Announcements • Troubles with Assignments… – Assignments are 20% of the final grade – Exam questions very similar (30%) • Deadline extended to 5pm Fridays, if you need it. – Place in my mailbox (rm 217), or under my door (rm 222) – Any later than that will not be graded • Come and talk to me if you need help. • mid-term: Thursday, October 27th Adam Riess, Brian Schmidt and Saul Perlmutter Lecture 10 Overview • • • • Transistors (continued) The common-emitter amplifier Amplifier parameters Black box amplifier models Summary of useful equations • Basic DC operating conditions: I E I B IC I C I B I E I C 1 I C I S eVBE / VT VC VCC I C RC • Add a small signal: ic IC vbe g m vbe VT IC gm VT vbe r ib gm vbe 1 re ie gm gm Using small signal models e.g. Tmodel 1) Determine the DC operating conditions (in particular, the collector current, IC) 2) Calculate small signal model parameters: gm, rπ, re 3) Eliminate DC sources: replace voltage sources with shorts and current sources with open circuits 4) Replace BJT with equivalent small-signal model. Choose most convenient depending on surrounding circuitry 5) Analyze re = 1 gm Voltage gain with small signal model ic + RC vc ic eliminate DC sources and apply T-model + vbe re vbe - vc ic RC Find the gain using a small signal model: ie RC ie RC vbe ie re Voltage gain re gm vc ie RC R C vbe ie re re 1 gm v so voltage gain c g m RC vbe gm = IC VT How to build a Real Common emitter amplifier • Why bother with 2 voltage supplies? • Use a voltage divider R2/R1 to provide base-emitter voltage to correctly bias the transistor. DC condition: the voltage divider VBE VB VE 0.6V • The voltage divider VB I E RE 0.6V should provide sufficient R1 voltage to place the now VB VBB R1 R2 transistor in active mode (base-emitter forward R1 I R 0. 6 V so E E biased): R1 R2 VBB • Current through resistors should be >10 times base current for stability Vbb IC 10 R1 R2 Amplifier specifications • What other parameters of an amplifier do we care about? – Voltage gain – Dynamic range – Frequency response (bandwidth) – Input impedance – output impedance Voltage Gain • Voltage gain • Use small signal model (short Voltage sources and capacitors) ground vb ie (re RE ) vc ic RC ie RC since 1 αie voltage gain vout vc RC vin vb re RE usually re<<RE vc RC vb RE • Voltage gain is only defined by resistors RC and RE ground Frequency response (Bandwidth) • Normally interested in providing a small, AC signal to the base • Use capacitors to remove ("block") any low frequency (DC) component ("capacitively couple the signal to the base") which could affect the bias condition • C1 forms a high-pass filter with R1in parallel with R2 (Assuming the AC impedance into the base is large). • Cut off frequency ω0=1/RC, so to remove frequencies <fmin: C1 1 RR 2f min 1 2 R1 R2 Frequency response (Bandwidth) • Also worthwhile to place a capacitor on the output • C2 forms a high pass filter with RL. • Cut off frequency ω0=1/RC, so to remove frequencies <fmin: C2 1 2f min RL Dynamic Range • Maximum voltage output = Vbb • Minimum = 0 • Beyond this the signal becomes 'clipped' or distorted • To get the maximum possible voltage swing, both positive and negative, set VC=0.5 VBB • Maximum 'dynamic range' VC Input impedance • Consider the circuit without the voltage divider resistors. What's the small signal (AC) input impedance at the base, rb? vb ib v b = (re + Re )ie i ib = e b +1 vb rb = = (b + 1)(re + Re ) ib b + 1 » b; re << Re rb » bRe rb = rOUT rb • Including voltage divider resistors in parallel • Input signal sees a total input impedance rIN= R1 // R2 // rb RB Output impedance rOUT vo io RC RC io io ROUT rb • If RL=10kΩ and we want a low frequency cutoff of 20Hz, What is C2? • If VBB=15V and IC=2mA what is the output impedance? DC condition R1 I E RE 0.6V R1 R2 VBB Vbb IC 10 R1 R2 Frequency response C1 1 R1 R2 2f min R1 R2 C2 1 2f min RL Impedance rb » bRe rIN » R1 || rb || R2 rOUT vo io RC RC io io Gain/Dynamic range VC 0.5VBB vc RC vb RE Impedances • Why do we care about the input and output impedance? • Simplest "black box" amplifier model: ROUT VIN RIN AVIN VOUT • The amplifier measures voltage across RIN, then generates a voltage which is larger by a factor A • This voltage generator, in series with the output resistance ROUT, is connected to the output port. • A should be a constant (i.e. gain is linear) Impedances • Attach an input - a source voltage VS plus source impedance RS RS VS ROUT VIN RIN AVIN • Note the voltage divider RS + RIN. • VIN=VS(RIN/(RIN+RS) • We want VIN = VS regardless of source impedance • So want RIN to be large. • The ideal amplifier has an infinite input impedance VOUT Impedances • Attach a load - an output circuit with a resistance RL RS VS ROUT VIN RIN AVIN • Note the voltage divider ROUT + RL. • VOUT=AVIN(RL/(RL+ROUT) • Want VOUT=AVIN regardless of load • We want ROUT to be small. • The ideal amplifier has zero output impedance VOUT RL Operational Amplifier • Integrated circuit containing ~20 transistors Operational Amplifier • An op amp is a high voltage gain amplifier with high input impedance, low output impedance, and differential inputs. • Positive input at the non-inverting input produces positive output, positive input at the inverting input produces negative output. • Can model any amplifier as a "black-box" with a parallel input impedance Rin, and a voltage source with gain Av in series with an output impedance Rout. Ideal op-amp • Place a source and a load on the model RS + vout RL - • Infinite internal resistance Rin (so vin=vs). • Zero output resistance Rout (so vout=Avvin). • "A" very large • No saturation • iin=0; no current flow into op-amp So the equivalent circuit of an ideal op-amp looks like this: Schematics An amplifier will not work without a power supply. And a more complete diagram looks like the figure below, which also indicates the standard pin configuration. Function Pin 2 Inverting input 3 Non-inverting input 4 V- supply 6 Output 7 V+ supply Measuring Impedances RS VS ROUT VIN RIN AVIN VOUT RL • Assuming you can only vary RL and RS, how would you measure the input and output impedances of the amplifier? Measuring Impedances RS VS ROUT VIN RIN AVIN VOUT RL • With the black box model, it is simple to measure the input and output impedances of an amplifier • To measure the input impedance, vary RS until the output voltage has dropped to half ; then RS=RIN= input impedance •To measure the output impedance, vary RL until the output voltage has dropped to half ; then RL=ROUT= output impedance Cascaded Amplifiers • Easiest way to increase amplification is to link amplifiers together ROUT1 VIN1 RIN1 A1VIN1 ROUT2 VOUT1 VIN2 RIN2 A2 VIN2 • Ideal amplifiers; VOUT2=A1A2VIN1 • In reality, take account of voltage divider action due to input and output impedances VOUT2
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