SP.764, Practical Electronics Lecture 4: RC Circuits Topics: 1) RC circuit review 2) Equations for RC 3) Time Response of RC circuits 4) RC Filters Question: What factors contribute to the capacitance of a capacitor? Area A Material has a dielectric constant C =ε ε A d d q = CV If d becomes smaller, the capacitance goes up. Therefore, the capacitor can store more charge. ++++++++ Electric Field E E= _________ VCAP d For a given voltage, if one makes the separation d smaller, the electric field E increases. Why is this a problem? If the electric field gets too large, the insulator breaks down. A small spark will jump between plates and short the capacitor, which will not work anymore. In summary, if V gets too large or d gets too small, the dielectric material will “breakdown” and destroy the capacitor. In some cases the capacitor “pops”, or may not act properly anymore. A capacitor is specified by: - Capacitance value - Maximum voltage across capacitor - Polarized or non-polarized. What is polarized? - Polarized capacitors are marked with “+” lead and “-” lead. - If the positive and negative leads are connected the wrong way, the capacitor can blow up. If one is working with high voltages, then one would want to keep electric field small by putting plates apart, but one would loose some capacitance. RC circuit review: I (t) “A” “B” C VO Move switch from “A” to “B” at time t = 0. t − ⎛ ⎞ RC ⎟ ⎜ VCAP (t) = V0 ⎜1− e ⎟ ⎝ ⎠ RC = time constant [OHM][FARAD]=[SEC] R t , V − I (t) = 0 e RC R The current I(t) starts at peak value V0/R, and then drops off and dies out exponentially. The voltage across the capacitor VCAP(t) starts at 0, goes up rapidly, and asymptotically approaches V0. Time Response of RC circuits: RC circuits can do interesting things. Consider the following example: Problem: A square wave that goes between +/-Vo with period T is applied to the following RC circuit. R V0 C t − V0 T Question: Complete the following graphs. Plot VCAP(t) for: 1. RC ≈ T VCAP V0 2. RC << T 3. RC >> T − V0 t Hint: e-1 = 1/3 Solution: Assume that the square wave signal has been “ON” for a long period of time. At least one knows VCAP(t) will range between +V0 and –V0 (but will never reach those voltages). VCAP (t) + VO t − VO + VO RC ≈ T t − VO + VO RC << T t − VO + VO RC >> T t − VO RC Filters: Swing analogy If one pushes a person sitting in a swing slowly, the person swings smoothly, so the swing works fine. However, if one starts pushing too many times per second, the swing barely moves at all. The performance of the swinging motion depends on the frequency at which the person is pushing the swing. Frequency is just the inverse of the period T, f = 1 T When RC << T Æ f is “small” or “LOW” When RC >> T Æ f is “HIGH” (Signal gets through) (Signal is blocked) IF RC << T , SIGNAL PASSES Î LOWPASS FILTER On the upcoming Lab 5, groups will build one either a highpass or lowpass filter circuit with one of the RC values listed below, and measure plot the voltage gain at each of the frequencies listed below. C VO Highpass R VOUT R VO Plot Lowpass C VOUT vs. Frequency VO RC = 10ms 1ms or 0.1ms VOUT 3Hz 300Hz 30kHz 10Hz 1kHz 100kHz 30Hz 3kHz 300kHz 100Hz 10kHz

Lecture 4: RC Circuits

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