Impedance matching in a Vhf tuner circuit...

Thread Starter

Himanshoo

Joined Apr 3, 2015
265
Hi guys..

The following is an initial part of a transistor Vhf tuner involving a balun used to balance an unbalanced line and an impedance matching network consisting of the tapped coil L1 and shunting capacitor of 10 pf and 15 pf matching a 75Ω antenna impedance to the rest part of the circuit ..

My queries are..
1. How does this impedance matching network works..??..Why two capacitors are used instead of one..??
2. What is the role of 470 pf capacitors ??
 

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Ylli

Joined Nov 13, 2015
1,086
The 470 pF caps are there to prevent any DC on the lines from entering the tuner. And vice versa.

The two caps across the coil is very similar to adding a tap to the coil. It is often more convenient that actually trying to add the tap to a small coil, particularly if that coil already has one tap.
 

ian field

Joined Oct 27, 2012
6,536
The 470 pF caps are there to prevent any DC on the lines from entering the tuner. And vice versa.

The two caps across the coil is very similar to adding a tap to the coil. It is often more convenient that actually trying to add the tap to a small coil, particularly if that coil already has one tap.
75R impedance suggests consumer equipment.

Certainly in the case of TV sets that used series chain tube heaters, it was common practice to tie one of the mains wires to chassis - if you used a 2 pin mains plug; there was a 50% chance of the chassis being live. The early solid state sets were no better, the SMPSU would be pretty basic and at best the chassis would be at the negative end of the bridge rectifier.

A tube radio was more likely to have an isolating mains transformer, and I can't remember seeing any solid state mains radios that didn't.
 

Thread Starter

Himanshoo

Joined Apr 3, 2015
265
The two caps across the coil is very similar to adding a tap to the coil. It is often more convenient that actually trying to add the tap to a small coil, particularly if that coil already has one tap.
Literally I don't know the technicality of the circuit as how its working for each instantaneous value of the rf signal coming from the antenna to the later part of this circuit..all I know is that the tapped coil and the two capacitors forms a parallel LC impedance matching circuit and at some point of resonance the value of the impedance of the LC component goes so high so that no current flows through them and the voltage is directly coupled to the other part of the circuit.

So I would appreciate a detailed working of this matching circuit for each RF half cycle of input signal voltage..
 

Ylli

Joined Nov 13, 2015
1,086
Without knowing quite a bit more about the circuit, an in depth explanation would be rather tough. To start, I would determine the load on the network. Add the 15 pF cap in parallel with the load. Convert to a series network and determine the equivalent capacitance with the 10 pF in series. Convert back to a parallel network and put it across the inductor. Find the resonance frequency. Determine the tap location on the coil. Approximate the impedance transformation for the coil tap.

What else would you like to determine?
 
The impedance matching is all done in the BALUN. The unbalance match may not even be 75 ohms.

At one point the tube sets had filaments that added up to about 120 V, so you had 37 V filaments and neutral went to chassis ground. They were fitted with polarized plugs. An isolation transformer was a must when working on these sets.
 

Thread Starter

Himanshoo

Joined Apr 3, 2015
265
If we shift our discussion from balun to impedance matching network L1 acts as an auto transformer with turns ratio that is equal to the square root of impedance ratio.
In impedance ratio one impedance is 75 ohm and other is the input impedance of Q1, but the problem is that I don't know the value of input impedance hence without this value one cannot get the turns ratio, therefore without this other values of circuit elements cannot be determined....

So how can I be able to calculate the input impedance..in order to calculate the turns ratio ??
 
Last edited:

Ylli

Joined Nov 13, 2015
1,086
I'd measure it with a network analyzer. If you can come up with an appropriate model for the input transistor, you may be able to throw the circuit into simulation and come up with a close answer. Calculate it by hand? I haven't a clue.
 

Thread Starter

Himanshoo

Joined Apr 3, 2015
265
A full analysis is a bit complicated and beyond my ability to explain it. Do a search on "tapped capacitor impedance matching" and you will get pages such as http://analog.intgckts.com/impedance-matching/tapped-capacitor-matching/ that will take you into the math.
The network is 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.

Here is the complete analysis
Aim is to match a source resistance of 75 ohm to a load resistance of 3.3k and calculating the value of the capacitors involved in impedance transformation...
As assumed...it is given...

QL=4.9 (loaded Quality factor)
Qul = 100 (unloaded Quality factor)
fc =100Mhz (center frequency)
B.W =20 mHZ (Bandwidth)
Rs=75Ω (source impedance)
RL=3.3k (load 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.......................................(i)


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......................................(ii)


Substituting value of Rp from equation (i) in (ii)


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

thanks....
 
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