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2.4 Circuits with Resistors and Capacitors • response of a series connected resistor and capacitor to a dc (steady) voltage • response of a series connected resistor and capacitor to an ac (varying) voltage • decibels and Bode plots • high-pass electronic filter • band-pass electronic filter • using a low-pass filter for signal-to-noise enhancement 2.4 : 1/10 Response of an RC Circuit to DC Consider the circuit at the right. When the switch is connected to the battery, current will flow and charge the capacitor. The charging will continue until the voltage across the capacitor equals the voltage of the battery. switch i R + V ! C For V = 9 V, R = 10 kΩ and C = 1 mF, the current can be computed as a function of time by tracking the total capacitor charge. time 0s 1 2 4 C ∞ 2.4 : 2/10 capacitor charge 0 mC 0+0.9 0.9+0.81 1.71+0.73 C 9 mC VC 0 mC/1 mF = 0 V 0.9 mC/1 mF = 0.9 1.71 mC/1 mF = 1.71 2.43 mC/1 mF = 2.43 C 9V VR 9V 8.1 7.29 6.57 C 0V i 0.9 mA 0.81 0.73 0.66 C 0 mA Response of an RC Circuit to DC (2) V = VC + VR Q V = + iR C dV 1 dQ di = +R =0 dt C dt dt t t 1 di = − ∫ i RC ∫ dt 0 0 1 voltage (V) The figure at the right shows the voltage across the resistor (blue) and capacitor (red) when a square voltage pulse is applied the circuit on the previous slide. The time dependence of the current is derived as follows. RC = 20 0 1 100 200 The time dependence of the two voltages is given by the following. Ri = Ri0e−t VR = Ve −t RC RC VC = V − VR = 2.4 : 3/10 0 100 time (s) Impedance of an RC Voltage Divider What is the impedance of an RC voltage divider when measuring across the capacitor? Start with resistance and capacitive reactance, R ~ V C VC XC − j 2π f C = R + X C R − j 2π f C solve for the impedance, Z RC = − j 2π f C j 2π f C 1 = 2 R − j 2π f C R + j 2π f C 1 + ( 2π RCf ) and finally, define a characteristic frequency, _______________. Z RC = 2.4 : 4/10 1 1 + ( f f3dB ) 2 The impedance is frequency dependent with the equation constant depending upon RC. Voltage Across the Capacitor 1.2 G ≡ VC V = Z RC 1 RC = 1 f3dB = 0.16 gain 0.8 as f → 0, G → 1 as f → ∞, G → 0 when f = f3dB 0.6 0.4 0.2 0 0 0.25 0.5 0.75 1 1.25 1.5 1.75 frequency When voltage is measured across the capacitor, the series combination of a resistor and capacitor is called a low-pass filter. Note that the gain of the circuit does not decrease rapidly. At frequencies larger than f3dB the functional form approaches _____ . 2.4 : 5/10 2 Phase Shift Between V and VC • put the voltage divider expression into the standard form, a + jb XC − j 2π f C 1 2π RCf = = −j 2 R + X C R − j 2π f C 1 + ( 2π RCf )2 1 + ( 2π RCf ) • derive an expression for the phase angle using the arctangent in the complex plane ⎛ −2π RCf ⎜ 2 + 1 2 RCf π ( ) Im ⎛ ⎞ φ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎜ 1 ⎝ Re ⎠ ⎜⎜ 2 RCf π + 1 2 ( ) ⎝ ⎞ ⎟ ⎟ = tan −1 ( −2π RCf ) = ⎟ ⎟⎟ ⎠ when f = 0, φ = 0°; when f = f3dB, φ = -45°; as f → ∞, φ → -90° • the phase shift is close to 0° at frequencies up to ____f3dB • the phase shift is close to -90° at frequencies above ____f3dB • the phase shift varies significantly from 0.1f3dB to 10f3dB • phase shifts distort the shape of the signal 2.4 : 6/10 Decibels and the Low-Pass Bode Plot • graphs of amplitude or power gain versus frequency are usually in decibels • A is called attenuation for G < 1 and amplification for G>1 • for a large range of frequencies a Bode plot is easier to interpret than a linear plot • one plot represents all low-pass filters • A is flat below ___f3dB and drops by 20 dB per decade above ___f3dB 2.4 : 7/10 2 ⎛ Vout ⎞ R⎞ ⎟⎟ = ⎟ = 10log ⎜⎜ 2 ⎠ ⎝ Vin R ⎠ 10 -0.06° -3dB 0 -0.6° -5.7° -45° -10 -20dB/decade A(dB) ⎛ Pout A ( dB ) ≡ 10log ⎜ ⎝ Pin -20 -84° -30 -40 -89° -50 -89.9° -60 -3 -2 -1 0 log(f/f3dB) 1 2 3 RC High-Pass Filter 10 -3dB 0 -10 -20 dB/decade R V VR A(dB) ~ -20 -30 C -40 -50 -60 -3 -2 -1 0 1 2 log(f/f3dB) for a high-pass filter voltage is measured across the resistor * ⎛ R ⎞⎛ R ⎞ Z = ⎜ ⎟⎜ ⎟ R + X R + X C ⎠⎝ C ⎠ ⎝ as f → 0, G → 0; as f → ∞, G → 1; when f = f3dB, G = 1 2.4 : 8/10 2 3 RC Band-Pass Filter A band-pass filter can be created by feeding the output of a lowpass filter into a high-pass filter. R R Vout Vin GBP = GLP GHP ABP ( dB ) = C C bandpass RC filter -10 -20 gain (dB) The values of R and C do not need to be the same for the two parts of the filter. This permits the creation of a "flat" gain between the low-pass and high-pass 3dB frequencies (not shown). 0 -30 -40 -50 -60 -2 -1 0 1 log(f/f0) 2.4 : 9/10 2 3 Signal-to-Noise Enhancement RC = 10 ms signal + raw data 1 0 200 400 600 time (ms) 800 RC = 20 ms This is an everpresent trade in instrument design. signal 1 0.5 0 200 2.4 : 10/10 400 600 time (ms) As the RC time constant is increased, the _____ decreases and _________ increases. 800 signal 0.5 0.5 0 200 400 600 time (ms) 800 RC = 50 ms 1 signal signal 1 0.5 0 200 400 600 time (ms) 800