Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.chapter 10electronics fundamentalscircuits, devices, and applicationsTHOMAS L. FLOYDDAVID M. BUCHLAElectronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Sinusoidal response of RC circuitsWhen both resistance and capacitance are in a series circuit, the phase angle between the applied voltage and total current is between 0 and 90, depending on the values of resistance and reactance.RVRCVR leads VSVC lags VSI leads VSIVSVCElectronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Impedance of series RC circuitsIn a series RC circuit, the total impedance is the phasor sum of R and XC.R is plotted along the positive x-axis. RXC is plotted along the negative y-axis. XC1tanCXRq-� �=� �� �ZIt is convenient to reposition the phasors into the impedance triangle.RXCZZ is the diagonalqqElectronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Impedance of series RC circuitsR = 1.2 kXC = 960 Sketch the impedance triangle and show the values for R = 1.2 k and XC = 960 .( ) ( )2 21.2 k + 0.96 k1.33 kZ = W W= W10.96 ktan1.2 k39q-W=W= �Z = 1.33 k39oqElectronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Analysis of series RC circuitsOhm’s law is applied to series RC circuits using Z, V, and I. V VV IZ I ZZ I= = =Because I is the same everywhere in a series circuit, you can obtain the voltages across different components by multiplying the impedance of that component by the current as shown in the following example.Electronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Analysis of series RC circuitsAssume the current in the previous example is 10 mArms. Sketch the voltage phasor diagram. The impedance triangle from the previous example is shown for reference.VR = 12 VVC = 9.6 V The voltage phasor diagram can be found from Ohm’s law. Multiply each impedance phasor by 10 mA. x 10 mA =R = 1.2 kXC = 960 Z = 1.33 k39oVS = 13.3 V39oq qElectronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Variation of phase angle with frequencyPhasor diagrams that have reactance phasors can only be drawn for a single frequency because X is a function of frequency. As frequency changes, the impedance triangle for an RC circuit changes as illustrated here because XC decreases with increasing f. This determines the frequency response of RC circuits.qZ3XC1XC2XC3Z2Z112312fff3Increasing fqqRElectronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.q(phase lag)ff(phase lag)VApplicationsRVRVoutCVoutVinVinVoutVinFor a given frequency, a series RC circuit can be used to produce a phase lag by a specific amount between an input voltage and an output by taking the output across the capacitor. This circuit is also a basic low-pass filter, a circuit that passes low frequencies and rejects all others.Electronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.qq(phase lead)(phase lead)VApplicationsRVCVoutCVoutVinVinVoutVinReversing the components in the previous circuit produces a circuit that is a basic lead network. This circuit is also a basic high-pass filter, a circuit that passes high frequencies and rejects all others. This filter passes high frequencies down to a frequency called the cutoff frequency.Electronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.ApplicationsAn application showing how the phase-shift network is useful is the phase-shift oscillator, which uses a combination of RC networks to produce the required 180o phase shift for the oscillator.AmplifierRfRRRC C CPhase-shift networkElectronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.10 V dcVoutVin100 1 Fm10 V dc010 V dc0Frequency Response of RC CircuitsWhen a signal is applied to an RC circuit, and the output is taken across the capacitor as shown, the circuit acts as a low-pass filter.As the frequency increases, the output amplitude decreases.Plotting the response:Vout (V)9.988.461.570.790.11 10 20 100f (kHz)9876543211ƒ = 1 kHz8.46 V rms10 V rms100 Fm1.57 V rms10 V rms1ƒ = 10 kHz100 Fm0.79 V rms10 V rms1ƒ = 20 kHz100 FmElectronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Vin10 V dc0Vout0 V dc10 V dc 1001 FmFrequency Response of RC CircuitsReversing the components, and taking the output across the resistor as shown, the circuit acts as a high-pass filter.As the frequency increases, the output amplitude also increases.Plotting the response:ƒ = 100 Hz0.63 V rms10 V rms100 1 Fmƒ = 1 kHz5.32 V rms10 V rms100 1 Fmƒ = 10 kHz9.87 V rms10 V rms100 1 FmVout(V)f (kHz)9.875.320.6300.01 0.1 11098765432110Electronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Impedance Phase angleBandwidthRC Lag CircuitThe total opposition to sinusoidal current expressed in ohms.Selected Key TermsThe range of frequencies passed from input to the output of a circuit.The phase difference between source voltage and total current in a reactive circuit.A phase shift circuit where the output voltage lags the input voltage by its phase angle. The output voltage decreases as the input frequency increases.Electronics Fundamentals 8th edition Floyd/BuchlaChapter 10Chapter 10© 2010