How do tapped coils and capacitors match 75Ω impedance in VHF tuners?

[tech99]

You can model L2 and L3 as a transformer with a low coupling coefficient m.
You can model L4 as an auto-transformer, or two parallel windings, with a high coupling coefficient M = 1.0
Those two transformers share or blend the coupling in proportion to their impedance.
If I had the time I would model the circuit with Spice.

The circuit does not look tunable, so there does not seem much point in having two sorts of coupling in order to maintain constant coupling during tuning. Also, not sure if L2 and L3 are a transformer, and I think that separate coils in cans would be more predictable.
Regarding the value for L4, XC2 is about 300 Ohms and XC3 is about 600 Ohms. SQRT 300x600 = 424 Ohms. Suppose we want to cover about 90 to 100 MHz, requiring a Q of 10, so that kQ=1 and k=0.1. Then the bottom coupling reactance provided by L4 needs to be 424 x 0.1 = 42 Ohms. This is quite a reasonable value.
It is interesting to notice that if L4 is a very high impedance, then the circuit would behave as a pi-network between the two stages. The two tuned circuits would be blended into one, as Baluncore mentions. I think this may be what Davenn had in mind.

 

[Baluncore]

Also, not sure if L2 and L3 are a transformer, and I think that separate coils in cans would be more predictable.

Then how do you interpret the T₂ designation between L₂ and L₃ like all other transformers on the schematic.

 

[davenn]

The circuit does not look tunable, so there does not seem much point in having two sorts of coupling in order to maintain constant coupling during tuning. Also, not sure if L2 and L3 are a transformer, and I think that separate coils in cans would be more predictable.

Then how do you interpret the T₂ designation between L₂ and L₃ like all other transformers on the schematic.

exactly

and again look at L2 within the T2 designation ... it is clearly marked as a tuneable inductor

 

[tech99]

If you compare it with the other cases in the diagram it is different. The inductor spacing is greater and the windings are reversed. But I agree it could be, although it just seems to make things harder during manufacture.

 

[brainbaby]

The circuit does not look tunable, so there does not seem much point in having two sorts of coupling in order to maintain constant coupling during tuning.

What I realized is that I made a little mistake in elaborating what I mean by constant k..actually k tells us about the extent of coupling...and what I actually meant is bandwidth..so what i visualize here.. is that two resonant circuit and each having its different resonant frequency ...and when they are coupled they collectively are resonant at different frequency owing to the mutual inductance...so the circuit would provide output at the two initial resonant frequencies (when the circuits are not coupled) and the third resonant frequency (after coupling)...so it can be just visualised as ...more resonant frequencies, more output...and more bandwidth...

It seems that you provide me some clue about relating Lm with bandwidth...as more the value of Lm more is the mutual impedance and more is the bandwith..
agreeing with Baluncore

Those two transformers share or blend the coupling in proportion to their impedance.

so i need to find certain things...
its not L4 that will keep k constant (though which i earlier thought) rather it control the mutual impedance which in return regulates bandwidth...exactly how??

 

[tech99]

What I realized is that I made a little mistake in elaborating what I mean by constant k..actually k tells us about the extent of coupling...and what I actually meant is bandwidth..so what i visualize here.. is that two resonant circuit and each having its different resonant frequency ...and when they are coupled they collectively are resonant at different frequency owing to the mutual inductance...so the circuit would provide output at the two initial resonant frequencies (when the circuits are not coupled) and the third resonant frequency (after coupling)...so it can be just visualised as ...more resonant frequencies, more output...and more bandwidth...

It seems that you provide me some clue about relating Lm with bandwidth...as more the value of Lm more is the mutual impedance and more is the bandwith..
agreeing with Baluncore

so i need to find certain things...
its not L4 that will keep k constant (though which i earlier thought) rather it control the mutual impedance which in return regulates bandwidth...exactly how??

We couple two resonant circuits together, both tuned to the same frequency. Each has a bandwidth to the -3dB points of 1/Q. If the coupling is small, the two responses (measured on a dB scale) are added and so provide about the same bandwidth but better skirt rejection. With strong coupling, the response splits into a double peak. With "critical coupling" when k=1, there is a flat response which is about 40% wider than each circuit and with good skirt rejection.
It is possible to tune the circuits to different frequencies, but the end result is the same, assuming suitable coupling is used.
When you said constant k, I thought you meant that k must remain constant as the circuits are varied in frequency, but the tuner shown seems to be fixed frequency.

 

[Baluncore]

When you said constant k, I thought you meant that k must remain constant as the circuits are varied in frequency, but the tuner shown seems to be fixed frequency.

The bandwidth of the RF amplifier needs to cover the whole VHF TV band as it is not tuned with channel changes. The coupling coefficient needs to remain reasonably constant over the entire band. The factory tuning is there to adjust the response at the edge of the VHF TV band.

 

[davenn]

If you compare it with the other cases in the diagram it is different. The inductor spacing is greater and the windings are reversed.

The significant difference between T2 ( L2 and L3) and T3 is that T2 is air cored and T3 is indicating an iron/ferrite core

 

[brainbaby]

If I had the time I would model the circuit with Spice.

Really I am eager to find out an example which verify constant k over entire frequency range...

As Lm provides common impedance coupling, so for a certain frequency there is a certain inductive reactance XL of Lm (i.e shared) and for different frequencies, XL is also different...when reactance change it also change the value of Lm...and it can be seen that by substituting value of Lm for different-2 frequency in formula k = Lm + M / √ (Lp + Lm)(Ls + Lm) will result in different values of k...which is against our aim...so i find myself not worth going anywhere...??

 

[Merlin3189]

I can't get pictures in the conversation, so I've put it here. (With apologies to other thread readers.)

What I'm trying to show is that the mutual impedance becomes part of both tuned circuits and the tuned circuits are linked by that shared impedance.
The bottom diagram is purely an analogy, showing how the size of a shared impedance can affect transfer between loops.

 

[brainbaby]

What I'm trying to show is that the mutual impedance becomes part of both tuned circuits and the tuned circuits are linked by that shared impedance.The bottom diagram is purely an analogy, showing how the size of a shared impedance can affect transfer between loops.

Thanks for the figure ...now I am able to visualize exactly how the signal is coupled, earlier I was thinking that the two resonant circuits (L2, C2 and L3, C3) are coupled by mutual induction..hell I was wrong..

If I am right then equivalent circuit of your last figure would be like this

Now I am solving like this
for second case R=1 and A= 0.08
Now finding Equivalent resistance
1/Req1 = 1/R1 + 1/2R
= 1/10 + 1/2
=10*2/10+2
Req1 = 5/3
Req2 = 5/3 (for other half of the circuit)
Now these resistance are in series so

Req = 5/3+5/3 = 10/3 ohm

But the total resistance should be 125 ohm
as per
V= I .Req
Req = 10/ 0.08
= 125 ohm...

Where I am mistaken...??

 

[Merlin3189]

Yeah, that's my fault introducing the splitting of R. Turned out not to be helpful !

I said
$R1\ in\ series\ with\ (R || R2) = R1 + \frac{R R2}{R + R2} = 10 + \frac{1\times10}{1 + 10} = \frac{120}{11}$

## \rm {So } I_R1= 10\times\frac{11}{120} = 0.917 ##

## \rm {and } V<sub>R1[/SUB ]= 9.17 ##

## \rm {making } VR=0.83 ##

## \rm {So } IR2= \frac {0.83 }{R2} = 0.083 ##

(apologies for the Latex. It's playing up today and ignoring it on some lines!?)

In plain text
R1 in series with (R || R2) = R1 + R* R2/(R + R2) = 10 + 10/11 = 120/11
IR1 =10 * 11/120 = 0.917
VR1 = 10 * 11/12 = 9.17
So VR = 10 - 9.17 = 0.83
And IR2 = 0.83 / 10 = 0.083</sub>

 

[brainbaby]

In plain text
R1 in series with (R || R2) = R1 + R* R2/(R + R2) = 10 + 10/11 = 120/11
IR1 =10 * 11/120 = 0.917
VR1 = 10 * 11/12 = 9.17
So VR = 10 - 9.17 = 0.83
And IR2 = 0.83 / 10 = 0.083

Thanks for your wonderful elaboration and making me understanding the exact fundamental idea of common impedance coupling employed in the circuit.
Your theory seems to be unified with most of what Baluncore have said earlier, and that's a good omen that we are on the right track and have almost understood the circuit and its operation , however some little things do need further clarification such as...

Baluncore post 30# you seem to comply with what he has said in it...
"Those two transformers share or blend the coupling in proportion to their impedance."..absolutely correct!
But in his post 28" you seem to be in little contradiction for what he said...
"The explanation is that T2 is only a closely coupled transformer." and in your post 40# your figure states that "L1 and L2 (i.e L2 and L3 in original circuit) are not magnetically linked..so no transformer action..."
What you have to say on this...!

 

[Merlin3189]

Baluncore was presumably speculating, as was I at the time, as to how the circuit functioned. But having read Mr Gulati further, I saw on page 434 of your reference,

Impedance Coupling
Another form of coupling, that employs a common impedance Zm for mutual coupling between two tuned circuits, is shown in Fig. 22.5 (a). This method is also called band-pass coupling because it usually has the bandwidth of an overcoupled double tuned transformer. In Fig. 22.5(b) coil Lm is common to both the tuned circuits with L1, C1 and L2, C1. As a result, signal voltage across Lm is coupled from one tuned circuit to the other. There is no direct coupling between L1 and L2 and both L1, C1 and L2, C2 are pretuned by means of brass slugs to the same centre frequency. The bandwidth is controlled by the value of Lm. The greater the value of Lm the greater the mutual impedance and greater is the bandwidth. This type of coupling is commonly employed between the RF amplifier and the mixer. Values of Lm and damping resistances are chosen to provide a band pass of approximately 7 MHz.

The original circuit you posted, fig 22.17 says,

The RF amplifier output circuit is double-tuned and mutually coupled through common inductor L4. Note that capacitors C2 and C3 are effectively in parallel with L2, L4 and L3, L4 respectively to provide tuning. The cores of L2 and L3 are adjusted to obtain a broad tuning with peaks at the picture and sound carrier frequencies.

He does not say here that L2 and L3 are not coupled. They could be, but in view of his earlier description of the same layout, I now think they are not.
The coupling can be set by L4 and there is no reason to add transformer coupling.

As for the general issue of coupling varying across a band, as has been mentioned earlier, I now understand that this is referring to cases where L1 and L2 (or L2 & L3 in some diagrams) are linked variable capacitors being used to tune across the band. Here the coupling will vary because the fixed Lm (or L4) will not be a constant proportion of the tuned circuit capacitance. This is then mitigated by providing another form of coupling which varies in the opposite way.
In 22.17 and 22.5 the C1 and C2 are fixed as are L1 and L2 (maybe trimmed during setup), so L4 remains a constant proportion at all frequencies across the band and provides fixed coupling coef't. It is this coupling coef't, along with the tuning and damping of the two tuned circuits, which determines the overall bandpass characteristic.

 

[Baluncore]

There are three RF amplifier applications for an inductively-coupled double-tuned circuit. The frequencies chosen;
1. Close in frequency, tracked with a ganged capacitor for a tunable narrow band FM or AM channel.
2. On either side of one TV channel. Optimised for the approximate 6 MHz channel, from the vestigial sideband and AM carrier to the FM sound carrier.
3. Far apart and fixed for an entire TV band of channels, eg VHF1, VHF2 or UHF.

I believe the circuit under discussion here is number 2. TV sets once used turret tuners to select the channel. The set of inductors needed for each channel was wound on a strip of bare phenolic card, or a plastic strip or rod.
http://www.modip.ac.uk/sites/modip/images/large/BXL974.jpg

The schematic provided does not show that turret or the wiper contacts, but it does show a fine tuning control, R3 on the LO. So where is the coarse tuning? That must have been done by a channel selection turret. Everything prior to the IF was within that turret tuner module. The active elements were fixed and shared between channels, so only the tuned circuits were on the channel cards. Each card would have had the five inductors from L1 to L5. The T2 coupling for each channel was adjusted by selecting the relative positions of the L2 – L3 windings on that card. This is consistent with the schematic showing no values for the inductors, and giving only a vague hint of proximity coupling for T2.

 

[tech99]

There are three RF amplifier applications for an inductively-coupled double-tuned circuit. The frequencies chosen;
1. Close in frequency, tracked with a ganged capacitor for a tunable narrow band FM or AM channel.
2. On either side of one TV channel. Optimised for the approximate 6 MHz channel, from the vestigial sideband and AM carrier to the FM sound carrier.
3. Far apart and fixed for an entire TV band of channels, eg VHF1, VHF2 or UHF.

I believe the circuit under discussion here is number 2. TV sets once used turret tuners to select the channel. The set of inductors needed for each channel was wound on a strip of bare phenolic card, or a plastic strip or rod.
http://www.modip.ac.uk/sites/modip/images/large/BXL974.jpg

The schematic provided does not show that turret or the wiper contacts, but it does show a fine tuning control, R3 on the LO. So where is the coarse tuning? That must have been done by a channel selection turret. Everything prior to the IF was within that turret tuner module. The active elements were fixed and shared between channels, so only the tuned circuits were on the channel cards. Each card would have had the five inductors from L1 to L5. The T2 coupling for each channel was adjusted by selecting the relative positions of the L2 – L3 windings on that card. This is consistent with the schematic showing no values for the inductors, and giving only a vague hint of proximity coupling for T2.

I think this circuit is probably for a VHF Band II tuner rather than a TV tuner, as no indication of channel switching is shown and the circuit could not be broadbanded enough to cover the entire width of Bands I or III.

 

[Baluncore]

The schematic comes from here: Monochrome and Colour Television. R.R. Gulati - 2009 isbn = 8122416071
" Transistor VHF Tuner. The circuit of a commonly used transistorized VHF tuner is shown in Fig. 22.17. "

"The cores of L2 and L3 are adjusted to obtain a broad tuning with peaks at the picture and sound carrier frequencies. "

Edit; Here is the OCR text extract.

Monochrome and Colour Television. R.R. Gulati - 2009 isbn = 8122416071

Transistor VHF Tuner.

The circuit of a commonly used transistorized VHF tuner is shown in Fig. 22.17. It employs BF196 for the RF amplifier and mixer stages, and BF194-B in the local oscillator circuit. All the three transistors operate in grounded emitter configuration. The 75 ohm impedance of the balun on antenna side is matched to the input of Q1 (RF amplifier) by a tap on coil (L1) and the impedance transforming network formed by the two capacitors (10 pF and 15 pF) which shunt the coupling coil. Forward AGC voltage is fed at the base of BF196 via 1 K resistance and a decoupling network. The 47 ohm resistance inserted in the base circuit helps to suppress any parasitic oscillations. The RF amplifier output circuit is double-tuned and mutually coupled through common inductor L4. Note that capacitors C2 and C3 are effectively in parallel with L2, L4 and L3, L4 respectively to provide tuning. The cores of L2 and L3 are adjusted to obtain a broad tuning with peaks at the picture and sound carrier frequencies.

The output from the RF amplifier is coupled to the base of mixer transistor Q2 through the voltage divider and matching network formed by capacitors C4 and C5. The oscillator output is also fed at the base of Q2 through capacitor C6. The mixer output circuit is double tuned and feeds the first IF amplifier. The 150 ohm resistor across the secondary winding of T3 is a damping resistance to achieve broad tuning.

In the local oscillator circuit, L5 in parallel with a 2 pF capacitor constitutes the tank circuit. The oscillator output frequency is adjusted by varying L5. A varactor diode D is used for fine tuning. As shown in the circuit, it is connected in parallel with the tank circuit. Reverse bias voltage to the diode is fed through the voltage divider formed by R2 and R3. The potentiometer R3 is provided on the front panel of the receiver. It serves as fine tuning control because varying R3 changes the reverse bias voltage and hence the capacitance across the tank circuit. The capacitor C7 blocks dc voltage shorting to ground.

 

[brainbaby]

Here the coupling will vary because the fixed Lm (or L4) will not be a constant proportion of the tuned circuit capacitance.

Does constant proportion means that voltage at point (XY) or Z will be constant at all frequencies??

This statement seems to say that Lm is acting as a resonant circuit, but since Lm is providing impedance coupling by shared impedance, then what the use of it being in resonant with circuit capacitance.??

What does tuned circuit capacitance involves here, Is it the collective capacitance of C1,C2,C3,C4…??

This is then mitigated by providing another form of coupling which varies in the opposite way.

Please illustrate this anticoupling "another form of coupling"

 

[tech99]

Does constant proportion means that voltage at point (XY) or Z will be constant at all frequencies??

This statement seems to say that Lm is acting as a resonant circuit, but since Lm is providing impedance coupling by shared impedance, then what the use of it being in resonant with circuit capacitance.??

What does tuned circuit capacitance involves here, Is it the collective capacitance of C1,C2,C3,C4…??
Please illustrate this anticoupling "another form of coupling"

We do not need resonance of L4 to obtain the required mutual coupling. We can use a resistor in this position if required, or a capacitor. Any type of impedance can provide the required coupling coefficient. However, if the circuits were adjusted by a variable capacitor, the Q would rise towards the HF end, and so if kQ is to be 1 at all frequencies, we would prefer a type of mutual impedance which reduces with frequency, rather than one which increases, as does a inductor. A combination of coupling impedances is sometimes used to obtain the desired k as the tuning capacitor is adjusted, and for this purpose mutual inductance can be of either sign, adding or opposing the coupling provided by L4.

 

[brainbaby]

This is then mitigated by providing another form of coupling which varies in the opposite way.

Hi friends feeling great to be back again...

After research of many days I was able to break down the obscurity of the coupling that has been employed in this circuit...
Here goes the story beside this coupling...

When two tuned circuits are mutually coupled, at higher frequencies particularly at resonance the impedance of the primary circuit is higher which causes the impedance of the secondary circuit to be also high causing increase in gain. The requirement is initially to control this high frequency gain desiring the gain to be constant for all frequencies. Hence to achieve this special techniques are being employed and one of them is involving mutual inductance to manipulate the gain in predetermined way. This discussion is mainly of effective mutual inductance.

In mutual coupling of two tuned circuit the impedance is reflected from primary circuit to secondary.

This transfer of impedance from one circuit to other is mainly because of equivalent mutual inductance denoted by m. There is a difference between mutual inductance M and m.

Mutual inductance M, is generally exist when two coils are in close vicinity to each other and tells us that a change in current in first circuit produces an emf in other coil. The gain of the circuit largely depends upon M and to make gain constant over the whole broadcast range M has to be altered simultaneously. The only way to alter M is either by rearranging the location of the two coils in space or tuning condenser. But moving any two parts within a circuit to alter M for constant gain was not feasible, so a provision was found out so that M can be altered automatically with frequency just by only use of the tuning condenser and this was found out by incorporating equivalent mutual inductance m.

However m is the sum total of various planned and unplanned coupling prevalent in the primary circuit. The modification of primary circuit impedance can cause a desired change in circuit parameters of the amplifier in question.

The eq. mutual inductance seeks to maintain its value constant around every frequency of the vhf broadcast band. This is possible because m has two components 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 maintaining gain constant over the whole frequency band.

This action is what I think so Merlin3189 have talked about...

Hoping that I have correctly justified the topic..