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Okay so I'm looking at RF designs and I've come across a couple of designs which need 2p2 caps.

I've seen

C1.C2/C1+C2

being plugged into a function (the tank part of the circuit) which gives a lower capacitance for the circuit i.e. C1 && C2 == 27pf (gives 13.5pf if tap taken between C1 and C2)

My question is this:

Lets say I have + 10pf ->|-> 10pf -> 10pf

Where | is the tap does that mean I can somehow get a lower capacitance by tapping in a series of caps? (please show me the equation if poss :))

Paul Sullivan
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2 Answers2

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Capacitors in series are like resistors in parallel, to get the total capacitance add the reciprocals, then take the reciprocal of the result.
For example, if we have 20pF, 10pF, and 5pF caps, the total capacitance would be:

1 / ((1 / 20pF) + (1 / 10pF) + (1 / 5pF)) = 2.857pF

For two capacitors you can use the (C1 * C2) / (C1 + C2) but this won't work for 3 or more. 

                 20pF 10pF 5pF
Connect here -> -||---||---||- <- and here

                    2.857pF
Equivalent to:  ------||------

Note that putting multiple caps in series probably won;t work very well at such a low value (there will be extra unwanted inductance which will lower the resonant frequency)
2.2pF is actually a very common value, for example Farnell has 46 options under RF caps and 94 options under ceramic caps. I advise picking one of these.

Oli Glaser
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  • ... So are you telling me that by just putting them in series actually lowers the combined capacitance Oli? i.e. where do I tap or do I just take it off 'the end' + C1 -> C2 -> C3 - ... do I tap? – Paul Sullivan Aug 26 '12 at 22:15
  • Yes, putting them in series will lower the total capacitance. You connect at either end (i.e. free lead of C1, and free lead of C3) – Oli Glaser Aug 26 '12 at 22:20
  • @PaulSullivan - see bad attempt at ascii art, hopefully it's clear enough ;-) – Oli Glaser Aug 26 '12 at 22:26
  • Hahah thanks Oli - your ascii art is a much appreciated hark back to my c64 and earlier days - marked as answer and see I need to go back to text books. It's a bit counter intuitive really as I expected greater dielectric surface area to = more capacitance. – Paul Sullivan Aug 26 '12 at 22:40
  • AHHHH! Now I see there are only so many charge carriers to each side of each part of the capacitor so thts why its reduced... obvious when I get my brain in gear! (I think...) Why does this happen?!? – Paul Sullivan Aug 26 '12 at 22:48
  • If you think about it, the way to get greater surface area is to put them in parallel (in parallel = bigger total plate area) Have a look at the edit for source of 2.2pF caps, I'd go for a single cap for RF purposes. – Oli Glaser Aug 26 '12 at 22:49
  • You could try some in series and see how it goes (might as well while you're waiting for the others to arrive), just be aware it's not the "ideal" solution. How it affects things depends on the circuit, it may be fine. – Oli Glaser Aug 26 '12 at 22:58
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You already provided the formula to calculate the capacitance of two capacitors in series: $$C_{1,2}=\frac{C_1 \cdot C_2}{C_1+C_2}$$

When you want to apply the formula to three capacitors (\$C_1, C_2, C_3\$) in series, think it this way: Combine the capacitors in groups of two. First calculate $$C_{1,2} = \frac{C_1 \cdot C_2}{C_1+C_2}$$ then calculate as if you have a capacitor with capacitance \$C_{1,2}\$ and \$C_3\$ in series: $$C_{1,2,3} = \frac{C_{1,2}\cdot C_3}{C_{1,2}+C_3}$$

So for your example of 3 capacitors each 10pF: $$C_{1,2} = 5pF = \frac{C_1 \cdot C_2}{C_1+C_2} = \frac{10pF \cdot 10pF}{10pF+10pF}$$ $$C_{1,2,3} = 3.33pF = \frac{C_{1,2}\cdot C_3}{C_{1,2}+C_3} = \frac{10pF \cdot 5pF}{10pF+5pF}$$

PetPaulsen
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