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Why the inrush current at startup is about 500mA

When I use a power supply, (3.0Volts-60mA limitation) with the DevKit boards nRF51 or nRF52 the input voltage is clamped at around 2.0 V and the current is 60mA : the SoC doesn’t start.

By increasing the current limitation, step by step, the SoC start, with the current threshold of 80mA.

After the start up, the current of the devKit is low as expected. (With the software ble_peripheral>ble_app_hts s110 for nRF51 or s132 for nRF52).

I've checked the inrush current with a scope. I've seen a pulse of 500mA and 30µs width .

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  • I measured the startup_current on a nRF52QFAABB chip. See the below image.

    The measurement was done using an Agilent N6705B power analyzer, with a sampling interval of 5us. The power analyzer is used as a power supply and has close to zero internal resistance. I get a maximum peak of 34 mA.

    What are you using as power supply? And which chip revision do you have?

    The nRF52 internal power supply charges up caps, and if your power supply has zero internal resistance you might get a high peak, because nothing is limiting the inrush current when the chip starts up. But from what I know zero internal resistance is not a realistic scenario except when you are using high performance lab equipment.

    You say that you can see 500 mA over a period of 30us. That is some narrow spike. I think that adding a series resistance of a couple of Ohm at the input would even out this.

    image description

    EDIT

    So, the startup current in the above plot is the decoupling caps drawing current.

    The development kit has a decoupling cap at 4.7uF at VDD_nRF. And with close to zero internal resistance of the power supply (which is the case with high performance lab equipment, and some batteries), nothing is holding back the inrush current when the caps are charging up.

    New measurements I did with the development kit (using a 4.7uF decoupling cap) was showing an inrush current of 453mA over 30us. This gives a total charge of 13.6uC. 13.6uC over 3V corresponds to a capacitance of 4.5uF, which is not too far off.

    You can try to remove the 4.7uF cap (C9). However, there are still more decoupling caps (100nF) at VDD_nRF and DEC1, so you will probably still see a spike, but it will be much lower. Adding a series resistance at the input would reduce the current spike even more.

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  • I measured the startup_current on a nRF52QFAABB chip. See the below image.

    The measurement was done using an Agilent N6705B power analyzer, with a sampling interval of 5us. The power analyzer is used as a power supply and has close to zero internal resistance. I get a maximum peak of 34 mA.

    What are you using as power supply? And which chip revision do you have?

    The nRF52 internal power supply charges up caps, and if your power supply has zero internal resistance you might get a high peak, because nothing is limiting the inrush current when the chip starts up. But from what I know zero internal resistance is not a realistic scenario except when you are using high performance lab equipment.

    You say that you can see 500 mA over a period of 30us. That is some narrow spike. I think that adding a series resistance of a couple of Ohm at the input would even out this.

    image description

    EDIT

    So, the startup current in the above plot is the decoupling caps drawing current.

    The development kit has a decoupling cap at 4.7uF at VDD_nRF. And with close to zero internal resistance of the power supply (which is the case with high performance lab equipment, and some batteries), nothing is holding back the inrush current when the caps are charging up.

    New measurements I did with the development kit (using a 4.7uF decoupling cap) was showing an inrush current of 453mA over 30us. This gives a total charge of 13.6uC. 13.6uC over 3V corresponds to a capacitance of 4.5uF, which is not too far off.

    You can try to remove the 4.7uF cap (C9). However, there are still more decoupling caps (100nF) at VDD_nRF and DEC1, so you will probably still see a spike, but it will be much lower. Adding a series resistance at the input would reduce the current spike even more.

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