Help: vhf front end

[hamopp]

Hi All.
I have a circuit of a 2m rx front end which works well, the output is fed
into a 9mhz if stage.

What i would like, is to be able to have this circuit modified to work in the
4m(70Mhz) and 6m(50Mhz) bands, I know that some components will have
to be changed and i have marked the parts with a * .

The front end coils in the 2m one are from Coilcraft (146-04J08SL), i also
have these coils from them:
2x 146-06J08SL 6.5t
2x 146-07J08SL 7.5t
2x 146-08J08SL 8.5t
2x 146-10J08SL 10.5t

data at https://www.coilcraft.com/uni7.cfm

Can anyone help with the values of the parts that need to be changed.

Regards
Howard

 

[MikeMl]

For starters:

NewCap = OldCap*(OldFreq/NewFreq)

NewInd = OldInd*(NewFreq/OldFreq)

This will keep the Qs about the same.

To get the specified inductances, start with coilcraft's data sheets. If you have to wind them yourself, an inductance meter is indespensible...

 

[hamopp]

For starters:

NewCap = OldCap*(OldFreq/NewFreq)

NewInd = OldInd*(NewFreq/OldFreq)

This will keep the Qs about the same.

To get the specified inductances, start with coilcraft's data sheets. If you have to wind them yourself, an inductance meter is indespensible...

Hi

NewCap = OldCap*(OldFreq/NewFreq)

Do you mean for NewCap, divide 145.000Mhz(OldFreq) by 70.000Mhz(NewFreq) = 2.0714285
then multiply OldCap(3.3pF) by 2.0714285 = NewCap(6.835714) or 6.8pF.

NewInd = OldInd*(NewFreq/OldFreq)

Do you mean for NewInd, divide 70.000Mhz(NewFreq) by 145.000Mhz(OldFreq) = 0.4827586
then multiply OldInd(.146uH) by 0.4827586 = NewInd(.0704827) or .071uH

Is that Right ???

Regards
Howard

 

[MikeMl]

Yes on the the capacitor. I screwed up the inductance change. It should increase, too.

For a two to one decrease in frequency, both the inductor and the capacitor must increase by a factor of two. Square root of (2*2) =2...

The corrected inductance change is:
NewInd = OldInd*(OldFreq/NewFreq)

 

[hamopp]

Yes on the the capacitor. I screwed up the inductance change. It should increase, too.

For a two to one decrease in frequency, both the inductor and the capacitor must increase by a factor of two. Square root of (2*2) =2...

The corrected inductance change is:
NewInd = OldInd*(OldFreq/NewFreq)

Hi
Here are my calc's as to the Formula's you gave:

My Calc's.txt

Do these look right ??

Regards

​

 

[MikeMl]

Looks ok,

You might have to increase some of the coupling caps that couple in the LO injection.

 

[hamopp]

Looks ok,

You might have to increase some of the coupling caps that couple in the LO injection.

Hi
Well last night i got the 70Mhz(4m) version going, seems to be ok

Howard