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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: Add a small signal:

Using small signal models 1) Determine the DC operating conditions (in particular, the collector current, I C ) 2) Calculate small signal model parameters: g m, r π, r e 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 e.g. Tmodel

Voltage gain with small signal model Find the gain using a small signal model: vbe ic re RC ic vc eliminate DC sources and apply T-model

Voltage gain So, we have a voltage-to-current amplifier (a voltage controlled current source) To convert the voltage controlled current source into a circuit providing voltage gain, connect a resistor to the collector output and measure the voltage. So, small signal voltage gain:

How to build a Real Common emitter amplifier Why bother with 2 voltage supplies? Use a voltage divider R 2 /R 1 to provide base-emitter voltage to correctly bias the transistor.

DC condition: the voltage divider The voltage divider should provide sufficient voltage to place the transistor in active mode (base-emitter forward biased): Current through resistors should be >10 times base current for stability

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) voltage gain usually r e <<R E Voltage gain is only defined by resistors R C and R E ground αi e

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 R 1 in parallel with R 2 (Assuming the AC impedance into the base is large). Cut off frequency ω 0 =1/RC, so to remove frequencies <f min :

Frequency response (Bandwidth) Also worthwhile to place a capacitor on the output C 2 forms a high pass filter with R L. Cut off frequency ω 0 =1/RC, so to remove frequencies <f min :

Dynamic Range Maximum voltage output = V bb Minimum = 0 Beyond this the signal becomes 'clipped' or distorted To get the maximum possible voltage swing, both positive and negative, set V C =0.5 V BB Maximum 'dynamic range' VCVC

rbrb r OUT Input impedance Consider the circuit without the voltage divider resistors. What's the small signal (AC) input impedance at the base, r b ? Including voltage divider resistors in parallel Input signal sees a total input impedance r IN = R 1 // R 2 // r b RBRB

rbrb R OUT Output impedance

If R L =10kΩ and we want a low frequency cutoff of 20Hz, What is C 2 ? If V BB =15V and I C =2mA what is the output impedance? DC condition Frequency response Impedance Gain/Dynamic range

Why do we care about the input and output impedance? Simplest "black box" amplifier model: Impedances R IN R OUT V IN AV IN V OUT The amplifier measures voltage across R IN, then generates a voltage which is larger by a factor A This voltage generator, in series with the output resistance R OUT, is connected to the output port. A should be a constant (i.e. gain is linear)

Attach an input - a source voltage V S plus source impedance R S Impedances R IN R OUT V IN AV IN V OUT Note the voltage divider R S + R IN. V IN =V S (R IN /(R IN +R S ) We want V IN = V S regardless of source impedance So want R IN to be large. The ideal amplifier has an infinite input impedance VSVS RSRS

Attach a load - an output circuit with a resistance R L Impedances Note the voltage divider R OUT + R L. V OUT =AV IN (R L /(R L +R OUT ) Want V OUT =AV IN regardless of load We want R OUT to be small. The ideal amplifier has zero output impedance R IN R OUT V IN AV IN V OUT VSVS RSRS RLRL

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 R in, and a voltage source with gain A v in series with an output impedance R out.

Ideal op-amp Place a source and a load on the model Infinite internal resistance R in (so v in =v s ). Zero output resistance R out (so v out =A v v in ). "A" very large No saturation i in =0; no current flow into op-amp - + v out RLRL RSRS 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. Pin Function 2Inverting input 3Non-inverting input 4V- supply 6Output 7V+ supply

Measuring Impedances Assuming you can only vary R L and R S, how would you measure the input and output impedances of the amplifier? R IN R OUT V IN AV IN V OUT VSVS RSRS RLRL

Measuring Impedances With the black box model, it is simple to measure the input and output impedances of an amplifier To measure the input impedance, vary R S until the output voltage has dropped to half ; then R S =R IN = input impedance To measure the output impedance, vary R L until the output voltage has dropped to half ; then R L =R OUT = output impedance R IN R OUT V IN AV IN V OUT VSVS RSRS RLRL

Cascaded Amplifiers Easiest way to increase amplification is to link amplifiers together R IN1 R OUT1 V IN1 A 1 V IN1 V OUT1 R IN2 R OUT2 V IN2 A 2 V IN2 V OUT2 Ideal amplifiers; V OUT2 =A 1 A 2 V IN1 In reality, take account of voltage divider action due to input and output impedances