lectureslides11

The current jumps to an initial value and then decays

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t = 0: vC (t = 0) = VCi ￿ dvC ￿ VS − VCi ￿ iC (t = 0) = C = ￿ dt t=0 R After a long time, (t → ∞ ) vC (t → ∞) = VS iC (t → ∞) = 0 In between, the capacitor voltage changes, in an exponential fashion, from VCi to VS. The current jumps to an initial value and then decays exponentially to zero. Since this all happens within a given timespan, these are known as transient effects. EE 442 RC and RL transients – 9 Plots of capacitor voltage and current for a circuit with VS = 6 V, VCi = 2 V, R = 5 kΩ, C = 1 µF (RC = 2 ms). EE 442 RC and RL transients – 10 Things to consider Note the form of the equation. ￿ t vC (t) = VS − [VS − VCi ] exp − RC ￿ vC(t) = final value – [voltage swing]·exponential factor. The quantity RC has units of time. (Check it.) It shows up in the exponential factor and determines the relative rate of change of the capacitor voltage. Thus, it is known as the RC time constant. It gives the time scale of how long the transients last. After a time span corresponding to a few time constants, most of the voltage change has occurred. For example, after t = 5·RC (5 time constants), exp(-t/RC) = 0.00674. So after 5 time constants, more than 99% of the transition has occurred. If RC is on the order of 1 ms, then the “switching time” will take a few ms. If the product is on the order of 1 ns, then the switching time will be a few ns. EE 442 RC and RL transients – 11 In our example leading up to the capacitor transient voltage equation, we assumed that the source voltage was higher than the initial voltage of the capacitor, resulting in the capacitor being charged up to the source voltage after the switch was closed. However, there was nothing in the analysis that was dependent on the source being at a higher voltage than the capacitor initially. In fact, the equation works just as well if the capacitor was initially at a higher voltage than the source. In that case, that the capacitor discharges current to the source, and the capac...
View Full Document

This document was uploaded on 04/05/2014.

Ask a homework question - tutors are online