# P.L.L as F.M Detector

Discussion in 'General Electronics Chat' started by omerysmi, Nov 25, 2014.

1. ### omerysmi Thread Starter Member

Oct 10, 2014
48
0
Hello, I need help in understanding the operation of PLL as FM Detector.

This is the graph of the output voltage in LPF as a function of the frequency of the input signal.

Suppose that:
- the amplitude of the input signal is 4v.
- f0 is 100kHz.

As we can see when the frequency of the input signal is equal to f0 we will get in the output of the LPF 0V.

My question is, if we will change the frequency of the input signal to the lock frequency (you can see where is it in the graph) , the voltage in the output of the LPF will be 4V? (I know that after fLock1 and fLock2 the voltage will be zero of course as we can see in the graph).

Last edited: Nov 25, 2014
2. ### wayneh Expert

Sep 9, 2010
13,638
4,433
The output voltage is proportional to the offset, not just 0V or 4V, so that this voltage can drive the VCO.

The VCO can only operate within the range shown in the graph; at some point it can no longer make a linear change in frequency in response to a change in voltage from the output.

Oh sorry, it's the output voltage that is "right" within a limited frequency range.

3. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
429
the voltage will also be determined by the oscilator tuning. the voltage in the output of the lpf will not go to zero. vco's dont work that way.

4. ### omerysmi Thread Starter Member

Oct 10, 2014
48
0
there is a way to calculate the voltage when f=fLock2 for example? as you can see is the max voltage in the graph

5. ### wayneh Expert

Sep 9, 2010
13,638
4,433
I would look for this information in the specifications.