# Help:vhf front end

Discussion in 'Wireless & RF Design' started by hamopp, Apr 11, 2013.

1. ### hamopp Thread Starter Active Member

May 13, 2009
68
1
Hi All
Could you please have a look at this attachment, it is for a 2m front end
it works ok.
I would like to make the same circuit work on 6m and 4m, i know that
there are components to change, i think that these parts need changing:
L101 and L102 (Coilcraft coils)
C101,C102,C103 and C104
But not sure of the values to use, I would like to use Coilcraft Coils.

TIA
Regards
Howard

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2. ### vk6zgo Active Member

Jul 21, 2012
677
85
Have a look at other circuits for those frequencies & steal their ideas!

Also,ask the same question on QRZ.com-----lots more Hams there.

3. ### Slarsen New Member

Apr 29, 2013
12
4
Capacitors C101 and C103 are roughly connected in parallel (when the antenna is connected) and they resonate with inductor L101. To confirm this we use the LC resonance formula:

f = 1 / (2*Pi*SqrRoot(L*C))

C = 3.3pF + 6.8pF = 10.1pF
L = 110nH

Plugging in and solving for frequency f, we get 151 MHz. That is very close to the 2 mt band and this confirms what I suspected.

If you do the same procedure with C104, C105, and L102, you will find that they too resonate at the 2 mt band.

To make the two LC networks resonate in the 4 mt band, you need to double the denominator of the resonance formula. (And triple for the 6 mt band.) Here you have a choice of raising the capaitances, inductances, or both. For example, double C101 to 6.6pF and C103 to 13.6pF, and double inductor L101 to 220 nH.

Of course you have to pick part values that are available. I would start by picking an inductor close to 220nH. Then use the resonance formula to determine the correct total capacitance that resonates with that inductance. Finally pick two capacitors that add up close to the correct amount.

If you can't get an Coilcraft inductor that high, then increase the capacitances more-so than the inductance.

As for capacitor C102, that is just a DC decoupling capacitor I believe. I would double its value for 4 mt and triple it for 6 mt. That way its reactance remains the same in all three cases. BTW, the reactance formula for capacitors is

X = 1 / (2*Pi*f*C)

where X is the capacitive reactance in ohms, f is the frequency, and C is the capacitance. So as you can see, if you wish the reactance to remain the same at 4 mt (half the frequency), the capacitance must be doubled.