RC时间常数为什么单位是s?

lukeluck · 发表于 2018-4-24 10:51

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今天看了老外的分析，豁然开朗。

At first glance, this would seem to be very strange. How can the product of a resistance in ohms and a capacitance in farads possibly give us a time in seconds? To understand how this is possible, we go back to the basic definitions and some dimensional analysis.

Resistance

Resistance opposes the flow of current through a circuit. By Ohm's Law, R = E/I. Thus, 1 ohm may also be expressed as 1 volt/ampere.

Current

Current is a measure of the amount of charge flowing through a circuit in a given amount of time. By primary definition, 1 ampere is equal to 1 coulomb/second.

Capacitance

Capacitance is the capacity to hold an electrical charge. A capacitance of 1 farad will exhibit a change of 1 volt if 1 coulomb of charge is moved from one plate to the other. Hence, 1 farad can be expressed as 1 coulomb/volt.

Putting these three basic definitions together we get the following progression:

RC	=	ohms	×	farads
	$ s( ^% W, d+ \|
	=	volts	×	coulombs

		amperes		volts

	=	volts	×	coulombs

		coulombs/seconds		volts

	=	volts × seconds	×	coulombs

		coulombs		volts

	=	seconds

Thus, we see that the RC product is indeed a measure of time, and can properly be described as the time constant of this circuit. This in turn means that this curve can be used to determine the voltage to which any capacitor will charge through any resistance, over any period of time, towards any source voltage. It is the general curve describing the voltage across a charging capacitor, over time.

Theoretically, C will never fully charge to the source voltage, E. In the first time constant, C charges to 63.2% of the source voltage. During the second time constant, C charges to 86.5% of the source voltage, which is also 63.2% of the remaining voltage difference between E and vC. This continues indefinitely, with vC continually approaching, but never quite reaching, the full value of E. However, at the end of 5 time constants (5RC), vC has reached 99.3% of E. This is considered close enough for practical purposes, and the capacitor is deemed fully charged at the end of this period of time.