I'm on a (very) long and arduous journey to build a RF sweep generator for tuned circuit filters. After much experimentation with free running LC oscillators, I conclude that for this application (and keeping things analogue), some form of AGC to control the amplitude is key. This is the context that finds me reviewing circuit diagrams for the venerable Wavetek 1067 and I'm quite perplexed by what I find there.

Wavetek 1067 User Manual - Circuit diagram on pg.73

Here's an annotated snippet from the main sweep oscillator board:

enter image description here

I'm guessing that section B (Blue box) is the main sweep oscillator with L8 - L10 and varactors CR9 - CR11 forming the main tuning components. L11 isn't coupled to anything (that I can see) and so forms a tapped inductor, where the centre tap is at the potential of either CR1/CR2 or CR3/CR4 (it isn't a fullwave bridge although it's drawn that way!)

So, it seems that CR1 to CR4 form a kind of amplitude limiting circuit (2 diode drops) whose DC bias is controlled by R6. The signal is then coupled to L2 and here's where is falls off the cliff for me. What exactly is the purpose of section A? I can see that the signal exits through the centre tap of L3 to the rest of the circuit - but section A doesn't seem to go anywhere??

My guess (for what it's worth) is that the amplitude limiting circuit in section C produces a lot of harmonic content (I know this because I've tried this kind of clipping myself and the resulting sine wave is nasty) and section A is there to improve the wave shape. But that's a wild guess.


1 Answer 1


Indeed, that's an odd diode arrangement.

It must be that L11 is coupled to the other 3 coils in that block otherwise L11 serves no purpose. The coupling might not be indicated in the schematic but I think that it must be coupled.

I am not that familiar with but did think: a diode mixer maybe ?

And indeed, see here

enter image description here

This is a double-balanced mixer and very similar to section C.

So I think it is a double-balanced mixer.

Section A: combining the double-balanced mixer with section A then A could generate the RF signal. Since it only has a supply connection that would make sense.

Right of C looks like an amplifier (signal goes to base of Q2) so that could be the IF output.

  • \$\begingroup\$ Ah, OK. So, section A is delivering a signal (fixed?) and section B is delivering a ramped signal. Then section C mixes them together. Does that suggest that the output is some kind of IM product? \$\endgroup\$
    – Buck8pe
    Commented Nov 22, 2017 at 14:23
  • \$\begingroup\$ I suppose I'm wondering why they did things this way?? \$\endgroup\$
    – Buck8pe
    Commented Nov 22, 2017 at 14:24
  • \$\begingroup\$ Yes, the output of C (going to Q2) will contain the frequency sum and differences, that's what a mixer does. Since some filtering is there (R8, C4) the sum of the frequencies is probably filtered out and only the difference is used. I think doing it like that means that the frequency range of the sweep gen. can be more limited. For example, if it can do 100 - 200 MHz (easy to design) and we mix with 90 MHz we can get 10 MHz to 110 MHz. A 10 MHz to 110 MHz sweepgen would be a challenge to design! 200MHz/100MHz = 2 (easy) 110MHz/10MHz = 11 (hard) \$\endgroup\$ Commented Nov 22, 2017 at 14:32
  • \$\begingroup\$ So, can say that mixing the signals like this produces a better wave shape when the signal ranges from low to high? I'm not sure I understand why a 100-200MHz is easier to design than one from say 1MHz to 100MHz?? \$\endgroup\$
    – Buck8pe
    Commented Nov 22, 2017 at 14:39
  • 1
    \$\begingroup\$ Excellent! BTW I love these old-style, only using discrete transistors, not using any ICs except the odd opamp designs of the old days :-) \$\endgroup\$ Commented Nov 22, 2017 at 14:56

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