by **R.G.** » 22 Jan 2008, 18:22

Let me try.

A capacitor has no frequency response of its own. All it has is a "resistance" to AC that decreases as frequency increases. So a cap may "resist" by the equivalent of 100 ohms at one frequency. If you double the frequency, the same cap now "resists" by 50 ohms. If you half the frequency, the same cap now "resists" by 200 ohms.

I say "resists" in quotes because it's not really resistance the same way a resistor does it. The work "impedance" was coined because the guys who messed with electricity first had used up "resistance" for what resistors do, and they needed another word for getting in electricity's way but different in some respects to a resistor.

Meanwhile, back at the capacitors. If you keep lowering the frequency you put through a cap, the "resistance" (impedance) to current flow goes up and up, until finally as the frequency goes to 0 - that is, it's DC, not AC - the impedance is infinite, and it totally blocks current flow. We use caps a lot to block DC but let AC flow through, and that's how they do it.

But if we have a resistor and a cap in series, what happens?

At DC, the capacitor blocks all current flow, so the resistor has nothing to say about it. It's blocked and the resistor doesn't matter.

If we use very low frequency AC, the "resistance" (impedance) of the capacitor is still very large, so the capacitor only lets a trickle of AC current through, and the resistor matters very little.

You can see where this is going, right? As we raise the frequency, the cap "resists" less and less. At very high frequencies, the cap's "resistance" (impedance) is so much smaller than the resistor that the current is determined entirely by the resistor, and the cap conducts so well at the high frequency that it might as well be a wire.

The frequency where the cap and resistor both matter is what the EEs take as the "rolloff frequency" of the combination.

And now we have to get to the - very little - math. You have to be able to calculate the frequency from the resistor and capacitor.

Trust me on this: the impedance of a capacitor is Xc = 1/(2*pi*F*C).

Xc is used to mean Impedance (X instead of R, cause they're a little different) subscript c (meaning, belonging to the capacitor, not something else).

"pi" is the number 3.14159; it's just a constant number that happens to make the calculations come out to the right numbers. Just memorize it. 3.14 is generally good enough for calculations.

F is the frequency in cycles per seconds, or Hertz, same thing.

C is the capacitance in Farads; a microfarad is one one-millionth of a Farad. A pico farad is a millionth of a microfarad.

Since we have to calculate the frequency where the impedance of the cap equals the resistance, we just say

R= 1/(2*pi*F*C) - I apologize, this bit of math is critical

and the frequency is then

F = 1/(2*pi*R*C) and that's how you calculate it.

You multiply the Resistance in ohms times the capacitance in farads, then multiply by 2, and finally multiply by 3.14, and divide 1 by the result. The number you get is the frequency in cycles per second or Hertz.

An example: you have 10pF cap and a 100K feedback resistor. What's the frequency where the cap starts letting more current through than the resistor?

It's the frequency where the cap's impedance is equal to the resistance. So

it's 1/(2*pi*100K*10pF).

10pF is 10 times ten to the minus twelfth power, or 1x10^-11, the "^" meaning "to the power of"

so the frequency is 1/( 2* 3.14* 100,000* 1x10^-11) and if I punched the calculator buttons right, it's 159,235 Hz, or a bit over 159kHz.

(a) did that help?

(b) OK, where did I mystify you?

- For this message the author R.G. has received thanks:
- nzCdog (24 Jan 2011, 05:55)