Common Base BJT Amplifier Common Collector BJT Amplifier

ESE319 Introduction to Microelectronics

Common Base BJT Amplifier

Common Collector BJT Amplifier

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Common Collector (Emitter Follower) Configuration

Common Base Configuration

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Small Signal Analysis

Design Example

ESE319 Introduction to Microelectronics

Basic Single BJT Amplifier Features

CE Amplifier CC Amplifier CB Amplifier

Voltage Gain (A_V) moderate (-R_C/R_E) low (about 1) high

Current Gain (A_I) moderate ( ) moderate ( ) low (about 1)

Input Resistance high high low

Output Resistance high low high

 1

CE BJT amplifier => CS MOS amplifier

CC BJT amplifier => CD MOS amplifier

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Common Collector ( Emitter Follower)

Amplifier

In the emitter follower, the output voltage is taken be-tween emitter and ground.

The voltage gain of this ampli-fier is nearly one – the output “follows” the input - hence the name: emitter “follower.”

ESE319 Introduction to Microelectronics

Split bias voltage drops about equally across the transistor

V_CE (or V_CB) and V_Re(or V_B). For simplicity,choose: R₁₌R₂ For an assumed = 100: R_B=R_1∥R₂₌ R1 2 =1 R_E 10 ≈10 RE R₁₌R₂₌₂₀ R_E V_B=VCC 2 R_E=VE I _E = V_CC/2−0.7 I _E ⇒

Then, choose/specified I_E, and the rest of the design follows:

Vb i B i C i E vout 

As with CE bias design, stable op. pt. => R_{B≪1}R_E , i.e.

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Typical Design

Choose: I _E=1mA

V_CC=12V

And the rest of the design follows immediately:

R_E=VE I _E =

12_/2−0.7

10−3 =5.3k 

Use standard sizes! R₁₌R₂₌₁₀₀k _ R_E=5.1k _ vout Vb i B i C i E

ESE319 Introduction to Microelectronics

Equivalent Circuits

ESE319 Introduction to Microelectronics

Multisim Bias Check

Identical results – as expected! <=> Rb + -VRb V_Rb=I_BR_B= IE 1 RB=0.495V i B Rb vout _vout

ESE319 Introduction to Microelectronics

Small signal mid-band circuit - where C_B has negligible reactance (above _min). Thevenin circuit consisting of R_S and R_B shows

effect of R_B negligible, since it is much larger than R_S.

Emitter Follower Small Signal Circuit

Mid-band equivalent circuit:

v_sig'₌ RB R_BR_S vsig= 50 50.05 vsig≈vsig R_TH=R_S∥R_B= 50 50.05 RS≈RS  Rb vout

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Follower Small Signal Analysis - Voltage Gain

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Small Signal Analysis – Voltage Gain - cont.

v_out v_sig = R_E R_Sr_ 1 RE Since, typically: R_Sr_ 1 ≪RE

A

v

v

R

R

ESE319 Introduction to Microelectronics

i_b= vbg

r_{1}R_E

Use the base current expression:

To obtain the base to ground resistance of the transistor: This transistor input resistance is in parallel with the 50 k

R_B, forming the total amplifier input resistance: R_{i n=}R_SR_B∥r_bg≈R_B∥r_bg= 515 5155050 k=45.6 k  v_bg=r_i_bR_Ei_E=r_{1}i_b vbg + -R_in r_bg=vbg i_b =r1RE≈1RE=101⋅5.1k=515k  Rb R_B=50k≫RS R_S=50

Blocking Capacitor - C

- Selection

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C

– Selection cont.

R_{i n≈46} k_

=100

_min=220≈125=1.25⋅102

Assume the lowest frequency is 20 Hz:

C_B≥ 10⋅10−7

1.25_⋅0.46≈1.73 F

Choose C_B such that its reactance is ≤ 1/10 of R_in at _min:

C_{B ≥} 10 minRi n 1 C_B= R_{i n} 10

Pick C_B = 2 F (two 1 F caps in parallel), the nearest standard value in the RCA Lab. We could be (unnecessarily) more precise and include R_s as part of the total resistance in the loop. It is very small compared to R_in.



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Final Design

2.0 uF

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Multisim Simulation Results

20 Hz. Data

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Of What value is a Unity Gain Amplifier?

To answer this question, we must examine the

output impedance of the

amplifier and its power gain. vout i b i e

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Emitter Follower Output Resistance

v_x i_x R_out 0 i_b ix=−ib−ib=−1ib⇒ib= −i_x 1 v_x=−i_bRs_r_= RSr 1_ ix R_out=vx i_x = R_Sr_ 1_ ≈ r_ 1_ Assume: I_C=1mA⇒r_=VT I _B = V_T I_C =2500 =100 R_S=50 R_out≈ 2550 100 =25.5 vout

R_B=50k≫R_S Rout is the Thevenin resistance looking

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Multisim Verification of R

Multisim short circuit check

( = 100, v_out = v_sig): R_out= voc i_sc = A_Vv_sigrms i_scrms = 1 0.0396=25.25

Thevenin equivalent for the

short-circuited emitter follower. If was 200, as for most good NPN transistors, R_out would be lower - close to 12 .    Rout Av*vsig A_V = 1 <=> i_sc=i_x i_sc=i_x Rin i_x=−1i_x v_sig= R_Si_br_i_b R_out= AV_ivsig x = RSr 1 R_in=RSr1 RE∥RL≈1 RL + -v_oc=AVv_sig

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Equivalent Circuits with Load R

vout i_b i_e RL + - vload RL||Re + -R_out=vsigrms i_scrms = 1 0.0396=25.25 <=> Rin Z_in=vsig i'e R_in=R_Sr_{1} R_E∥R_{L≈1}R_L Rout Av*vsig

ESE319 Introduction to Microelectronics

Emitter Follower Power Gain

Consider the case where a R_L = 50 load is connected through an infinite

capacitor to the emitter of the follower we designed. Using its Thevenin equivalent: v_load= RL AV vsig R_LR_out = 50 75 vsig= 2 3 vsig i_load= AVvsig R_outR_L= v_sig 75

p_load=v_load i_load= 2

225 vsig 2 i_sig=i_b=vsig R_in ≈ v_sig 1 R_E∥R_L≈ v_sig 101_⋅50 p_sig=v_sigi_sig≈ 1

5000 vsig 2  + - vload -vth=G vsig Rout≈25  RL≈50 +

50 load is in parallel with 5.1k

R_E and dominates:  C=∞ A_V≈1 iload Rin vsig Av*vsig G_pwr= pload p = 2_5000 225 =44.4≫1 isig

ESE319 Introduction to Microelectronics

The Common Base Amplifier

Voltage Bias Design Current Bias Design

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Common Base Configuration

Both voltage and current biasing follow the same rules as those applied to the common emitter amplifier.

As before, insert a blocking capacitor in the input signal path to avoid disturbing the dc bias.

The common base amplifier uses a bypass capacitor – or a direct connection from base to ground to hold the base at ground for the signal only!

The common emitter amplifier (except for intentional R_E

feedback) holds the emitter at signal ground, while the common collector circuit does the same for the collector.

ESE319 Introduction to Microelectronics

We keep the same bias that we established for the gain of 10 common emitter amplifier.

All that we need to do is pick the capacitor values and calculate the circuit gain.

vout

470 Ohm 4.7 k Ohm 47k Ohm

5.1k Ohm

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Common Base Small Signal Analysis - C

Determine C_IN:

Find a equivalent impedance for the input circuit, R_S, C_IN, and R_E2:

i'_b I'_c i'_e i_sig Z_in 4.7 k Ohm 470 Ohm

ideally for ≥min

1 minCIN ≪R_SR_E2∥r_e⇒ 1 minCIN = RSre 10 ⇒CIN= 10 minRSre v_Re2= RE2∥re R_E2∥r_eR_S 1 j C_IN v_sig v_Re2= RE2∥re R_E2∥r_eR_S vsig (let ) C_B=∞ ∞ i_b i_c i_e r_e= r 1_ r_e

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Determine C

cont.

C_IN= 10

125.6_⋅75≈1062  F ! Not too practical!

Must choose smaller value of C_IN.

1. Choose: _minC_INR_Sr_e=1

or

ESE319 Introduction to Microelectronics

Small-signal Analysis - C

i'_{b i'}

c

i'_e

Note the reference current reversals (due to v_sig polarity)! v_sig=R_Si_e' _









ESE319 Introduction to Microelectronics

Determine – C









ESE319 Introduction to Microelectronics

Determine - C

cont.









ESE319 Introduction to Microelectronics

Small-signal Analysis – Voltage Gain





ESE319 Introduction to Microelectronics

Multisim Simulation

ESE319 Introduction to Microelectronics

Multisim Frequency Response

20 Hz. response