Frequency Oscillator

[YAN-1]

Hello everyone. I'm trying to desing a 315 MHz oscillator and I am required to use the MC145151-2 PLL IC by Motorola as a starting stage. In general, the stages are the PLL, a VCO (Voltage-Controlled-Oscillator), a frequency multiplier, a filter, and an amplifier. Can anyone suggest a VCO compatible with the Motorola PLL? I'm having trouble trying to figure out whether the VCO produces a square wave or a sine wave. The final outcome of my design should be a sine wave at 315 MHz. Your help is much appreciated.

Nichola V. Abdo

 

[Nigel Goodwin]

For RF use a VCO will normally produce a sinewave - it would usually be a discrete LC oscillator with a varicap diode for the voltage control. You would normally use frequency dividers from the VCO to the PLL, rather than multipliers up from a lower frequency, but it depends somewhat on what you're trying to do?.

 

[JimB]

Have a look here:



One of these may do just what you want.

JimB

 

[YAN-1]

Well my problem is that I am required to use the MC145151 PLL IC but I'm not sure how to choose the VCO. I should choose depending on the voltage output of the PLL but I'm not sure what that is. Please take a look at the datasheet. I will be connecting the Phase Detector output (PD) pin of the PLL to the VCO's input (as in page 6 of the datasheet) through a low-pass filter. But I can't figure out what VL and VH for the PDout are in the diagram of page 17. I need to know that in order to find a matching VCO. Also, I too thought that the VCO produces a sinewave but how come all waveforms in the datasheet are square? I mean it's weird but I think it makes sense since the PLL is essentially a digital cct. Please help! Thanks a lot.

Nichola

 

[Hero999]

This circuit will work but you need to subsitute the variable capacitor for a varicap diode with a capacitor in series, the voltage bias on the diode will vary the frequency.

http://circuitos.cl.tripod.com/schem/r102.gif

 

[YAN-1]

PLEASE does anyone know the waveform of the output signal of the JTOS-535 VCO? I mean in terms of the voltage level of the output sine (the VL and VH between which the sine oscilaltes). Thanks a lot.

Nichola

 

[Hero999]

Can't you find a propper datasheet?

They look rather expensive for what they are.
**broken link removed**

It says the output is 2dB, I don't know what this means, are they talking about dBm, dBV?

Anyway the output isprobably sinewave.

 

[JimB]

Looking at the datasheet in the reference from Hero999.

The output power is stated as +9.5dBm (dBm is decibels releative to 1mW), so the output is just a touch under 10mW.

The datasheet also tells us that the harmonics are typically -28dBc (dBc is decibels relative to the carrier - the wanted output), so the power in the harmonics is 0.15% of the output.

Conclusion the output is a reasonable sinewave.

JimB

 

[Hero999]

10mW, driving what impedance?

I doesn't say.

 

[JimB]

10mW, driving what impedance?

99.9% safe to assume 50ohm.

JimB

 

[YAN-1]

So let me get this straight.. If the output is a sine wave and its power is 9 dBm, this means that the output power is at 7.94 mW. Now if the load is at 50 ohms, then this means that the rms voltage of the sine is 0.63V and the peak is 0.9 V. Now I am required to get an output at 0dBm and 50 ohm so I implement a simple T attenuator with a 50 ohm impedance. Is this correct up to this point? And another question, how can I know by how much the peak voltage will be attenuated after the attenuator circuit and before adding the load?

Thanks a lot.
Nichola

 

[JimB]

YAN

Your maths is correct.

I assume (I hope!) that MiniCircuits have specified the output power when the oscillator is terminated with a 50ohm load. (If they have not, then they are very bad boys).

So, to get 0dBm all you need to do is put a 9dB attenuator between the oscillator and the load.

If you disconnect the load from the attanuator, the voltage coming out of the attenuator will be twice the voltage you get when tha attenuator is terminated. That is simple ohms law, however in practice, the voltage may vary due to loading effects on the output of the oscillator.

JimB