Help with PLL circuits

[Geekyguy]

Hi guys,

How are you all? I am trying to build to a PLL circuits. As part of the circuit, I will need an offset adder, integrator, rectifier, and another stage of offset adder.

I havent done circuits for very long time. So please bear with me if Im crapping.

Appreciate your help.

I have designed the first stage offset adder and integrator.

Please see attached diagrams. Schematic 1 is the circuit for offset adder which basically a summing amplifier, and then the second stage op-amp is the integrator.

The AC source with 1 Vac, is passed through the R3 and R1, R3 will later be a variable resistor when I transfer this to PCB.

From the diagram, I dont understand how the 5V added by V2 becomes 2.5V at the output of U1 (first Op-amp). I tried calculating it out, but could not get the exact point. Please advice.

Then, at the second stage. If you see the frequency response diagram. The 3dB is approximately at 200 kHz. But by using 1/2\[\prod\]RC formula, R=R3 and C=C1, which are 1 kohm and 50 pF, respectively, the cutoff should be approximately 3.1 MHz. So I dont really get it. Please advice if Im in the right direction, or the equivalent circuit calculation is wrong.

Please point where Im going wrong. I have also attached the time domain graph.


Thank you so much.

 

[crutschow]

The output will be 2.5V if R4 = 500Ω. If R4 = 1kΩ then the output should be -5V.

What is the frequency response (GBW) of the op amps you are simulating? It should be at least 10MHz for a 3MHz response.

The output of your circuit integrator op amp, as shown, will go to the rail, since it will integrate the -2.5Vdc at its input.

 

[erikl]

... I dont understand how the 5V added by V2 becomes 2.5V at the output of U1 (first Op-amp).

Obviously R4=0.5kΩ . Gain(U1, inv.) = -(R4/R2) = -0.5 ; 5V*( -0.5) = -2.5V

 

[Geekyguy]

Thanks guys. I fixed it, its the variable resistor. Thanks again.

Between, I have found the XR-2211A that could do the job :

See attached datasheet.

https://www.farnell.com/datasheets/7277.pdf"]https://www.farnell.com/datasheets/7277.pdf"]http://www.farnell.com/datasheets/7277.pdf

The datasheet provides the circuit construction for tone detection, and I need that. But however, it is unclear how could I tap out the VCO output, cause that is what I need, I need the locked VCO output.

Please guys, advice me, I really dont understand.

 

[crutschow]

You might try a CD4046. That has an output for the VCO signal.

 

[Geekyguy]

Thanks for the suggestion crutschow, but its coming with a mixer altogether, I cant use that. Thanks anyway.

I have built the circuit in practical. Its working fine. But I have a small problem, please see the attached schematic.

So the signal goes through a summing amp, integrator, full wave rectifier and final summing amp. All the sections are working fine. But after full wave rectifier (Op amp U4), there a is a DC offset. I dont want that DC offset. Because I would like to add DC through the final stage of summing amp from 0V to any arbitrary +V. So how can I remove the DC, I dont want any inline capacitor. Im thinking of buffer, will that be the correct move?

Please advice.

 

[crutschow]

You need to determine where the offset is coming from before a cure can be suggested. How much is the offset?