0
\$\begingroup\$

I have a question with regard to the differences between a PLL on a PCB and on a breadboard.

In this test, I have connected the output of the amplifier (LMH6503) to both the PLL on the PCB and the PLL on the breadboard.

A simple connection in blockscheme

Another picture to make it clear:

enter image description here

Then, I have set the following parameters in the function generator:

  • fcarrier = 15.75 kHz
  • fmod = 1 kHz
  • frequency deviation = 3 kHz
  • amplitude = 305 mVp

Then I obtained the results, which will be listed here.

The result between the VCO output of both the PLL on the PCB and the breadboard:

enter image description here

In this figure, the yellow channel and the blue channel are the VCO outputs of the PLL on the breadboard and the PCB respectively.

The result between the outputs of the analog filter between the PLLs:

enter image description here

The yellow channel and the blue channel represent the output of the analog filter of the PLL on the breadboard and the PCB respectively.

If I zoom in on this figure, the PLL on the breadboard does demodulate the modulated signal (1 kHz) whereas the PLL on the PCB does not do anything. Furthermore, since this NE564 operates on a single power supply, it should have a particular DC-voltage, which is not zero, at the output when the VCO is on the free running frequency, but the one on PCB does not have an 'offset DC-voltage.'

How can it be that both differ from the output of the analog filter of the PLL even if both of the PLLs are locked on 15.7 kHz? (Both VCO outputs tracks the input signals.)

My circuit of the PLL on the PCB:

enter image description here

The PLL built on the breadboard is exactly the same as the circuit on the PCB. Besides, this circuit is found in the datasheet of the NE564 with the frequency capacitor 39 nF to obtain a free running frequency of approximately 16.1 kHz.

\$\endgroup\$
9
  • 1
    \$\begingroup\$ If the circuits are supposed to be the same but work differently, then the first suspicion is a difference in construction - wired wrong or different parts used. \$\endgroup\$
    – JRE
    Jul 30, 2021 at 13:43
  • \$\begingroup\$ if you talk about difference in construction, do you mean inside the ic-chip? \$\endgroup\$
    – Kwok
    Jul 30, 2021 at 14:01
  • 2
    \$\begingroup\$ I mean in the circuit. You built the circuit on the breadboard, and you made the PCB. When you build a circuit, it is very easy to make separate copies different even though you meant for them to be identical. \$\endgroup\$
    – JRE
    Jul 30, 2021 at 14:03
  • \$\begingroup\$ I have checked the breadboard and the pcb wiring connections already, also with some other colleague. However, it still turns out that the breadboard version works properly whereas the pll pcb version does not display the modulating signal. Is there maybe another suspicion? \$\endgroup\$
    – Kwok
    Jul 30, 2021 at 14:09
  • \$\begingroup\$ Are you picking up the demodulated output from pin 14 for both circuits? \$\endgroup\$
    – JRE
    Jul 30, 2021 at 14:20

1 Answer 1

Reset to default
0
\$\begingroup\$

Check VCO bias pot voltage on p3.! It is too low.

Also check p2 ph comp. bias pot. For loop gain. It will be equal to some Vbe.

You may also adjust the tuning cap on PCB by 10% or more to raise the center frequency and retest. Then Vvco should increase where the gain in linear and will increase.

Also excessive DC loading on pin 11 VCO out may affect tuning f from internal emitter follower.

\$\endgroup\$
2
  • \$\begingroup\$ I will try that. \$\endgroup\$
    – Kwok
    Jul 30, 2021 at 14:29
  • \$\begingroup\$ I have measured the p2 and p1. Both on breadboard and PCB seems like similar results. Furthermore, I removed the 39 nF tuning cap and changed to 82 pF, but results remains the same. Then, I let pin 14 be floating and other pins of the NE564 are connected to the PCB. From there, Pin 14 is working. \$\endgroup\$
    – Kwok
    Aug 6, 2021 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.