ECES 352 Winter 2007Ch. 7 Frequency Response Part 31 Common-Base (CB) Amplifier *DC biasing ● Calculate I C, I B, V CE ● Determine related small signal.

Presentation on theme: "ECES 352 Winter 2007Ch. 7 Frequency Response Part 31 Common-Base (CB) Amplifier *DC biasing ● Calculate I C, I B, V CE ● Determine related small signal."— Presentation transcript:

1 ECES 352 Winter 2007Ch. 7 Frequency Response Part 31 Common-Base (CB) 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 Input at emitter, output at collector.

2 ECES 352 Winter 2007Ch. 7 Frequency Response Part 32 CB Amplifier - DC Analysis (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 - I C R C - (β +1) I B R E = 4.4 V (okay since not close to zero volts). R 1 = 10K R 2 = 2.5K R C = 1.2K R E = 0.33K

3 ECES 352 Winter 2007Ch. 7 Frequency Response Part 33 *Construct small signal ac equivalent circuit (set DC supply to ground) *Substitute small signal equivalent circuit (hybrid-pi model) for transistor *Neglect all capacitances ● Coupling and emitter bypass capacitors become shorts at midband frequencies (~ 10 5 rad/s) «Why? Impedances are negligibly small, e.g. few ohms because C C1, C C2, C E ~ few μF (10 -6 F) ● Transistor capacitances become open circuits at midband frequencies «Why? Impedances are very large, e.g. ~ 10’s M Ω because C π, C μ ~ pF (10 -12 F) *Calculate small signal voltage gain A Vo = V o /V s CB Amplifier - Midband Gain Analysis High and Low Frequency AC Equivalent Circuit

4 ECES 352 Winter 2007Ch. 7 Frequency Response Part 34 CB Amplifier - Midband Gain Analysis Equivalent resistance r e VeVe +_+_ IπIπ rere βIπβIπ Voltage gain is less than one !

5 ECES 352 Winter 2007Ch. 7 Frequency Response Part 35 What Happened to the CB Amplifier’s Midband Gain? *Source resistance R s = 5K is killing the gain. ● Why? R s >> r e = 0.0051 K so V e /V s <<1 *Need to use a different signal source with a very low source resistance R s, i.e. ~ few ohms *Why is r e so low? ● V s drives formation of V e ● V e creates V π across r π ● V π turns on dependent current source ● Get large I e for small V e so r e =V e /I e is very small. VeVe +_+_ rere Voltage gain is now much bigger than one !

6 ECES 352 Winter 2007Ch. 7 Frequency Response Part 36 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 and base capacitors C C1, C C2, C B ● 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 the corresponding pole frequency *Determine the dominant (largest) pole frequency *Calculate the final low pole frequency using

7 ECES 352 Winter 2007Ch. 7 Frequency Response Part 37 Common Base - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique *Base capacitor C B = 12 μF ViVi + _ R xC B RiRi IπIπ Low Frequency AC Equivalent Circuit VπVπ VxVx VoVo IxIx

8 ECES 352 Winter 2007Ch. 7 Frequency Response Part 38 Common Base - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique Input coupling capacitor C C1 = 2 μF VeVe IeIe IπIπ VeVe +_+_ rere VπVπ VoVo IxIx RsRs VxVx

9 ECES 352 Winter 2007Ch. 7 Frequency Response Part 39 Common Base - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique *Output coupling capacitor C C2 = 3 μF *Low 3dB frequency VoVo RLRL VXVX RCRC Dominant low frequency pole is due to C C1 !

10 ECES 352 Winter 2007Ch. 7 Frequency Response Part 310 *What are the zeros for the CB amplifier? *For C C1 and C C2, we get zeros at ω = 0 since Z C = 1 / jωC and these capacitors are in the signal line, i.e. Z C  at ω = 0 so V o  0. *Consider R B in parallel with C B *Impedance given by *When Z’ B  , I π  0, so g m V π  0, so V o  0 *Z’ B   when s = - 1 / R B C B so pole for C B is at Common Base - Low Frequency Zeros IπIπ

11 ECES 352 Winter 2007Ch. 7 Frequency Response Part 311 Common Base - Low Frequency Poles and Zeros Magnitude Bode Plot

12 ECES 352 Winter 2007Ch. 7 Frequency Response Part 312 Common Base - Low Frequency Poles and Zeros Phase Shift Bode Plot

13 ECES 352 Winter 2007Ch. 7 Frequency Response Part 313 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, C C2, C B 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

14 ECES 352 Winter 2007Ch. 7 Frequency Response Part 314 Common Base - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique High frequency AC equivalent circuit NOTE: We neglect r x here since the base is grounded. This simplifies our analysis, but doesn’t change the results appreciably.

15 ECES 352 Winter 2007Ch. 7 Frequency Response Part 315 Common Base - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique *Equivalent circuit for Z e ZeZe ZeZe Replace this with this. VeVe +_+_ Parallel combination of a resistor and capacitor.

16 ECES 352 Winter 2007Ch. 7 Frequency Response Part 316 *Pole frequency for C π =17pF Common Base - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique Turn off signal source when finding resistance seen by capacitor.

17 ECES 352 Winter 2007Ch. 7 Frequency Response Part 317 Common Base - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique *Equivalent circuit for Capacitor C μ = 1.3 pF *Pole frequency for C μ =1.3pF *High 3 dB frequency Dominant high frequency pole is due to C μ ! R s || R E || r π = 0

18 ECES 352 Winter 2007Ch. 7 Frequency Response Part 318 Common Base - High Frequency Zeros *What are the high frequency zeros for the CB 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 output will be shorted to ground thru C μ. *Similarly,we get a zero from C π when when ω   since Z C π = 1/sC π  0, so the voltage V π  0. *Both C π and C μ give high frequency zeros at ω  !

19 ECES 352 Winter 2007Ch. 7 Frequency Response Part 319 Common Base - High Frequency Poles and Zeros Magnitude

20 ECES 352 Winter 2007Ch. 7 Frequency Response Part 320 Common Base - High Frequency Poles and Zeros Phase Shift

21 ECES 352 Winter 2007Ch. 7 Frequency Response Part 321 Comparison of CB to CE Amplifier CE (with R S = 5K) CB (with R S = 5Ω) Midband Gain Low Frequency Poles and Zeros High Frequency Poles and Zeroes Note: CB amplifier has much better high frequency performance!

22 ECES 352 Winter 2007Ch. 7 Frequency Response Part 322 Comparison of CB to CE Amplifier (with same R s = 5 Ω) CE (with R S = 5 Ω) CB (with R S = 5Ω) Midband Gain Low Frequency Poles and Zeros High Frequency Poles and Zeroes Note: CB amplifier has much better high frequency performance!

23 ECES 352 Winter 2007Ch. 7 Frequency Response Part 323 Conclusions *Voltage gain ● Can get good voltage gain from both CE and CB amplifiers. ● Low frequency performance similar for both amplifiers. ● CB amplifier gives better high frequency performance ! «CE amplifier has dominant pole at 5.0x10 7 rad/s. «CB amplifier has dominant pole at 7.1x10 8 rad/s. *Bandwidth approximately 14 X larger! *Miller Effect multiplication of C  by the gain is avoided in CB configuration. *Current gain ● For CE amplifier, current gain is high A I = I c / I b ● For CB amplifier, current gain is low A I = I c / I e (close to one)! ● Frequency dependence of current gain similar to voltage gain. *Input and output impedances are different for the two amplifiers! ● CB amplifier has especially low input resistance.