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Para_1

The following discussion is about type of coupling involved to couple rf amplifier with the mixer. The coupling involved is impedance coupling in which Lm provides a shared mutual impedance between two stages and provides coupling.The higher the Lm higher is the mutual impedance and higher is bandwidth.

Now one requirement that needs to be fulfilled in Vhf tuner circuit is that the gain should remain constant at all frequencies of the entire vhf band(that need to be coupled). According to the figure L2//C2 and L3//C3 is not coupled directly so there is no transformer action between them. They are only linked via Lm.

Para_2

So to make gain constant for whole passband and to achieve this special techniques are being employed and one of them is involving mutual inductance to manipulate the gain in predetermined way. The equivalent mutual inductance m is the sum total of various planned and unplanned coupling prevalent in the primary circuit.

The eq. mutual inductance m, seeks to maintains its value constant around every frequency of the vhf broadcast band. This is possible because m has two component one is fixed and other is variable with frequency. This variable component is higher at high frequency. This variable second component is subtracted from the fixed one so m is less at high frequencies and at lower frequencies vice versa making m a little high. So in both of these cases m has been observed to put its value apparently constant thus maintaing gain constant over the frequency range in demand.

The how does this is possible to maintain gain constant in the below circuit if the circuit is not mutually coupled..??

Figue_1

Figure_2

Figure_3

enter image description here

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  • \$\begingroup\$ Where does it say it is not coupled? Otherwise it just becomes a HPF. But this has nothing to do with the AGC. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 1 '17 at 3:48
  • \$\begingroup\$ I am saying that L2//C2 and L3//C3 are coupled with L4 via impedance coupling, but they are not coupled magnetically (transformer coupling), so what is the provision here is to maintain the gain constant for the vhf passband.. \$\endgroup\$ – partykid Mar 1 '17 at 6:12
  • \$\begingroup\$ L4 does not provide any coupling, L2 must be coupled to L3 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 1 '17 at 6:42
  • \$\begingroup\$ Yes L2 must be mutually coupled to L3 for para_2 to be valid..but the text which I am referring doesn't agree to this.... \$\endgroup\$ – partykid Mar 1 '17 at 6:50
  • \$\begingroup\$ I have added the text which I have referred in original post.. \$\endgroup\$ – partykid Mar 1 '17 at 7:33
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This schematic ought to be shown in a more familiar LC filter arrangement to show the response more clearly as I have a done below.

enter image description here

The centre shunt coil independently controls the 1st pole when its L impedance is lower. While the outer loop of series L's and shunt C's controls the other pole. Q is controlled mainly by source impedance ( such as a current source with load R.) while load impedance ratio affects slope of skirts.

Rather than call it Inductive coupling, it is more accurately impedance ratios of series/shunt elements that control the response.

A simple FM filter is shown that I quickly designed. (not optimal)

enter image description here

Note above the FM filter requires both series L and shunt caps to be the same to tune the upper pole for equal peaks and the centre shunt L tunes the 1st pole is 1/4 of the series L elements.

Your title of this question is misleading.

None of this filter arrangement provides "Maintaining constant gain through impedance coupling". This is done by AGC bias control in the preceding active stage. You might want to edit your question to reflect the details in the question such as LC double tuned filter without mutual coupling.

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  • \$\begingroup\$ Yes you are right that gain is controlled by AGC...however bandwidth is made constant by using certain coupling circuits ..please refer frank.yueksel.org/other/RCA/… page 420... for more info....and about my title it should be "Maintaining constant bandwidth through impedance coupling" \$\endgroup\$ – partykid Mar 5 '17 at 4:57
  • \$\begingroup\$ @TonyStewart..What I didnt got in the article on page 420..Fig 9.16B is that with frequency the reactance of Cm will change...changing the mutual inductance..thus changing the coupling ..so how bandwith does not change with frequency.....What are your suggestions..? \$\endgroup\$ – partykid Mar 5 '17 at 5:08
  • \$\begingroup\$ I believe it refers to the limited tuning range and does not mean you can scale it up 10 decades in frequency and get the same BW. But these are different from the filter in question. Also did you understand my explanation of the circuit in your question from my two examples? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 5 '17 at 6:08
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This circuit is complex. The transistors add their own frequency dependent loadings to the reactive circuit, there are coil losses and a couple of resistors, too. The resonant circuits are surely NOT having the same resonant frequency. Fortunately at least the local oscillator seems to be behind a substantial resistor.

It can be well possible that as the coupling increases due the reactace increase of L4 as the frequency goes higher, it will be compensated by increased losses and detuning.

The fixed resonance circuits, input circuit and LO construction are typical for normal consumer type FM receivers around 88...107 MHz from the era when frequency synthesizers were high end stuff. That's not especially wide band for thislike system.

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  • \$\begingroup\$ @Himanshoo added to the answer "as the frequency goes higher". I mean: This is an equalizer that compensates the growth of the losses when the freguency goes higher. \$\endgroup\$ – user287001 Mar 1 '17 at 10:01
  • \$\begingroup\$ Yes there is an equilizer...and that equalizer is the Equivalent mutual inductance as I have discussed in Para_2, and for that equalizer to work the L2//C2 and L3//C3 should be magnetically coupled...which ofcourse they are not.....This is exactly my query.. \$\endgroup\$ – partykid Mar 1 '17 at 10:21
  • \$\begingroup\$ @Himanshoo no inductive coupling is needed. Any coupling is enough. Read this. frank.yueksel.org/other/RCA/… BTW see the equivalent circuit of an AC transformer. By adding L4 the transformer function is replaced here by its equivalent circuit. L4 generates the coupling emf in two LC current loops just like the mutual inductance does. More complex coupling is needed for flat freq.response. But here an equalizer is surely welcome. \$\endgroup\$ – user287001 Mar 1 '17 at 10:48
  • \$\begingroup\$ Good article ....can you explain me how gain and bandwidth is made constant by circuit shown in fig 9.16B and 9.16C...Xm should remain constant over whole frequency range..how it is achieved...how does the equalizer action works.. \$\endgroup\$ – partykid Mar 1 '17 at 15:30
  • \$\begingroup\$ @Himanshoo NOTE: 9.16 B-C are for wide band systems, such as old radios where one freq. range covered 0.5 to 1.5MHz or 5 to 15 MHz. Impossible to give short word explanations. Simulate them. Your system is narrow band. One coupling method is enough. Your circuit is not a bandpass filter but a matching circuit that transfers the power in the 88-107 MHz band well enough to the mixer and its attenuation curve compensates the attenuation curve in the preamp and the mixer. \$\endgroup\$ – user287001 Mar 1 '17 at 15:51

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