How does a Capacitor remove DC bias?

Question

If you put it at the base of a transistor amplifier, it removes the DC bias somehow. Can you show me the equation?

What? Is everyone on this site an IDIOT or what?

HOW? You can't just say "becaise"

Answer 1

caps dont let dc pass. no equation needed.
We graduated, maybe you are the idiot..
MIKE

Answer 2

A capacitor is made up of two conductive plates (with a lead attached to each plate) which are seperated by a thin dielectric material (Insulator)

There isn't a direct electrical connection between the plates because of the insulating properties of the dielectric. It is possible to arc across the dielectric if a high enough voltage is applied. (Dielectric Breakdown)

The amount of capacitance is directly proportional to the surface area of the plates and inversely roportional to the distance between the plates.

So how does a capacitor conduct.....

If you apply a DC voltage across a capacitor there will be an initial surge of current as charge is built up on the plates. This current flow lasts as long as it takes for the charge to build up to equal the applied voltage. Once the charge has built up the current flow stops.

With voltage applied to the cap, one plate is positive while the other is negative. Electrons from teh battery try to complete a circuit to the positive battery terminal. An electron entering the capacitors negative plate soon discovers its path blocked by the dielectric. But due to its close proximity to the other plate the electron "SEES" positive charges on the other plate and due to opposite charges attracting it tries to go across the dielectric (but its blocked). Also at the same time as this electron is pushed onto the negative plate by the battery a positive charge is being pushed onto teh positive plate.

Since equal charges repel any electrons on the positive plate see a build up of electrons on the negative plate. The electrons on the negative plate tends to push any electrons on the positive plate away.

This pushing and pulling of charges contnues until the charge buils up on the plates to such an extent that the charges see an equal pressure from the opposite plate and the battery. This is when the charge equals the applied voltage and the movement of charges (current) stops.



Current can also flow out of the capacitor, so if i reverse the DC polarity. The original built up charge will leave the capacitor (current flow) and a new charge which matches the new polarity will build up on the plates. Again this charge build up until it matches the input voltage then stops.. I can also get current flow if I change the level of applied voltage. The charge on the cap will change so as to match the applied voltage.

If you Alternate the switching of the applied voltage to the capacitor (AC) you will have some current flow every time the polarity reverses. Do this fast enough ( higher frequency) and you will reach a point where current is constantly flowing into and out of the cap.

When AC is biased with DC the two voltages combine (add together) Add enough DC to an AC signal and you can have a varying voltage which never reverses direction. (Varying DC)


Capacitive reactance in Ohms(Xc)
= 1/(2*pi*Freq* Capacitance)

Answer 3

I'm assuming you mean a capacitor in series with the base. Caps are a DC block in series. The equation that justifies this statement is the equation for the impedence of a cap. Xc=1/2pi(f)C. Since the frequency of DC is 0, then the equation becomes infinity, therefore a cap won't pass DC. As the frequency increases, the impedence becomes lower and lower eventually approximating a short circuit, so AC will pass.