# VHF tuner circuit....

1. Nov 16, 2016

### brainbaby

Hi friends,

I want to frame out how 75Ω impedance is matched by using tapped coil and two shunt capacitors connected in parallel. I think so for that I have to calculate the load value, and if we take the entire circuit out to be as a load then the most acute hurdle would to be to calculate the load value of the circuit.

Thanks

2. Nov 16, 2016

### Baluncore

The 75 ohms is matched by the tap low down on L1. The top of L1 is much higher impedance.
As shunting capacitors, the 10pF in series with 15pF will peak resonance of L1 in VHF band. They then form a capacitive voltage divider that applies RF to the base of Q1.

You should think of the 10 pF as the top of a Pi matching network, with L1 and the 15pF as the two legs. That network matches the high impedance of L1 to the base of Q1.

3. Nov 17, 2016

### brainbaby

If you are talking about resonance as what every tuner circuit meant out to be ...then how can we think of that the capacitors would divide voltages...
At resonance the imaginary part of the impedance of coil L1 and capacitors ( 10pf and 15 pf) will cancels each other resulting no current to flow across it causing the voltage from antenna side to perfectly couple to the base of Q1...hence they shouldn't acts as a voltage divider...

also...

4. Nov 17, 2016

### Baluncore

Energy circulates between L1 and the capacitors. A current is flowing in the circuit. The voltage that appears across the inductor also appears across the two capacitors. The two capacitors are in series. The series combination of two capacitors is resonant with L1. Where two capacitors are in series they form a voltage divider.That helps match the LC network to the BJT base.

L1 is an inductor. It is also a transformer because it has a low impedance tap near the bottom, so it can be analysed as an autotransformer. It must also be an impedance transformer. You can analyse the turns ratio to work out the voltage, current and impedance ratios.

5. Nov 18, 2016

### brainbaby

This is the main problem of the query. According to text its only 75 ohms impedance that is to be matched, but I don't know the other value of the impedance to which it should be matched i.e the input impedance of Q1, with out knowing this value I cannot determine the turns ratio. As turns ratio equals the square root of the impedance ratio...

So is there any way to determine the input impedance...!!

Last edited: Nov 18, 2016
6. Nov 18, 2016

### Baluncore

The circuit is all there. Model it with spice from antenna to the collector of Q1, then read the theoretical input impedance.
Does this receiver exist? If so, count the turns. If not, redesign L1 based on the spice model.
What type of signals and what frequency range is this VHF front-end supposed to cover?

7. Nov 19, 2016

### tech99

To manually calculate this conversion I would proceed as follows. Firstly, there is an auto transformer formed by the tapped coil. Find the turns ratio N and square it and so find the new parallel resistance across the coil, RLP = 75 N^2. Now take the reactance of the coil and find the loaded Q = RLP/XL approx. Now find the equivalent series resistance of the coil (in loaded condition) from RLS = RLP / Q^2 approx.
Now find the reactance of the 15pF shunt capacitor X15 (this capacitor operates in a parallel resonance with the series combined reactance of L and the 10pF capacitor, but you do not need to work this out). X15 has a series resistance equal to RLS as they are in a series circuit. Now find the reactance of X15. Find the loaded Q of X15 from Q = X15 / RLS. Now find the parallel equivalent across X15, given by RP = RLS x Q^2 approx. This is the resistance presented to the amplifier.
It is laborious and easy to make a mistake, as I probably have myself!

8. Nov 19, 2016

### Baluncore

Then all you now need to know is the L1 inductance, L1 turns ratio and the frequency of operation.

The resistive loading of the RF tuned circuit by the 1k AGC injection makes this look like a low-Q front-end, designed for wide-band TV reception, probably circa 1980.

Where does the schematic diagram come from ?

9. Nov 19, 2016

### tech99

You also need to know the reactance of the 15pF shunt capacitor, because this dictates the output resistance. And you are right that we also need to take into account the 1k shunt resistors, which I had not noticed.

10. Nov 19, 2016

### Baluncore

15pF, Xc is a simple function of the frequency of operation.

11. Nov 19, 2016

### Baluncore

brainbaby. Here are a couple of files for LTspice to get you going.
Remove the .txt extension to run the simulation.

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• ###### VHF_1.plt.txt
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12. Nov 20, 2016

### brainbaby

The circuit is taken from a book...
Monochrome and color telivision by rr gulati...
and BTW thanks for the files .....I'll get back to you soon....

13. Nov 22, 2016

### brainbaby

Hi Baluncore...
I have analysed your simulation files and it has been found very informative...but I want to find out the impedance at point 1,2 and 3....since the resultant graphs are in db/Hz.....I want something to be in V/hz for voltages and amperes for currents...

14. Nov 22, 2016

### Baluncore

The impedance at point 3 will be zero, at point 2 will be 75 ohm and at point 1 will be higher.
Here are two more LTspice files.
VHF_2 is improved model with cuttable link to see resonance.
VHF_3 is the resonant circuit only, with volts and amps.

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• ###### VHF_3.asc.txt
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15. Nov 23, 2016

### Merlin3189

Thanks for the helpful reference to the source. I can see how you're not too satisfied with his explanation for this part of the circuit. (Though the earlier chapter on RF stages helps a bit in general terms.) I notice it is also in "TV and Video Engineering" by A.M.Dhake (P.236), so it must be a common circuit.
At least we know now what range of frequencies it might cover (band 1 and maybe band3) and possibly what bandwidth is required (probably a single channel at 7MHz.)

As I said, I'm no expert in this area, just a suck it and see, codge and bodge amateur, so my thoughts for what little they may be worth:

Since the 10pF and 15pF in series is equivalent to 6pF,
at 50MHz to 70MHz L1 needs to be about 1600nH to 850nH
at 175MHz to 240MHz L1 needs to be about 145nH to 73nH

So if we took around 60MHz and 1200nH, and we need about 7MHz bandwidth, then we need a Q of about 8.5, which means a parallel resistance of about 3700Ω. (And this resistance will not change at other frequencies.)

Now I'm probably more in the dark than you about the input impedance of Q1, but we know that there is a maximum about 1kΩ resistance in parallel with the tuned circuit due to the DC biasing circuit. So I don't think they can achieve that Q. Maybe the double tuned collector circuit provides the necessary bandwidth shaping and they just put up with intermodulation effects in the RF stage?

Anyhow, if I guess they are looking at an impedance of around 1kΩ at the tuned circuit, and step up the input to that level.
Tapping L1 at 2:8 would give 1:16 impedance ratio, matching the 75Ω input with around 63Ω, or 2:7 tap giving 82Ω. (Or looking at it the other way, stepping 75Ω up to 1200Ω or 920Ω.)

I'd have thought, your 4nH:400nH tap seems a bit high (though I'm not sure how you've calculated this.) I'd have thought, this would match the 75Ω into about 7500Ω. That would be great if the other side of the circuit could match it, but otherwise seems unnecessarily high. (Or again looking at it the other way, would transform a low Q circuit down to too low an impedance to be a good match for 75Ω.)

I'm not into Spice at all, so I'll await your results on this. I wonder whether you can simply drive the Q1 circuit directly, leaving L1 and the two capacitors out of the equation for now? From your point of view (wondering about impedance matching and tuned circuit tapping) it might be interesting to know the raw load, then see how attaching a tuned circuit and tapping in different ways affects it.

If you like Spice, maybe even forget this circuit and simply set up a tuned circuit with a resistive load and see how driving and loading it via taps affects things. It would take away any variation of Q1 input impedance which might obscure things.

16. Nov 23, 2016

### Baluncore

Here is the tuned circuit, load 1k with BW = 4.5MHz.
Driven by a current of 1 amp, the input voltage is 75V, therefore Zin = 75 ohms.

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• ###### VHF_6.asc.txt
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17. Nov 24, 2016

### Baluncore

Better approximates reality by lowering turns ratio of inductor. Now fixed the centre frequency, bandwidth, loading and input impedance.

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• ###### VHF_9..plt.txt
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18. Jan 7, 2017

### brainbaby

Hi....guys

Moving on to the other part of the circuit let us quickly discussed what we have discussed so far.....

My initial query was to discuss the role of C1 and C2 and tapped coil L1 ..which is now clear to me..
Its just a kind of resonant circuit which resonates and provides selectivity and C1 and C2 acts as a impedance transformer which fools the load impedance and make it see the source impedance equal to its value.
Some were the missing values which was not present in the text..however I assume some of them from suggestions to carry forward my analysis...
Aim is to match 75Ω to 3.3k load resistance.....
As assumed...

Qul = 100 (unloaded Quality factor)
fc =100Mhz (center frequency)
B.W =20 mHZ (Bandwidth)
Rs=75Ω (source impedance)
Ns=2 (primary turns)
Np=8 (secondary turns)

Rint = Rs(n/n1)^2

= 75 X (8/2)^2

= 1200Ω

RL= Rint (1+Cx/Cy)^2

3333 = 1200(1+Cx/Cy)^2

C1= 0.66 C2

Qul = Rp/Xp

Rp = 100Xp

QL= Rtotal / Xp

Rtotal = Rp//Rs//RL / Xp (Rs-source impedance, RL-load impedance, Rp- parallel resistance of inductor)

4.9 = Rp // 882.32/Xp

4.9 = Rp.882.32 / (Rp+882.32) Xp

Substituting value of Rp

Xp = 171.24Ω

Xp = 1/wCT

171.24 = 1/2pi(100 x 10^6)CT

CT= 9.29 pf

CT = Cx.Cy/Cx+Cy

9.29 = 0.66 C2. C2/0.66C2+C2

Cx=23.36 pf

Cy=15.41 pf

Now moving further in the tuner circuit...

I want to know what is the function of L2,L3,L4 and C2 and C3.....???

The text says that is a double tuned mutual coupled via L4 network....but how can it be mutual coupled since L4 is not physically separated from the circuit in order to be mutually coupled....??

Special thanks to Baluncore and Merlin3189!!

19. Jan 8, 2017

### tech99

The coils are presumed to be spaced apart so there is no mutual inductive coupling between them.
Then we find that the currents in the two coils L2 and L3 both flow through L4, which provides "bottom coupling" by providing a common impedance.
The inductance L4 acts the same as the mutual inductance between two magnetically coupled coils. However, its inductance is added to both L2 and L3, which alters the resonant frequency.
Assuming L2=L3, then Coupling Coefficient is L4 / (L3+L4). Radio Engineers Handbook, Terman, p 164.
For critical coupling, kQ = 1. This give the flattest response.

20. Jan 8, 2017

### brainbaby

Apart from the book which you have mentioned ...from where else I can get an elaborative explanation of common impedance matching....

21. Jan 8, 2017

### tech99

22. Jan 8, 2017

### davenn

I don't really agree with that
L2 and L3 are part of a transformer T2 and therefore are most likely to have inductive coupling

don't really agree with that either
It is standard practice to put inductors in the DC lines feeding a stage. L4 along with C1 decouple the DC rail from the RF signal that it feeds
This is to stop RF (AC) from going out ion the DC rail .... If allowed to, it would cause havoc as it caused oscillations on the DC rail

Dave

23. Jan 9, 2017

### tech99

Of course, we do not know if L2 and L3 have mutual inductive coupling, but from a design point of view it is undesirable. It would require close mechanical tolerances, or adjustment by moving the coils little-by-little. It is better to have shielded and isolated coils and use a known value of mutual coupling provided by L4. It is possible to use a combination of mutual and bottom end coupling, but there seems no point in doing this.
L4 is not used as a decoupling component. It is undesirable to use an inductor for this purpose here because of unwanted inductive coupling. Further, if it were used for decoupling purposes, the capacitor C4 would be placed at the bottom of L4, where its effect would be greater. The 390 Ohm resistor is a decoupling component.

24. Jan 9, 2017

### davenn

because they are both part of a transformer ..... it's 99% probability there is inductive coupling between them

pointless comment ..... the tuning adjustment IS PROVIDED .... L2 is adjustable

absolute rubbish .... coupling between stages using inductors like that is a common practice, there are a number of variations used for interstage coupling. This is just one of them

..... maybe you are not a RF tech ?

again not true ... it's standard practice

seriously .... for the 3rd time ... Its standard practice

Dave

PS ... I am happy to be proved wrong by some one more knowledgeable in RF than either of us

Last edited: Jan 9, 2017
25. Jan 9, 2017

### davenn

some examples of you so called ineffective decoupling that work very well

note particularly C4 and C6 ^

Note location of C4, C5, C6, C7 like the circuit above, they are on the DC side of the RFC inductor

OK ... I think I have made my point