variable cap question

[p.larner]

i have a 1000pf air variable cap,its min capacitance is to high for my needs,i was thinking of removing some plates to reduce it,its a nice cap so seems a pity to butcher it,if i put say a 175pf air variable in series with it,would that lower the overall min c?

[Kim Christensen]

Figure out it's minimum value, and the minimum value you require. Then do the math, to figure out the value of a 2nd series connected capacitance. Then do the math again for max variable value and see if that works for your application.

[p.larner]

that doesent answer my question,ie if cap 1 has a min of 50pf in series with a cap that on its own has say a 20pf min,if they are both connected in series will the total min c be less than the first caps min on its own ie 50pf.?

[bdunham7]

i have a 1000pf air variable cap,its min capacitance is to high for my needs,i was thinking of removing some plates to reduce it,its a nice cap so seems a pity to butcher it,if i put say a 175pf air variable in series with it,would that lower the overall min c?

The answer is "yes" but you'd have to tell us what your desired minimum capacitance is and what the minimum capacitance of the variable cap is to know if you'll get down to the number you need.

The math for series capacitors is the reciprocal of the sum of the reciprocals.  So for 50pF and 20pF, you get 1/(1/50 + 1/20) or 14.3pF.

[Vovk_Z]

i have a 1000pf air variable cap,its min capacitance is to high for my needs,i was thinking of removing some plates to reduce it,its a nice cap so seems a pity to butcher it,if i put say a 175pf air variable in series with it,would that lower the overall min c?

You don't really need another variable cap in series. You may use just a regular constant capacitance small cap (IDK, e.g. NP0/C0G ceramics).
For example, with another 1000 pF in series, you'll have 1000/2 = 500 pF maximum capacitance. And minimum - depends on the minimum of your variable cap.
For, example, if your air variable cap has a 100 pF minimum capacitance then in series with this new 1000 pF it will be about 91 pF in series (the new minimum).
With a 510 pF additional cap in series with your 1000/100 pF cap, you'll have 338 pF maximum and 84 pF minimum.
With a 330 pF additional cap in series with your 1000/100 pF cap, you'll have 248 pF maximum and 77 pF minimum.
With a 100 pF additional cap in series with your 1000/100 pF cap, you'll have 91 pF maximum and 50 pF minimum.
1/Cnew = 1/C1 + 1/C2.

[wofritz]

The problem with a fixed serial capacitor is that it reduces Cmax much more than may be desired.

Example: Cmax = 1000 pF, Cmin = 100 pf. If you want Cmin to be 50 pF, you need a serial C of 100 pF.

That gives Cmax of only 91pF. Rule of thumb: Resulting C of capacitors in series is always smaller than the smallest of the single capacitors.
 

[EPAIII]

In a word, YES. Two capacitors in series will have a value that is less than either of them individually. The formula is similar to the one for parallel resistors:

C = 1 / ((1/C1) + (1/C2))

Spoken verbally it is the reciprocal of the sum of the reciprocals.

Using your example values,

C = 1 / ((1/20pF) + (1/50pF))

C = 1 / ((0.05 pF^−1) + (0.02 pF^−1))

C = 1 / (0.07 pF^−1)

C = 14.29 pF

that doesent answer my question,ie if cap 1 has a min of 50pf in series with a cap that on its own has say a 20pf min,if they are both connected in series will the total min c be less than the first caps min on its own ie 50pf.?

[p.larner]

my most used band is 80m,should i get a match there on my lowest used freq and measure the c of the cap to get an idea of the max c i need and go from there?.

[Andy Chee]

To provide an answer, we need to know the inductance of your L as well.

[p.larner]

have no way handy to measure the rollercoaster its about 30 turns 2.5 inch dia,i would imagine its around 30uh?,the main cap value tunned to my most used 80m frequency is 275pf when its a 1-1 match,if that helps,i guess i need a series cap that gives me say 280pf max and test it on the higher bands,what value series cap would do that?,cheers.

[Andy Chee]

You can use your VNA to find the resonant frequency of your L/C. 

Given you already know C, and now you just measured f, you can calculate L using the formula f = 1/(2*pi*sqrt(L*C)).

[Vovk_Z]

,i guess i need a series cap that gives me say 280pf max and test it on the higher bands,what value series cap would do that?

Have you read my answer? 🤔
I guess every radio-DIYer must have a set of 1 pF - 1000 pF NP0 ceramic capacitors.

[p.larner]

so i have a 20kv 500pf dorknob and a 1kv 100pf silver mica would the mica cap in paralell with the 500pf dorknob both in series with the variable be near what i need?

[p.larner]

using an online calculator if i have the variable set half mesh  so 500pf  with the 100pf silver mica cap in parelell with a 500pf doorknob,with that pair in series with the main cap,it is coming back with the c as 272pf,that would be with the main variable at half mesh,does that sound ok?,or do i remove half the plates and switch in the 500pf doorknob when needed?.

[807]

For info, an easier way to calculate 2 capacitors in series is...Product over sum

i.e. (C1 x C2)/(C1 + C2). For 20pF & 50pF that would be:-

1000/70 = 14.3pF.