LP5907 - About an acceptable inrush current

None of this falls inside the ensured operating range,.

Best that can be done here is a rough estimate, or best guess, so I will make a guess at the 250mA load behavior being resistive, and not a 250mA step function at some output voltage threshold.

Rload = 4.5V/250mA = 18 ohms

Peak current is determined by the LP5907 current limit circuit. The typical current limit value (Isc) is claimed to be 500mA. Current limit is NOT precision, it is for functional circuit protection. This can, and will, vary with junction temperature, Vin, and Vin-Vout.

To charge a 200 uF cap, in parallel with an 18 ohm load, from 0V to 4.5V at a 500mA rate will take ~2.5 milli-sec.

So, the best guess for 'Pulse Width' is ~2.5 milli-sec, and the best guess for 'Peak Current' is ~500 milli-Amp.

Of course, if the load behavior is different that what I guessed at, this is all null and void.

Also, if you are using ceramic capacitors for the 200uF, the actual capacitance may be dropping, perhaps rapidly, as the applied voltage is increasing. This would cause the pulse width duration to be shorter.

Plus at the very start, when Vcap=Vout=0V, the dissipation is at the maximum (2.25W) so Tj is changing fast at the beginning, while Vin-Vout is going down as the 200uF cap charges up to 4.5V.

Hi Donald-san,

I greatly appreciate your cooperation!

And, I have additional question.

I guess that the internal MOSFET is broken if this inrush current is very large.

Because, this MOSFET has the absolute maximum of the pulsed drain current.

Is my understanding correct?

If yes, could you please let us know this value (peak current and plus width).

They knew the large inrush current(around 1A) by checking the Figure5.

Therefore, They are thinking that this is an extremely important point.

I’d greatly appreciate your verification.

Regards,

Kanemaru

At this point my very best suggestion is for your customer to build up a circuit with the intended load and Cout.

They can then measure the actual in-rush current, and the duration, for their specific application.

Hi Donald-san,

Thank you for your response.

We understood your advice.

However, they have to use the large Cout(around 200uF) to decrease the noise of the output.

Therefore, they would like to know the condition that this device is broken by the in-rush current.

If the customers measured the actual in-rush current and the duration, will you confirm whether this device may be broken?

If you contact to below my e-mail address or let me know your e-mail address,

We can send the customer information.

I’m looking forward to hearing from you.

[my e-mail address]

Kanemaru-h@clv.macnica.co.jp

Regards,

Kanemaru

For this we will need to know exactly how the LP5907 will be started, and the exact the Cin and Cout values.

       Cin= 47uF, and Cout= 200uF ?

Will the Enable pin be tied to directly to Vin, or will the Enable pin be held low (off) for some period of time, then taken high (on)?

Note that if the Enable pin is tied to Vin then the in-rush current will include charging the input capacitor and possibly some partial charge of the output capacitor before the Isc circuitry becomes active.

The current needed to charge the Cin capacitor is directly related to the rise time and value of Vin.

       I = (C * (dV / dt))

      where : C= Cin; dV= input supply voltage; and dt= input supply rise time

To charge a 1.0 uF capacitor from 0V to 3.7V in 5.us requires 740mA.

If the Enable pin is tied directly to Vin, what is the input supply rise time?

Note that if the input supply rise time is slow enough, there will not be any significant in-rush current.

Also note that if there is a chip inductor at the LP5907 input to improve PSRR, the L, DCR and Cin values will certainly reduce, possibly eliminate, the in-rush current.