Chapter 4 Bipolar Junction Transistors. Outline Basic operation of the npn Bipolar Junction Transistor Load-line analysis of a common-emitter amplifier The pnp bipolar junction transistor Small-signal equivalent circuits The common-emitter amplifier
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Chapter 4 Bipolar Junction Transistors Outline • Basic operation of the npn Bipolar Junction Transistor • Load-line analysis of a common-emitter amplifier • The pnp bipolar junction transistor • Small-signal equivalent circuits • The common-emitter amplifier • The emitter follower (common-collector amplifier) • The common-base amplifier
4.1 Basic operation of the npn transistor • A npn BJT consists of a thin layer of p-type material between two layers of n-type material. • Two pn junctions are formed in the device. • The current flowing across one junction affects the current in the other junction. It is this interaction that makes the BJT useful as an amplifier. Figure 4.1 The npn BJT
4.1 Basic operation of the npn transistor • Basic operation in the active region • The common-emitter configuration • Active region (normal operation) • Base-emitter junction is forward biased • The base-collector junction is reverse biased The emitter region is doped heavily. The base region is very thin. Figure 4.2 An npn transistor with variable biasing sources (CE configuration)
4.1 Basic operation of the npn transistor • Amplification by the BJT Figure4.4 CE characteristics of a typical npn BJT A small change in vbe can result in an appreciable change in ib, this causes a much larger change in ic, and in suitable circuits, it is converted into a much larger voltage change than the initial change in vbe.
4.1 Basic operation of the npn transistor • The common-emitter current gain • Device equations 10~1000 • The common-base current gain • Exercise: • Find the relationship between and • Estimate and for Figure4.4
4.2 Load-line analysis of a CE amplifier The power-supply voltages VBB and VCC bias the device at an Q point for which the amplification of the input signal is possible. • Exercise: • Write KVL for input loop. • Write KVL for output loop. P144 Example 4.2 Figure4.6 Common emitter amplifier
4.2 Load-line analysis of a CE amplifier Pay attention to Phase relationship between vin and vout
4.2 Load-line analysis of a CE amplifier (ref 4.4) • Active region model — amplification (1) • Saturation region model (2) • Cutoff region model (3) • Inverted (reverse) model (4) Figure 4.11 Amplification in the active region
4.2 Load-line analysis of a CE amplifier Distortion occurs? Figure Distortion illustration
4.3 The pnp bipolar junction transistor Example: Determine the diode states for the circuits shown below. Assume ideal diodes. Think about: 1. Circuit configuration 2. Active region — source polarity Figure 4.12 The pnp BJT
4.4 Large-signal DC analysis of BJT circuit Example: Find the Q-point (VBEQ, IBQ, VCEQ, ICQ) of the circuit (P151) • Think about: • Indicate the current flowing through RB and RC • Write KVL for input/output loop =100 Figure 4.13 The BJT VBEQ is given (about 0.7V for Si, and 0.2V for Ge) IBQ=(VCC-VBEQ)/RB ICQ= IBQ VCEQ=VCC-ICQRC
4.4 Large-signal DC analysis of BJT circuit Example: Find the Q-point of the circuit (P174) • Think about: • Indicate the current flowing through RB，RE and RC • Write KVL for input/output loop IBQ=(VB-VBEQ)/[RB+(1+)RE] ICQ= IBQ Figure 3.11 Half-wave rectifier with resistive load Figure 4.14 BJT with two voltage supplies VCEQ=VCC-IEQRE-ICQRC VCC-ICQ(RC+RE)
4.4 Large-signal DC analysis of BJT circuit Analysis of the four-resistor bias circuit Method 1 — Thevenin equivalent circuit + - Convert to Figure4.14 VBQ VCEQ Method 2 — Assume IR1>>IBQ, IR2>>IBQ R1and R2are in series VBQR2VCC/(R1+R2) Figure 4.15 The BJT with four resistors IEQ=(VBQ-VBEQ)/RE VCEQ=VCC-IEQRE-ICQRC VCC-ICQ(RC+RE)
4.5 Small-signal equivalent circuit Input resistance rbe rbe(r)=300+(1+)26(mV)/IEQ(mA) Tansconductancegm gm= /rbe Output resistance rce Usually large, negligible rce rce Figure 4.16 small signal equivalent circuits for the BJT
4.6 The common-emitter amplifier (P164) • DC supply voltage —— bias the device at a suitable operating point • AC signal ——what we want to amplify • The use of C1, C2 and Ce • Coupling—C1, C2 • bypass —Ce • Circuit configuration • What shall we do with this circuit? • DC bias circuit — determine Q point • Small signal equivalent circuit • Find Av, Rin and Rout C2 Ce Figure 4.28 Common-emitter amplifier of Example 4.9.
4.6 The common-emitter amplifier (P164) • Draw the DC bias circuit to find the Q point • Ac signal — short circuit • Capacitor —open circuit + - VCEQ VBQ Figure 4.28 Common-emitter amplifier of Example 4.9. Rule of thumb: VCEQ 0.3~0.7VCC
rce 4.6 The common-emitter amplifier (P164) • 2. Draw the AC circuit with BJT • DC signal — Connect to the GND • Capacitor —short circuit 3. Replace with the equivalent circuit Figure 4.28 Common-emitter amplifier of Example 4.9.
rce 4.6 The common-emitter amplifier (P164) Find Av=Vo/Vin, Rin=Vin/Iin andRout=Vo/Io|Vs=0 Think about: 1. the relationship between R1, R2 and rce 2. the relationship between rce, RC and RL 3. Write two equations for Vin and Vo
rce 4.6 The common-emitter amplifier (P164) Find the voltage gain Av=Vo/Vin Find the open circuit voltage gain Avo=Vo/Vin|RL=0 Find the source voltage gain Avs=Vo/Vs Find the input resistance Rin=Vin/Iin Find the output resistanceRout=Vo/Io|Vs=0 Rout
4.6 The common-emitter amplifier (P164) Analysis of the CE amplifier without CE DC operating point C2 Remains unchanged Ce AC performance Need more detailed analysis Exercise: 1. Draw the small-signal equivalent circuit 2. Find Av=Vo/Vin, Rin=Vin/Iin andRout=Vo/Io|Vs=0assumerce 3. Compare with the results for circuit with CE
4.6 The common-emitter amplifier (P164) Think about: 1. the currents flowing through rbe, RE and RL’ 2. Write two equations for Vin and Vo
4.6 The common-emitter amplifier (P162) Figure 4.27 Common-emitter amplifier. Think about: 1. Effects of RE1 on Q-point, Av,Rinand Rout 2. Effects of RE2 on Q-point, Av,Rinand Rout
4.7 The emitter follower (P166) • Draw the DC bias circuit to find the Q point Given R1=R2=100k, =200, VBEQ=0.7V find the Q point.
R1 RL 4.7 The emitter follower (P166) (b) AC circuit
4.7 The emitter follower (P166) Find Av=Vo/Vin, Rin=Vin/Iin andRout=Vo/Io|Vs=0 Think about: Write two equations for Vin and Vo Rout RS’ Rout’
4.8 The common-base amplifier (P299) • Draw the DC bias circuit to find the Q point
4.8 The common-base amplifier (P299) Find Av=Vo/Vin, Rin=Vin/Iin andRout=Vo/Io|Vs=0 Think about: Write two equations for Vin and Vo RC Rin Rout’ Rout
4.8 The common-base amplifier (P299) • Summary
The overall voltage gain of cascaded amplifier stages is the product of the voltage gains of the individual stages. Think over: If Avo1=100, Avo2=200, what is the overall open circuit voltage gain of cascaded amplifier? Avo=Avo1Avo2? (refer to p17) Review 1.5 Cascaded amplifiers
Review 1.5 Cascaded amplifiers • Applications calling for high or low input impedance • Applications calling for high or low output impedance • Application calling for a particular impedance • Refer to examples shown on page 26-27
4.9 Cascode amplifier (P300) • 共集－共射组合放大电路，不仅具有共集电极电路输入电阻大的特点，而且具有共射电路电压放大倍数大的特点； • 共射－共集组合放大电路，不仅具有共射电路电压放大倍数大的特点，而且具有共集电极电路输出电阻小的特点； • 共射－共基组合放大电路，共基极电路本身就有较好的高频特性，同时将输入电阻很小的共基极电路接在共射极电路之后，减小了共射极电路的电压放大倍数，使共射极接法的管子集电结电容效应减小，改善了放大电路的频率特性。因此，共射－共基组合放大电路在高频电路中获得了广泛的应用。该组合电路的电压放大倍数近似等于一般共射电路的电压放大倍数。
AC coupling versus direct coupling (P32) Figure 1.37 Capacitive coupling illustration AC coupling — The DC voltages of the amplifier circuits do not affect the signal source, adjacent stages, or the load. Amplifiers that are realized as integrated circuits are almost always DC coupled because the capacitors or transformers needed for ac coupling cannot be fabricated in integrated form. BACK
4.10 Frequency response for an amplifier • Frequency response (P30) • The complex gain: The ratio of the phasor for the output signal to the input signal • Bode plot (P271) • How circuit functions can be quickly and easily plotted against frequency? (straight line approximation & smart scale)
Review Logarithmic Frequency Scale A decade is a range of frequencies for which the ratio of the highest frequency to the lowest is 10. An octave is a two-to-one change in frequency.
Review: Passive low-pass filter Figure 8.1 Low-pass RC filter.
4.10 Frequency response for an amplifier • Question: • What is the voltage gain of the circuit? • Can you draw the frequency response of amplifier? • What kind of filter is it? • Do you remember the small signal equivalent circuit of the NPN transistor? • Any changes to (4) if the frequency is high?
4.10 Frequency response for an amplifier Cbc Mid-frequency Cbe The small signal equivalent circuit for the NPN transistor At high frequencies Usually, Cbe is about 10PF ~ 1000PF Cbc <10PF
Miller effect Question: (P296) If Zf=-j/C for a CE amplifier, how do you find Zin,miller and Zout,miller
(1+gmRL’)Cbc Cbc Cbe rbe ib Find the Thevenin equivalent resistance Rs’?
4.10 Frequency response for an amplifier Figure 8.34 Simplified equivalent circuit for the common-emitter amplifier. • What kind of filter is it? • What is the break (cut-off) frequency? • What factors affect the value of the break frequency? • What do we desire about an amplifier?
4.10 Frequency response for an amplifier Figure 8.36 High-frequency behavior of the common-emitter amplifier. Conclusion: The common-base amplifier achieves wide bandwidth, but its input impedance is very low. Consequently, the mid-band gain can be quite small due to loading of source. CE-CC cascade amplifier combines the advantages of both.
4.10 Frequency response for an amplifier Figure 8.47 Amplifier with coupling capacitors.