0
\$\begingroup\$

I have just realized that I actually do not have a through understanding of capacitors and hence, I am asking this basic question.

I was trying to understand how an astable multivibrator works using this source and I have come across the following part (diagram is in the link, and below):

enter image description here

...

The next significant thing which happens is when the voltage on the right hand plate of C1, and therefore TR2’s base, approaches +0.6V.

TR2 begins to conduct.

The collector voltage of TR2, which by now will have increased in value as C2 started to charge through R4, will fall and go down to +100mV.

The voltage across C2 can’t change instantaneously so when the voltage at its right hand plate goes down the voltage at its left hand plate goes down by the same magnitude. If TR2’s collector voltage had risen to, say, +3V, then when it reduces to +100mV (a negative-going change of 2.9V) the left hand plate of C2, and the base of TR1, will also go 2.9V more negative, but starting from +0.6V.

They will end up at –2.3V. This will cut TR1 off.

...

So, as far as I understand, the following happens:

\$ C = Q / V \$ equation must hold for capacitors at all times. When we consider a particular capacitor, \$ C \$ is constant. This means that only \$ Q \$ and \$ V \$ may change in order to make the equation hold at all times.

However, there is a quirk with the capacitors. In a capacitor, \$ Q \$ cannot change instantaneously. That is, it takes time to change \$ Q \$.

Hence, when the voltage at one plate of a capacitor undergoes a sudden change (while the voltage on the other plate remains untouched), this event creates a situation where the equation \$ C = Q/V \$ does not hold. The reason is, \$ V \$ has changed, while \$ C \$ and \$ Q \$ are the same. (\$ Q \$ is the same because charge on a capacitor cannot change instantaneously)

However, we know that the equation \$ C = Q / V \$ must hold at all times. Hence, the capacitor responds this change by altering the voltage at its other plate, putting the equation \$ C = Q / V \$ in balance once again.

Here comes my question: Since the capacitor has changed V at its other plate in response to a sudden voltage change on the opposite plate, the equation \$ C = Q / V \$ is once again satisfied. However, we observe that the capacitor starts charging or discharging after this event.

Since the equation \$ C = Q / V \$ has been put in balance by the voltage change at the opposite plate, the reason for (dis)charging cannot be to balance the equation.

So if a capacitor does not (dis)charge in order to satisfy the equation \$ C = Q / V \$, then why does a capacitor (dis)charge?

\$\endgroup\$
  • 1
    \$\begingroup\$ Capacitors (dis)charge because of current flowing through them. Your circuit starts to force current through the capacitor after the change. \$\endgroup\$ – Dmitry Grigoryev Apr 6 '16 at 11:44
  • \$\begingroup\$ @DmitryGrigoryev But why? The reason cannot be to re-balance the equation \$ C = Q / V \$. Because by the voltage change at the opposite plate of the cap (in response to the voltage change in the first plate), the equation has already been (re)balanced. Then what causes this charge flow, if not in order to satisfy the equation \$ C = Q / V \$? \$\endgroup\$ – Utku Apr 6 '16 at 11:51
  • 1
    \$\begingroup\$ Why? Because of Kirchhoff's current law. \$\endgroup\$ – Dmitry Grigoryev Apr 6 '16 at 11:56
  • \$\begingroup\$ @DmitryGrigoryev Could you elaborate with an answer please? \$\endgroup\$ – Utku Apr 6 '16 at 12:01
1
\$\begingroup\$

The basic rule is you can't instantly change the voltage ACROSS a capacitor.

If you suddenly change the voltage on ONE PLATE then to maintain the voltage across the capacitor the other plate must instantly rise by the same amount.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Yes, I understand this. And AFAIU, this is the reason why an astable multivibrator works. But the part I don't understand is, why does the capacitor charge in your second picture? Because in all pictures, the equation \$ C = Q / V \$ holds. So the reason the cap starts charging in the second picture cannot be to (re)balance the equation \$ C = Q / V \$. Then why does a capacitor (dis)charge? \$\endgroup\$ – Utku Apr 6 '16 at 11:47
  • 1
    \$\begingroup\$ The capacitor charges or discharges because the other plate is connected through a resistance to a different voltage. This voltage difference across the resistor will produce a current (I = V/R). In the case of the astable one side of the plate drops from +V to 0V so the other plate (connected to the base) must drop from 0.6V (clamped by the base-emitter junction) to - (V+0.6) volts. This plate is connected through a resistor to +V so it tries to charge up to that voltage. It gets up to 0.6V and turns on the transistor and gets clamped at that level. This turns OFF the other transistor. \$\endgroup\$ – JIm Dearden Apr 6 '16 at 11:56
  • \$\begingroup\$ So the capacitor charges because there is a current going into it due to Ohm's Law? \$\endgroup\$ – Utku Apr 6 '16 at 12:23
  • \$\begingroup\$ The flow of charge (current) is produced because there is voltage differnce across the ends of the resistor and the size of the current is calculated using Ohm's law. (I = V/R). As the capacitor charges, the voltage on the plate increases and the voltage across the resistor decreases, causing the charging current to decrease with time. (exponential curve). After about 5 time constants (T=CR) the capacitor is taken to be fully charged. Note that the plate on the capacitor NEVER fully reaches the final voltage. \$\endgroup\$ – JIm Dearden Apr 6 '16 at 12:33
  • 1
    \$\begingroup\$ Yes, Q cannot change instantly (this would require an infinite current which is clearly not possible) and that is why the voltage across the capacitor cannot change instantly as this would violate the equation Q = CV ( V being the voltage across the plates and not the voltage on one plate). \$\endgroup\$ – JIm Dearden Apr 6 '16 at 13:03
0
\$\begingroup\$

This statement: -

In a capacitor, Q cannot change instantaneously. That is, it takes time to change Q

Is contradicted by this: -

the voltage at one plate of a capacitor undergoes a sudden change

Because: -

You cannot change the voltage instantaneously without infinite current being sunk into the capacitor. If infinite current is sunk (or sourced) by the capacitor then Q will change instantaneously.

The remainder of your question is based on this incorrect (or contradictory) premise.

Q = CV and the rate of change of Q (dQ/dt) = C dV/dt.

This of course equals current because rate of change of charge IS current.

Therefore if voltage changes instantly, dV/dt is infinite therefore current flow is infinite.

\$\endgroup\$
  • \$\begingroup\$ Then how do we explain the sudden voltage drop (to negative voltages) in an astable multivibrator? \$\endgroup\$ – Utku Apr 6 '16 at 11:38
  • \$\begingroup\$ There is no sudden voltage drop - you are misled by your observations or, the practicality of a real capacitor having ESR and ESL, is causing you to misread the signs. \$\endgroup\$ – Andy aka Apr 6 '16 at 11:39
  • \$\begingroup\$ The source says: "The voltage across C2 can’t change instantaneously so when the voltage at its right hand plate goes down the voltage at its left hand plate goes down by the same magnitude. If TR2’s collector voltage had risen to, say, +3V, then when it reduces to +100mV (a negative-going change of 2.9V) the left hand plate of C2, and the base of TR1, will also go 2.9V more negative, but starting from +0.6V. They will end up at –2.3V. This will cut TR1 off." How does this happen then? \$\endgroup\$ – Utku Apr 6 '16 at 11:42
  • \$\begingroup\$ @Utku - I have no idea what circuit that quote is connected with. \$\endgroup\$ – Andy aka Apr 6 '16 at 11:46
  • \$\begingroup\$ Astable multivibrator circuit. It is explained in the link I have given and there is a circuit diagram in the link as well. \$\endgroup\$ – Utku Apr 6 '16 at 11:48
0
\$\begingroup\$

V is voltage across the capacitors plates, and that voltage can not be changed instantly. BUT potential of the plates can be changed instantly. When one plate of the capacitor, V1 ,is charged with some Q and because there is an electric field between plates, potential of other plate of the capacitor,V2, is instantly changed with same difference as first plate ( that makes V=V1-V2=Const.) When that process is done, the charge of the other plate V2 is going to flow through the resistor making a current with exponential form (because potential of that plate is falling down--that is discharge).

I hope this will help you to understand difference between voltage across capacitor and potentials of the plates, and what is discharge.

Charge of the capacitor is the same, if first plate of capacitor is connected to some voltage source via resistor (then you would have some current flow, and current is bringing charge to the plate of capacitor). At same time other plate is changing it potential making voltage across capacitor constant.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your answer. I understand that when the voltage at a plate undergoes a sudden change, the voltage on the other plate undergoes the same amount of change as well. However my question is, after these changes have occurred, why do we observe a current to or from the capacitor? \$\endgroup\$ – Utku Apr 6 '16 at 13:25
  • \$\begingroup\$ hmm, current flows whenever there is a difference between potentials (voltage difference). We need to observe this currents because there is no currents through capacitor (inside is dielectric), but the changes of potentials at the plates are forming currents to and from capacitor. Maybe I don't understand question wholly, sorry. \$\endgroup\$ – Haris778 Apr 6 '16 at 13:35
0
\$\begingroup\$

As Q=CV, the charge storage will depend upon the capacitance value which is factory built and specifies and also the voltage applied across its terminals.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.