LMR33630-Q1: PG pin current limit

LIN SUN

Part Number: LMR33630-Q1
Other Parts Discussed in Thread: LMR33630

Dear expert:

Can you help me understanding the follow paragraph of Chapter8.3.1 in DS, please?

thanks a lot!

Limit the current into the powergood flag pin to less than 5 mA D.C. The maximum current is internally limited to about 35 mA when the device
is enabled and, about 65 mA when the device is disabled. The internal current limit protects the device from any
transient currents that may occur when discharging a filter capacitor connected to this output.

over 5 years ago

0 Ankit Gupta95 over 5 years ago

TI__Expert 4555 points

Hello Lin Sun,

Can you please specify what exactly you didn't understand here.

Best Regards,

Ankit Gupta

0 Andy Chen over 5 years ago

TI__Mastermind 34875 points

Hi Lin,

The maximum open drain FET RDS(on) of the Power Good flag pin is 150ohm, PG pin will show 0.75V with 5mA pull up current, this maybe the maximum level for logic low accepted by downstream controller chip, so we suggest to use 100K pull-up resistor to limit the pull up current.

Some time customer will add capacitor like 0.1uF on PG flag trace to filter output noise, so LMR33630 PG pin is designed to limit the discharge current when PG pin is pull Low, otherwise the un-limited current may cause EOS damage to open drain FET.

You just need to consider the PG signal drop down slew rate (or the delay time to effective LOW) with the current limitation value and capacitor value on PG trace.

B R

Andy

0 LIN SUN over 5 years ago in reply to Andy Chen

Intellectual 620 points

Dear expert Andy,

One more question, if inductor is calculated to be 1.5uH as shown attached. What risk will happen if I choose 2.2uH?

0 LIN SUN over 5 years ago in reply to LIN SUN

Intellectual 620 points

0 LIN SUN over 5 years ago in reply to LIN SUN

Intellectual 620 points

0 Andy Chen over 5 years ago in reply to LIN SUN

TI__Mastermind 34875 points

Hi Lin,

2.2uH inductor will make 20% current ripple in your application, this is OK and no risk for your application.

Pls use ferrite core inductor as the Fsw is 2.1MHz.

B R

Andy

0 LIN SUN over 5 years ago in reply to Andy Chen

Intellectual 620 points

Dear expert Andy,

This 20% current is also calculated according to Formula(6), right?

Can you tell me why should ferrite core inductor be used?

Many thanks!

0 LIN SUN over 5 years ago in reply to LIN SUN

Intellectual 620 points

Dear expert Andy,

This 20% current is also calculated according to Formula(4), right? Can I always use this formula to calculate ripple current when choosing other inductor value?

As for ferrite core inductor and powered iron inductor, I used to pay no attention of their difference. Do core losses behaves critical in DCDC performance?

Many thanks for yuor professional work and kind help!

0 Andy Chen over 5 years ago in reply to LIN SUN

TI__Mastermind 34875 points

Hi Lin,

Yes, you can use formula 4) to calculate inductor with proper constant K, smaller K for lower output ripple and higher K for smaller solution size.

Powder iron core is normally used below 500KHz, it has quite high core loss and smaller effective inductance if working at 2MHz.

B R

Andy

0 LIN SUN over 5 years ago in reply to Andy Chen

Intellectual 620 points

Dear expert Andy:

Considering my inductor current is 3.2A (max output current 3A + 1/2 ripple current 0.2A).

When selecting power inductor, I think inductor with 3.5A saturation current (Isat is specified when the decrease of the initial inductance value at 30%) is enough?

Hope your correction and thanks!

0 Andy Chen over 5 years ago in reply to LIN SUN

TI__Mastermind 34875 points

Hi Lin,

The LMR33630 high-side current limit is 4.5A, normally we suggest to use inductor with saturation current =>4.5A in case there is output short circuit or big load transient conditions.

In your application, the current peak is 3.5A, so use 4.5A inductor will improve system level reliability with small cost adding.

B R

Andy

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LMR33630-Q1: PG pin current limit