms until it's fully charged. The peak power dissipated across the resistor is 8.6 W, and the total energy loss is 938 mJ.

In resistor datasheets, you usually find the permitted single pulse power. To convert the above simulation result into a single pulse load, you

• Take the simulated peak power as pulse load
• Calculate the single pulse duration by t = E / P

In our case, the pulse load is 8.6 W, and the duration is 938 mJ / 8.6 W = 109 ms. Compared to the datasheet of an arbitrary manufacturer (refer to Figure 4), you see that a case size of 2010 would be necessary to ensure the power capability. You could still optimize the schematic by playing around with the resistance or looking for other manufacturers and case types.

_{Figure 4 – Datasheet from an arbitrary manufacturer. The calculated single pulse load requirement is drawn into the graph using black lines.}