Physics 160Lecture 9R. Johnson“Bootstrap”• Clever technique to raise the input impedance of thebias network of an emitter-follower or common-ittlifiemitter amplifier.•Itreliesonthefactthatthegainoftheemitter•Itreliesonthefactthatthegainoftheemitter(relative to the base) is very close to unity.• Do not try to apply a “bootstrap”from a node withgain far from unity!!April 29, 2014 Physics 160 2Emitter-Follower with Simple Bias Network15.00V764.5uAExtra resistor serves noRSQ14.980mAC1R110kV2High source impedance of 10kohmserves no purpose at the moment…VRB4.7k10k5 009mA7.218VQ2N390429.08uA0V 7.355V735 5 A0VC210u0.1uF0VV215Vdc5.745mARE1.3k5.009mARLoad10Meg0AVS1Vac0Vdc0AR210k735.5uA00VInput impedance ~9.2k for ~1501April 29, 2014 Physics 160 3Hz170F1.0k2.921dB3fNo Source Impedance & No Bootstrapf170 H t df3dB~170 Hz as expectedApril 29, 2014 Physics 160 4With Source Impedance48.03.9103.9The low impedance of the biasThe low impedance of the bias network is killing more than half of our signal.April 29, 2014 Physics 160 5Follower with Bootstrap15.00V764 5uAInput impedance is boosted up to ~135 k!Note: the same trick will work with the RSR110k764.5uAC1Q14.980mAHigh source impedance of 10kohm5common-emitter amplifier.0VRS10k7.218VC3V215Vdc5.745mA7.355VRB4.7kVC210uC10.1uFQ2N390429.08uA0Vre~5 0RE1.3k5.009mAC30.4u355VRLoad10Meg0AVS1Vac0Vdc0A0VR210k735.5uA0V1Above ~80Hz the output feeds back through C3 and keeps the voltage hRBltApril 29, 2014 Physics 160 60Hz 80kΩ5μF4.021changes across RB close to zero, making RB look like a very high impedance.Ratio of Voltages Across RB99601300ER996.01305EErRkΩ780kΩ7.4994.011BRAt signal frequencies, the effective resistance of RBis April 29, 2014 Physics 160 7Bootstrapped Follower with 10k Source R93.013510135From this simulation we infer a signal input impedance of 135 kohm, dominated by the transistor base now, not by the bias network.April 29, 2014 Physics 160 8High Power Output Stage“Class A” AmplifierWhen the output swings high the NPN transistor sources the output current to the load. When it swings low, the NPN transistor turns off, and current from the load flows through the emitter resistor to thethrough the emitter resistor to the 30 V supply.With no signal the0.6 V0Ve.g. audio speakerWith no signal, the quiescent current is 30V/8 ohms = 4 amps.The quiescent power is 0 Vspeaker45 V4 A=180 W in just the output stage!April 29, 2014 Physics 160 9Push-Pull Output StageBias currents are near zero with no signal applied“Class B” AmplifierVCCQ1NPN emitter followerapplied.10VdcQ2N3904VV3VOFF=0VR11kQ2Q2N39060VFREQ = 1kVAMPL = 2VOFF 0VEE10VdcPNP emitter followerload0follower(The small 2N3904 and 2N3096 transistors in this For positive output swings the NPN transistor sources the tf ti iApril 29, 2014 Physics 160 10simulation are only for illustration. A real power amp would use big power transistors.)current; for negative swings the PNP transistor sinks the current.Severe “Cross-Over” DistortionMajor j“crossover” distortion!April 29, 2014 Physics 160 11Reducing Cross-Over DistortionR3510.1uADiodes ensure that at least one transistor is lbttth“Class AB” AmplifierFor stable bias tthQ1Q2N3904145.1uA18kalways on, but at the expense of using more power, even with no input signal.you want the current through the diodes to be much larger than the base current.R42QV3VCC10Vdc25.44mAD1D1N914pgThese improve temperature stability.Q3112 0uAVEE10VdcD2D1N914397.9uAV3FREQ = 1kVAMPL = 2VOFF = 032.88uAVR52R120065.98uAQ2N3906-112.0uA25.01mA25.41mA00R218kSize these resistors April 29, 2014 Physics 160 12according to the drive current needed.Cross-Over Distortion GoneWe will see that another way of greatly reducing crossoverreducing cross-over distortion is by negative feedback.April 29, 2014 Physics 160 13Differential AmplifierEt l di t t i it l t!•Extremely common and important circuit element!• When transmitting a small signal over a significant distance, noise can be rejected to a very high degree by encoding the signal into a difference of two voltages This typically translates into sending thedifference of two voltages. This typically translates into sending the signal as a current loop (e.g. the “LVDS” standard):Si lRcurrentVDiff-amp receiverSignaloutput• The two wires are typically twisted to avoid picking up magnetic noise in the current loop.•All voltage noise that is “common” to the two wires gets rejected by the ggjydifferential amplifier.• The “termination resistor” R is chosen to match the characteristic impedance of the transmission line (typically ~100 ohms for twisted i)Thi id fltiwires). This avoids reflections.April 29, 2014 Physics 160 14Differential AmplifierEittfll f1.706mARC7.5k945.8uA15.00VEmitter-follower for output drive capability (low output impedance)Differential Amplifier7.907VQ2934.7uAQ1952.0uAQ4Q2N390411.11uA-1.717mA7 229VV115Vdc3 604mADifferential-mode inputpv+vvout-661.3mV0VRLoad1Meg7.229uAQ2N39046.498uA-659.6mVQ2N39046.143uARE2100941.2uAVVdm1Vac0Vdc6.143uARE1100958.1uARE1.710mA0V7.229V3.604mA-755.5mV0RT1.899mA13k0VVcm0Vac0Vdc12.64uA75003025100750021121DMCeECRrRRGCommon-mode input“tail current”00VV215Vdc3.609mART7.5kDiff. mode gain: 30Common mode gain: 0.55.075007500221CMEETCrRRRGcurrentApril 29, 2014 Physics 160 15-15.00VCMRR=60 vvv 30outDiff Amp with Current Source in TailRC7.5k1.012mA15.00VEmitterFollower1 668mA7.410VQ21.001mANon-InvertingInputQ11.020mADiffAmpQ4Q2N390410.83uA1.668mA-1.679mAV115Vdc3.700mAInvertingInput0V-663.3mV-661.5mV RE21001.008mARLoad1Meg6.734uAQ2N39046.935uARE1.672mA0VQ2N39046.523uAVdm1Vac0Vdc6.523uA6.734VVRE11001.027mAInput0VR20-764.1mVRE13kQ32.035mAVcm0Vac0Vdc13.46uART1k2.047mA12k1.023mACurrentSource00VV215Vdc4.729mAR32.7k1.010mA-12.27VQ2N390412.39uA-12.95VGreatly improved CMRR ( 30 000)April 29, 2014 Physics 160 16-15.00VGreatly improved CMRR (~30,000) and CM range (~12V to