The maximum current flows when |X C | << R In other words, the capacitor couples the signal properly when |X C | << R The size of the coupling capacitor depends on the lowest frequency you are trying to couple, because |X C | increases when frequency decreases. We will use the rule T = RC (eqn. 2) at the lowest freq. (T = 1/f). (All other freq. will be well coupled.) Coupling and Bypass Capacitors Coupling Capacitor: -passes a signal from one ungrounded point to another ungrounded point. (Not practical for IC’s therefore IC amplifiers are usually direct coupled.) (eqn.1)

Coupling and Bypass Capacitors Example: To couple freq. in the range 20 Hz to 50kHz. The lowest freq. = 20 Hz  T = 1/20 = 0.05 sec If the total resistance in a one loop circuit is 10k , the coupling capacitor must satisfy T = RC = 10k X C  C = 0.05/10k = 5  F Another widely used rule is to keep |X C | < (1/10) R Bypass Capacitor: -similar to coupling capacitor except it couples an ungrounded point to a grounded point. Eqns. (1) and (2) still apply. In this case 

Coupling and Bypass Capacitors Example: The input signal in the Figure can have a frequency between 10Hz and 50khz. Find the size of the coupling capacitor. Example: We want an AC ground on point A in the Figure. Find the size of the bypass capacitor   24k | |12k = 8k and Thevenize cct. to the left of the capacitor R TH =2k for the lowest freq. We need at least 800  F

AC Equivalent Circuits and Frequency Response Up until now all of the small-signal AC equivalent circuits we have used have been mid-frequency band models. Low-frequency AC Equivalent Circuit CE Amplifier Fig p. 609  mid-frequency  low-frequency high-frequency

AC Equivalent Circuits and Frequency Response We will use a simplified analysis approach. (An exact analysis could be done using SPICE) To determine  L, the lower 3db frequency: First set V s to zero, C E and C C2 to infinity and find the resistance R C1 seen by C C1. R C1 = R S + [R B || (r x + r  )] Next set C C1 and C C2 to infinity and determine the resistance R’ E seen by C E Set both C C1 and C E to infinity and find the resistance seen by C C2 R C2 = R L + (R C || r o) An approximate value for the lower 3db freq. can now be found now that the equivalent RC time constants are known.

High–frequency model for CE Amplifier (a) High –Freq. Model (CE) (b) Simplified circuit (Thevenin theorem)

This circuit can be further simplified using Miller’s theorem. In our case and Thus Can be replaced by a capacitor connected from terminal B’ to ground. B’ Notice that the input forms a simple RC low-pass filter. The break freq. is given by where is the total capacitance in the input loop. Above f H the gain of the amplifier rolls off at 20 db/decade.

Example: CE Amplifier The hybrid pi parameters are: Also In the high-freq. equivalent circuit model we have:

SPICE simulation result