chapter 4 bipolar junction transistors
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Chapter 4 Bipolar Junction TransistorsOutline 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
Figure 4.1 The npn BJT
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.
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
Figure 4.2 An npn transistor with variable biasing sources (CE configuration)
4.1 Basic operation of the npn transistor
The emitter region is doped heavily. The base region is very thin.
Amplification by the BJT
4.1 Basic operation of the npn transistor
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.
The common-emitter current gain
Device equations
4.1 Basic operation of the npn transistor
B
C
ii
10~1000
CBE iii
The common-base current gain
E
C
ii
Exercise:1. Find the relationship between
and 2. 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:1. Write KVL for input
loop.2. Write KVL for output
loop.
P144 Example 4.2Figure4.6 Common emitter amplifier
4.2 Load-line analysis of a CE amplifier
Pay attention to Phase relationship between vin and vout
Figure 4.11 Amplification in the active region
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) Model B-E junction B-C junction
(1) forward reverse(2) forward forward(3) reverse reverse(4) reverse forward
Figure Distortion illustration
4.2 Load-line analysis of a CE amplifier
Distortion occurs?
Example: Determine the diode states for the circuits shown below. Assume ideal diodes.
Figure 4.12 The pnp BJT
4.3 The pnp bipolar junction transistor
Think about:1. Circuit configuration2. Active region — source polarity
4.4 Large-signal DC analysis of BJT circuit
Example: Find the Q-point (VBEQ, IBQ, VCEQ, ICQ) of the circuit (P151) Think about:
1. Indicate the current flowing through RB and RC
2. Write KVL for input/output loop
=100
VBEQ is given (about 0.7V for Si, and 0.2V for Ge)IBQ=(VCC-VBEQ)/RB ICQ= IBQ
VCEQ=VCC-ICQRC
Figure 4.13 The BJT
Figure 3.11 Half-wave rectifier with resistive load
Example: Find the Q-point of the circuit (P174)
4.4 Large-signal DC analysis of BJT circuit
Think about:1. Indicate the current flowing through RB , RE and RC
2. Write KVL for input/output loop
Figure 4.14 BJT with two voltage supplies
IBQ=(VB-VBEQ)/[RB+(1+)RE]
ICQ= IBQ
VCEQ=VCC-IEQRE-ICQRC VCC-ICQ(RC+RE)
Analysis of the four-resistor bias circuit
4.4 Large-signal DC analysis of BJT circuit
Method 1 — Thevenin equivalent circuit
VBQR2VCC/(R1+R2)
Figure 4.15 The BJT with four resistors
Convert to Figure4.14
R1 and R2 are in series
Method 2 — Assume IR1>>IBQ,
IR2>>IBQ
IEQ=(VBQ-VBEQ)/RE
VBQ VCEQ
+
-
VCEQ=VCC-IEQRE-ICQRC VCC-ICQ(RC+RE)
4.5 Small-signal equivalent circuit
Figure 4.16 small signal equivalent circuits for the BJT
Input resistance rbe
rbe(r)=300+(1+)26(mV)/IEQ(mA)
Tansconductance gm
gm= /rbe
Output resistance rce
Usually large, negligiblerce
rce
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
4.6 The common-emitter amplifier (P164)
Figure 4.28 Common-emitter amplifier of Example 4.9.
C2
Ce
4.6 The common-emitter amplifier (P164)
Figure 4.28 Common-emitter amplifier of Example 4.9.
1. Draw the DC bias circuit to find the Q point Ac signal — short circuit Capacitor — open circuit
VBQVCEQ
+
-
Rule of thumb: VCEQ 0.3~0.7VCC
2. Draw the AC circuit with BJT DC signal — Connect to the GND Capacitor — short circuit
4.6 The common-emitter amplifier (P164)
Figure 4.28 Common-emitter amplifier of Example 4.9.
rce
3. Replace with the equivalent circuit
Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0
4.6 The common-emitter amplifier (P164)
rce
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
Find the voltage gain Av=Vo/Vin
4.6 The common-emitter amplifier (P164)
Find the open circuit voltage gain Avo=Vo/Vin|RL=0
Find the source voltage gain Avs=Vo/VsFind the input resistance Rin=Vin/Iin
Find the output resistance Rout=Vo/Io|Vs=0
rce
Rout
Analysis of the CE amplifier without CE
4.6 The common-emitter amplifier (P164)
C2
Ce
DC operating point
Remains unchanged
AC performance
Need more detailed analysis
Exercise: 1. Draw the small-signal equivalent circuit
2. Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0 assume rce
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
Figure 4.27 Common-emitter amplifier.
4.6 The common-emitter amplifier (P162)
Think about: 1. Effects of RE1 on Q-point, Av,Rin and Rout
2. Effects of RE2 on Q-point, Av,Rin and Rout
4.7 The emitter follower (P166)
1. Draw the DC bias circuit to find the Q pointGiven R1=R2=100k, =200, VBEQ=0.7V
find the Q point.
4.7 The emitter follower (P166)
R1
RL
(b) AC circuit
4.7 The emitter follower (P166)
Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0
Think about: Write two equations for Vin and Vo
RS’ Rout
’
Rout
4.8 The common-base amplifier (P299)
1. Draw the DC bias circuit to find the Q point
4.8 The common-base amplifier (P299)
Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0
Think about: Write two equations for Vin and Vo
RC
Rin Rout
’Rout
Summary
4.8 The common-base amplifier (P299)
C-E amplifier C-C amplifier C-B amplifier
Av
Ri
Ro
inverting noninverting noninvertingamplification follower amplification
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=Avo1 Avo2 ? (refer to p17)
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
Review 1.5 Cascaded amplifiers
4.9 Cascode amplifier (P300)
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。
4.9 Cascode amplifier (P300)
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
Example
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)
4.10 Frequency response for an amplifier
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.
=
Review 2.8 Active Filter-High pass filter
4.10 Frequency response for an amplifier
Question:1. What is the voltage
gain of the circuit?2. Can you draw the
frequency response of amplifier?
3. What kind of filter is it?
4. Do you remember the small signal equivalent circuit of the NPN transistor?
5. Any changes to (4) if the frequency is high?
4.10 Frequency response for an amplifier
The small signal equivalent circuit for the NPN transistor
Mid-frequencyCbe
Cbc
Usually, Cbe is about 10PF ~ 1000PF
Cbc <10PF
At high frequencies
Miller effect
Question: (P296)If Zf=-j/C for a CE amplifier, how do you find Zin,miller
and Zout,miller
Cbc
ibrbe
Cbe
(1+gmRL’)Cbc
Find the Thevenin equivalent resistance Rs’?
Figure 8.34 Simplified equivalent circuit for the common-emitter amplifier.
1. What kind of filter is it?2. What is the break (cut-off) frequency? 3. What factors affect the value of the break
frequency?4. 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.
4.10 Frequency response for an amplifier
Figure 8.47 Amplifier with coupling capacitors.
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