2
\$\begingroup\$

In the datasheet of the nRF8001, they specify in the reference schematic (page 51) the LC circuit to use with the integrated DC/DC converter (see below).

LC circuit

Why is there a small inductor L5 in series with L4?

Is it related to the self resonant frequency of the inductors? That was my first thought, but the 10nF inductor specified in their integration note (Taiyo Yuden LBMF1608T100K) has a minimum self resonant frequency of 32MHz. The exact frequency of the nRF8001 buck converter is not specified, but it seems to be well above the frequency of any buck converters.

\$\endgroup\$
2
  • \$\begingroup\$ That self-resonant frequency is also a long way below the Bluetooth operating frequencies. So the second inductor may be required to increase impedance at those frequencies. \$\endgroup\$ – user_1818839 May 24 '16 at 20:46
  • \$\begingroup\$ @BrianDrummond Yes, indeed the SFR of L5 is close to Bluetooth frequencies, so i think it is the reason. Any idea why exactly it is needed? \$\endgroup\$ – Adrien Descamps May 25 '16 at 5:32
1
\$\begingroup\$

The only obvious answer would be that this is an RF blocking component. It should have a SRF close to the transmit frequency. Exactly why its needed is a more complex question.

\$\endgroup\$
1
  • \$\begingroup\$ Yes, i think you are right, it has a SRF close to the bluetooth frequency. \$\endgroup\$ – Adrien Descamps May 25 '16 at 5:25
0
\$\begingroup\$

During periods of high current consumption, the larger inductor will become saturated, leaving only the smaller inductor. This is a trick to keep the size small; otherwise you need a 15nH inductor capable of high DC current.

The output of the step-down converter drives a series-pass regulator, so it is possible that the circuit allows the voltage to rise during peak current periods when saturation occurs. The other possibility is that the frequency increases. The large inductor allows the converter to operate at a relatively low frequency during idle periods, keeping the converter efficiency high and reducing the battery current if a higher voltage battery is used.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.