LMX2594: LMX2594 Single Ended ref. clock connection

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Kind regards,
Lane

Hi Noel,

Thank you very much!

Can you help with review the Reference clock circuit for LMX2594?

We use the CDCLVC1104PWR buffer and pass the reference signal through it. We want to use this scheme:

Is it correct or not? Please recommend the correct option.

For the LMX2594 the ref. clock input voltage should be no more than 2Vpp according to sythesizer datasheet:

Therefore, I put a voltage divider from resistors 20/30 Ohms.

If we apply the option from Figure 10,

then the signal voltage swing will be 3.3 Vpp. It will not lead to failure of the LMX ref. input?

Perhaps I did not take into account something and am mistaken, tell me, please.

Hi Noel,

Thank you. About the dependence of the VOH drawdown on the load resistance is understandable.

But what swing / amplitude will the reference signal have in your recommended circuit (at the ref. PLL pin)?

Will there be an excess of 2Vpp? Or vice versa, drop are too low ... I do not understand how to calculate this, based on the data in section 6.5.

I measured the 40MHz output swing with 50ohm load on my bench using CDCLVC1104EVM. For VDD = 2.5V, the output amplitude is 1V. For VDD = 3.3V, the output amplitude is 1.6V.

Kind regards,
Lane

Hi Lane,

thank you very much for the info. Please clarify, did you use such a scheme?

Can you recommend it for CDCLVC1104 + LMX2594?

Lane,

Thank you very much!

Hi Lane!

I also wanted to clarify one question. I did not immediately notice that you wrote about the 1.6V AMPLITUDE. The total signal swing will be twice as large, i.e. 3.2V. For a LMX2594 synthesizer, a maximum value of only 2 Vp-p is permissible. The buffer power is fixed at 3.3V (in our case). Which circuit in this case would you recommend to use for connection CDC buffer with LMX?

Hi Artem,

You can use this circuit, it should reduce the swing to 1.6Vpp for 3.3V supply for CDCLVC1104, which is less than the 2Vpp input limitation.

Kind regards,
Lane

We had the opportunity and today we measured the signal using such a circuit - we observed is approximately 3Vpp (amplitude 1.5V). The synthesizer was not connected, but it has a high-impedance input, it should not affect...

That is strange, I measured using an equivalent load and got half the swing. So the swing is not attenuated when you use the resistor divider versus open circuit? I would expect the load to have some impact.

Perhaps you could try replacing the resistor divider with a 50ohm termination to ground and see if the swing is attenuated for double confirmation.

This is the typical method to attenuate a signal while maintaining signal integrity (50ohm to GND at receiver). If you would prefer a dual supply device that is more easily configured for 1.5V/1.8V output level, you might be interested in LMK00804B instead.

Kind regards,
Lane