capacitor in coupling & decoupling

Thread Starter

nouredine london

Joined Mar 31, 2018
4
good morning

I want to understand why we put coupling capacitors and decoupling capacitors, and what do they bring back to our circuit and how do these functions without the current going through them because the capacitor is a component that does not let the current flow
 

Jony130

Joined Feb 17, 2009
5,182
Only when capacitor is fully charged to the supply voltage the current won't flow .

But in general current in the capacitor is proportional the rate of voltage change across it (proportional to how quickly the voltage across the capacitor is changing). The faster the voltage change (frequency of an AC signal is high) the large the current flow through the capacitor.

I = C∗dV/dt


saGgG.png


This means that to sustain current through a capacitor the applied voltage must change. The more rapidly voltage changes the larger the current. On the other hand, if the voltage is kept constant no current will flow no matter how large the voltage. Likewise, if the current through a capacitor is found to be zero, this means that the voltage across it must be constant, not necessarily zero.

When the voltage on the capacitor rise, capacitor is charging ("absorbing" the current).
When the voltage on capacitor "drops", capacitor is discharging (supply the current).
https://forum.allaboutcircuits.com/threads/what-is-the-real-ac-voltage-phase-shift-across-a-capacitor.126190/#post-1024973
https://forum.allaboutcircuits.com/threads/capacitors-how-do-they-work.80390/#post-570788
https://www.allaboutcircuits.com/textbook/direct-current/chpt-13/capacitors-and-calculus/

 

danadak

Joined Mar 10, 2018
4,057
AC current flows in a capacitor,. not DC current (except for leakage).

The coupling use of a C is to remove any DC in the input which could alter
the bias conditions of the stage the signal is entering.

Same is true for the decoupling use, to remove the DC from the stage that
is used to process the signal, like an amplifier, so that it does not affect the
next stage being sent to entered.



In this case R1, R2, and Re set the bias conditions. Without C1 the stage bias would
be affected by input DC conditions. Same is true of C2, only it affects stages downstream.

Capacitors also in these instances ground reference the signals rather than referenced to
bias point, which in multistage designs where G of the overal stages, can make large bias
shifts in the bias point of stages causing them to distort the signal of interest.

Caps of of course have many other uses, bypassing (removing AC noise and transients from
power supplies), energy storage (function as a limited battery), filtering (causing specific AC
signals to be removed), impedance transformation...

Designing single stage amplifiers -

https://wiki.analog.com/university/courses/electronics/text/chapter-9

Google "single stage bipolar amplifiers", lots of references.

Basic capacitor uses -

https://en.wikipedia.org/wiki/Applications_of_capacitors


Regards, Dana.
 
Last edited:

nsaspook

Joined Aug 27, 2009
7,588
I love Dave's videos but this one above is one of a few that really suck for a proper explanation of inside the 'blackbox' electrical conduction and energy transfer in a circuit.
Variations in the electric field, which are imagined to be equivalent to a current. Displacement Current actually does not exist. The term 'current' here is an artifact of the theory of “aether” from the late 1800’s.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.205.6223&rep=rep1&type=pdf
In the special case of a steadily charging capacitor the magnetic field between the capacitor plates is entirely caused by the currents in the leads and in the plates. The displacement current itself then makes no contribution.
...
If these arguments are valid it would seem to follow that—in a vacuum—∂D/∂t (called by Rosser the ‘Maxwell term’) is no more and no less than what it appears to be: the rate of change of the electric field strength. Enormously important though it is, in no sense can it be described as an electric current. Does this mean that the concept, and even the term, ‘displacement current’ should disappear from physics? I will address this question first from a pedagogical point of view and then in terms of mathematical convenience. Undergraduates already find the concept of a displacement current in a vacuum puzzling since they think of a current as a transport of charges and there is no such transport here. The above analysis suggests that there is no theoretical foundation, either, for such a concept. It could, however, be argued that, since the law of Biot and Laplace is used with present values of the conduction currents in elementary calculations of the magnetic fields, one is then logically obliged to interpret ∂D/∂t as equivalent to an (imaginary) current. I have been persuaded by the arguments of a referee, however, that the tiny gain in calculational accuracy achieved by this would be more than offset by the perplexity caused to students. Perhaps, therefore, the term ‘displacement current’ should not even be mentioned to undergraduates?

Electromagnetism using sexy angels.:)

If you like the dreaded water analogy this is a capacitor. No water goes through, if it did then it could not store energy.

 
Last edited:

Marc Sugrue

Joined Jan 19, 2018
149
Coupling capacitors are normally used to block DC biasing of a circuit but allow a AC signal to pass through.

Decoupling capacitors are normally used to limit the current loop path for high frequencies. By placing the decoupling capacitors near the legs of an integrated circuit a low impedance voltage source is at the leg of the component. This means the effects of track inductance is Minimised (which can cause emi emissions due to resonances) if the ic is taking gulps of high frequency current as seen on logic or power supply ic’s. It also reduces susceptibility of the ic’s too for the same reasons which is why most IC’s recommend using them.

Hope this helps
 
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