Hi ADI, for testing purposes, how do I force pin 27 LKDET of the ADF4382, high or low?
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ADF4382
Recommended for New Designs
The ADF4382 is a high performance, ultra-low jitter, fractional-N
phased-locked loop (PLL) with an integrated voltage controlled
oscillator (VCO) ideally...
Datasheet
ADF4382 on Analog.com
Hi ADI, for testing purposes, how do I force pin 27 LKDET of the ADF4382, high or low?
Hi,
Pin 27 (LKDET) on the ADF4382 is a lock detect output, not an input. It reflects the PLL lock status (Digital Lock Detect or Analog Lock Detect, depending on the register configuration).
If your goal is simply to force the pin state for system testing:
The easiest method is to make the PLL lose lock, for example:
Remove the reference clock.
Program an invalid frequency.
Put the PLL into power-down.
Disable VCO calibration.
In the unlocked state, the Digital Lock Detect output goes LOW.
You must place the PLL into a valid locked condition:
Apply a valid reference.
Program a valid output frequency.
Wait for calibration and lock completion.
When the PLL is locked, the Digital Lock Detect output goes HIGH.
Hi Olim, thanks for the suggestions. I wanted to test the connection to the host controller from pin 27 whether or not the system was actually locked. Therefore I wanted to force the LKDET pin to a wanted state by configuring a suitable regeister. Obviously I can hold the lock detection system in reset, thus forcing pin 27 LOW, by setting the RST_LD bit.
But how can I force the LKDET pin high?
Can I use those reserved bits in reg 0x2D? An old version of the datasheet for the ADF4382A suggests that bit 0 and 1 are LD_O_CTRL. Does that help me? The old datasheet also suggests that bit 3 is PDET_POL. Does this help?
Thanks again
JP
Yes there is a debug mode whereby, if you set the 2 LSBs to binary ‘11’, this should force LD pin high. By default the value is ‘01’ for normal operation.