Can someone explain or point to some documentation that explains how the RF front-end works for BLE chips? Specifically, I am confused about why the VCO runs at 4.8Ghz and is then divided by 2. I have a background in EE but not much RF.

Reading here I get this but want to confirm if there are other things to consider:

"Typically, VCO performance is poor when the control voltage hovers close to the threshold voltage of the transistors in the Voltage-to-Current (V2I) converter and is also poor at high control voltages when the V2I current sources are pulledout of saturation. So, it may be beneficial to divide the VCO by 2 at low frequencies and divide by 1 at high frequencies to keep the control voltage in the VCO's "sweet spot"."

I see images similar to this online:

enter image description here

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    \$\begingroup\$ It looks like there is more than just dividing the frequency by 2, the divider provides both in-phase (I) and quadrature (Q) outputs for the mixers. \$\endgroup\$ – Elliot Alderson Feb 8 at 20:02
  • \$\begingroup\$ Your quote has very little to do with the divider's function here - as @ElliotAlderson says, part of a block that provides two local oscillator waves of the same frequency, but shifted by 90 degrees. Looks like their frequency here is one-half VCO frequency. \$\endgroup\$ – glen_geek Feb 8 at 20:14
  • \$\begingroup\$ @ElliotAlderson It looks like the VCO provides those two outputs at 4.8GHz, so I would guess the divider is actually two dividers working on two signals. \$\endgroup\$ – Hearth Feb 8 at 21:03

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