ECES 352 Winter 2007Ch. 7 Frequency Response Part 41 Emitter-Follower (EF) Amplifier *DC biasing ● Calculate I C, I B, V CE ● Determine related small signal equivalent circuit parameters «Transconductance g m «Input resistance r π *Midband gain analysis *Low frequency analysis ● Gray-Searle (Short Circuit) Technique «Determine pole frequencies ω PL1, ω PL2,... ω PLn ● Determine zero frequencies ω ZL1, ω ZL2,... ω ZLn *High frequency analysis ● Gray-Searle (Open Circuit) Technique «Determine pole frequencies ω PH1, ω PH2,... ω PHn ● Determine zero frequencies ω ZH1, ω ZH2,... ω ZHn High and Low Frequency AC Equivalent Circuit

ECES 352 Winter 2007Ch. 7 Frequency Response Part 42 EF Amplifier - DC Analysis (Nearly the Same as CE Amplifier) *GIVEN: Transistor parameters: ● Current gain β = 200 ● Base resistance r x = 65 Ω ● Base-emitter voltage V BE,active = 0.7 V ● Resistors: R 1 =10K, R 2 =2.5K, R C =1.2K, R E =0.33K *Form Thevenin equivalent for base; given V CC = 12.5V ● R Th = R B = R 1 ||R 2 = 10K||2.5K = 2K ● V Th = V BB = V CC R 2 / [R 1 +R 2 ] = 2.5V ● KVL base loop «I B = [V Th -V BE,active ] / [R Th +(β +1)R E ] «I B = 26 μA *DC collector current I C = β I B I C = 200(26 μ A) = 5.27 mA *Transconductance g m = I C / V T ; V T = k B T/q = 26 mV g m = 5.27 mA/26 mV = 206 mA/V *Input resistance r π = β / g m = 200/[206 mA/V]= 0.97 K *Check on transistor region of operation ● KVL collector loop ● V CE = V CC - (β +1) I B R E = 10.8 V (was 4.4 V for CE amplifier) (okay since not close to zero volts). R 1 = 10K R 2 = 2.5K R C = 0 K R E = 0.33K Note: Only difference here from CE case is V CE is larger since R C was left out here in EF amplifier.

ECES 352 Winter 2007Ch. 7 Frequency Response Part 43 EF Amplifier - Midband Gain Analysis Equivalent input resistance R i ViVi + _ RiRi VbVb + _ IπIπ DC analysis is nearly the same! I B, I C and g m are all the same. Only V CE is different since R C =0. VOVO NOTE: Voltage gain is only ~1! This is a characteristic of the EF amplifier! Cannot get voltage gain >1 for this amplifier!

ECES 352 Winter 2007Ch. 7 Frequency Response Part 44 Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique *Draw low frequency AC circuit ● Substitute AC equivalent circuit for transistor (hybrid-pi for bipolar transistor) ● Include coupling capacitors C C1, C C2 ● Ignore (remove) all transistor capacitances C π, C μ *Turn off signal source, i.e. set V s = 0 ● Keep source resistance R S in circuit (do not remove) *Consider the circuit one capacitor C x at a time ● Replace all other capacitors with short circuits ● Solve remaining circuit for equivalent resistance R x seen by the selected capacitor ● Calculate pole frequency using ● Repeat process for each capacitor finding equivalent resistance seen and corresponding pole frequency *Calculate the final low 3 dB frequency using

ECES 352 Winter 2007Ch. 7 Frequency Response Part 45 Emitter Follower - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique Input coupling capacitor C C1 = 2 μF RiRi ViVi IπIπ IXIX

ECES 352 Winter 2007Ch. 7 Frequency Response Part 46 Emitter Follower - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique *Output coupling capacitor C C2 = 3 μF *Low 3 dB frequency IπIπ VeVe IeIe IXIX rere So dominant low frequency pole is due to C C1 !

ECES 352 Winter 2007Ch. 7 Frequency Response Part 47 Emitter Follower - Low Frequency Zeros *What are the zeros for the EF amplifier? *For C C1 and C C2, we get zeros at ω = 0 since Z C = 1 / ωC and these capacitors are in the signal line, i.e. Z C   at ω = 0 so V o  0.

ECES 352 Winter 2007Ch. 7 Frequency Response Part 48 Emitter Follower - Low Frequency Poles and Zeros Magnitude Bode Plot

ECES 352 Winter 2007Ch. 7 Frequency Response Part 49 Emitter Follower - Low Frequency Poles and Zeros Phase Shift Bode Plot

ECES 352 Winter 2007Ch. 7 Frequency Response Part 410 Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique *Draw high frequency AC equivalent circuit ● Substitute AC equivalent circuit for transistor (hybrid-pi model for transistor with C π, C μ ) ● Consider coupling and emitter bypass capacitors C C1 and C C2 as shorts ● Turn off signal source, i.e. set V s = 0 ● Keep source resistance R S in circuit ● Neglect transistor’s output resistance r o *Consider the circuit one capacitor C x at a time ● Replace all other transistor capacitors with open circuits ● Solve remaining circuit for equivalent resistance R x seen by the selected capacitor ● Calculate pole frequency using ● Repeat process for each capacitor *Calculate the final high frequency pole using

ECES 352 Winter 2007Ch. 7 Frequency Response Part 411 Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique *Redrawn High Frequency Equivalent Circuit IeIe Z eq E IeIe z π =1/y π

ECES 352 Winter 2007Ch. 7 Frequency Response Part 412 Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique Modified Equivalent Circuit Replace this with this. ZB’ZB’ ZB’ZB’ z π =1/y π Looks like a resistor in parallel with a capacitor.

ECES 352 Winter 2007Ch. 7 Frequency Response Part 413 Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique R xCπ *Pole frequency for C π =17 pF

ECES 352 Winter 2007Ch. 7 Frequency Response Part 414 Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique *Pole frequency for C μ =1.3 pF

ECES 352 Winter 2007Ch. 7 Frequency Response Part 415 Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique *Alternative Analysis for Pole Due to C π E IeIe VxVx IxIx IπIπ I x -I π I e +g m V π We get the same result here for the high frequency pole associated with C π as we did using the equivalent circuit transformation.

ECES 352 Winter 2007Ch. 7 Frequency Response Part 416 Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique *Alternative Analysis for Pole Due to C µ VxVx IxIx IπIπ I x -I π I π +g m V π E We get the same result here for the high frequency pole associated with C µ as we did using the equivalent circuit transformation.

ECES 352 Winter 2007Ch. 7 Frequency Response Part 417 Emitter Follower - High Frequency Zeros *What are the high frequency zeros for the EF amplifier? *Voltage gain can be written as *When V o /V π = 0, we have found a zero. *For C μ, we get V o  0 when ω   since the node B’ will be shorted to ground and V π = 0. *Similarly, we get a zero from C π when y π + g m = 0 since we showed earlier that *Also, can see this from

ECES 352 Winter 2007Ch. 7 Frequency Response Part 418 Emitter Follower - High Frequency Poles and Zeros Magnitude

ECES 352 Winter 2007Ch. 7 Frequency Response Part 419 Emitter Follower - High Frequency Poles and Zeros Phase Shift

ECES 352 Winter 2007Ch. 7 Frequency Response Part 420 Comparison of EF to CE Amplifier (For R S = 5Ω ) CE EF Midband Gain Low Frequency Poles and Zeros High Frequency Poles and Zeroes Better low frequency response ! Much better high frequency response !

ECES 352 Winter 2007Ch. 7 Frequency Response Part 421 Conclusions *Voltage gain ● Can get good voltage gain from CE but NOT from EF amplifier (A V  1). ● Low frequency performance better for EF amplifier. ● EF amplifier gives much better high frequency performance! «CE amplifier has dominant pole at 5.0x10 7 rad/s. «EF amplifier has dominant pole at 1.0x10 10 rad/s. *Bandwidth approximately 200 X larger! *Miller Effect multiplication of C  by the gain is avoided in EF. *Current gain ● For CE amplifier, current gain is high  = I c /I b ● For EF amplifier, current gain is also high I e /I b =  +1 ! ● Frequency dependence of current gain similar to voltage gain. *Input and output impedances are different for the two amplifiers!