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I am trying to understand what kind of changes occur to a signal in the chips which sit between the tuner and, for example the FM audio demodulator.

Take the attached example - the TDA2546A. It takes 38.9 MHz IF from the tuner, which'll be downconverted only, but then it outputs a second IF which can be fed into its own FM demodulator, or to other demodulators in the set.

I understand what the right hand side of the chip is doing, but aside from the AGC - what is the left hand side doing? The datasheet doesn't really explain, and sure, it doesn't need to because this is standard stuff, but I don't know where to look!

For example, is it just downconverting the signal? performing AM demodulation? Thoughts appreciated.

TDA2546A

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It's helpful to include a link to the datasheet even if, as you note, it doesn't explain the circuit operation.

Why is the video carrier frequency 38.9 MHz but the audio 33.4 MHz? On-air the audio carrier would normally be 5.5 MHz higher than the video, not 5.5 MHz lower?

This indicates that the 1st local oscillator (LO) is 38.9 MHz above the video carrier radio frequency, rather than below it. So the sound carrier, which is 5.5 MHz above the video carrier in the RF signal, swaps to below it after the first down-conversion. (This web page lists 10 analogue TV systems, 8 of which have the sound IF below the video IF.)

what is the left hand side doing? ... For example, is it just downconverting the signal? performing AM demodulation?

No AM demodulation here. The left hand side downconverts the sound carrier from 33.4 MHz in the first IF to 5.5 MHz for the second IF.

The 1st sound IF signal at 33.4 MHz enters on the left of the diagram through 3 stages of gain controlled amplifiers to provide the automatic gain control feature. It is then applied to a frequency mixer and mixed with a 38.9 MHz signal, of which more later. The output emerges from pin 14. The difference signal (38.9 - 33.4 MHz) is selected by the 5.5 MHz filter and passes into the 2nd IF and demodulator stages.

In a conventional dual-conversion superhet design we would have a 2nd LO to generate the 38.9 MHz signal to drive the mixer.

However, rather than using a LO this circuit picks out the 38.9 MHz video carrier in the REF AMP circuit (presumably reference amplifier) with an external tank (LC) circuit tuned to 38.9 MHz. The tank circuit is specified (figure 2) to have a reasonably high Q of 60 so will be able to pick out the video carrier from the wider bandwidth video signal and also maintain the signal during dips in the modulated signal.

Recovering the video carrier in this way means we don't have two signals (from the video carrier and the 2nd LO) at about 38.9 MHz (but differing slightly) leading to the sound carrier being mixed twice to 5.5 MHz giving distorted sound.

The 5.5 MHz signal can be used on its own for mono audio. It can also be used to carry a stereo sum signal (left + right) for analogue stereo operation. In this case the 5.742 MHz signal (at 33.158 MHz in the IF input) is used to carry a stereo difference (left - right) signal. This will require a similar demodulator stage. The demodulated sum and difference audio signals can then be fed to a stereo matrix circuit to recover separate left and right channels.

(The particular split of 5.5 and 5.742 MHz is/was used for analogue stereo sound transmission in some countries. The same principle can be used with different frequencies for other secondary sound carriers in other countries.)

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  • \$\begingroup\$ Could we sum that up to say that it's a second down conversion? Of course with all of the considerations you've outlined. \$\endgroup\$ May 16 at 12:25
  • \$\begingroup\$ Thanks for that. Another thing that has me scratching my head based on your answer - why is the video carrier frequency 38.9 MHz but the audio 33.4 MHz? On-air the audio carrier would normally be 5.5 MHz higher than the video, not 5.5 MHz lower? \$\endgroup\$ May 16 at 14:26
  • \$\begingroup\$ @MatthewMillman I've reworked my answer to fold in my earlier comments, which I've now deleted. \$\endgroup\$
    – Graham Nye
    May 16 at 17:39
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    \$\begingroup\$ That is a super concise answer. Thank you \$\endgroup\$ May 16 at 18:02

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