LTC6946-1: using oscillators to generate reference frequency for local oscillator signal

Question

I have to use the LTC6946-1 (frequency synthesizer with inbuilt PLL, phase detector and programmable VCO) to generate a local oscillator signal for a radio transceiver. The output signal has to be 433.92 MHz. The thing is, I've never used PLLs before. This chip requires a reference frequency that meets the following requirements:

• frequency of 25 MHz
• If a CW signal, input power must be at least 6 dBm into 50 ohms
• If a square wave, then at least 0.5 Vp-p with a slew rate of at least 40V/microsecond.

The problem is, how do I chose an oscillator that provides these requirements? I prefer to use a square wave oscillator here. But I also haven't used oscillators before and when looking at their datasheets, I can't find any information that is helpful. For example, the datasheet here:

https://ecsxtal.com/store/pdf/ECS-TXO-3225.pdf

This is a component I would like to use, but how can I tell if this component needs external circuitry to drive it? How can I tell what circuitry is needed? And what the output levels are? None of that information is provided and the datasheets for other oscillators are the same.

Can't make heads or tails of this. Any help would be appreciated.

Answer 1

From the data sheet, the LTC6946 needs a \$V_{REF}\$ signal that's between \$0.5\mathrm{V}\$ and \$2.7\mathrm{V}\$ peak-peak, and it presents an impedance that looks like a \$3\mathrm{pF}\$ capacitor in parallel with an \$8\mathrm{k\Omega}\$ resistor. (This is in the section titled "Reference Inputs").

From the data sheet, the ECS-TXO-3225 is HCMOS compatible. Here, you're just supposed to know that means that it can drive a healthy (\$\pm20\mathrm{mA}\$, if they're telling the truth) current into a load, with a low of \$0\mathrm{V}\$ and a high of \$V_{DD}\$ (which is \$3.3\mathrm{V}\$ if you use the recommended voltage). They do actually state rise and fall times into a \$15\mathrm{pF}\$ load, from which you could compute currents if you were inclined.

So, you have an oscillator that'll produce up to \$3.3\mathrm{V_{P-P}}\$, going into a pin that can stand up to \$2.7\mathrm{V_{P-P}}\$.

I would make a simple voltage divider, followed by the DC blocking capacitor shown in the LTC9646 data sheet. Something like the enclosed schematic. The two resistors attenuate the signal down to 1.7V or so, which is comfortably in the range that the LTC9646 wants to see.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

Checking the data sheet shows that 25 MHz is within its range so thats not a problem. The low output is specified at 0.5V min. Let's assume its 0.4 volts. The high output is specified as 80% of the supply voltage which is nominally 3.3V. 80% of 3.3 is 2.64V. Therefore the output swing is from 0.4 to 2.64 volts or 2.24V. The rise time is specified as 10 ns maximum. Since the rise time is measured from the 10% value to the 90% value, the swing for rise time is .8X2.24 or 1.79V. Therefore the worst case slew rate is 1.79v/10ns which is equal to 179 volts/microsecond. Thus this oscillator meets that requirement. The load is specified as CMOS which should be good to go. Therefore, you can use this oscillator for your application.