Transcript Lecture 7 Overview

Announcements
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• mid-term: Thursday, October 27th
Adam Riess, Brian Schmidt and Saul Perlmutter
Lecture 10 Overview
•
•
•
•
Transistors (continued)
The common-emitter amplifier
Amplifier parameters
Black box amplifier models
Summary of useful equations
• Basic DC operating
conditions:
I E  I B  IC
I C  I B
I E  I C
 1


I C  I S eVBE / VT
VC  VCC  I C RC
• Add a small signal:
IC
ic  vbe  g m vbe
VT
vbe

r 

ib
gm
IC
gm 
VT
vbe 
1
re 


ie
gm gm
Using small signal models
e.g. Tmodel
1) Determine the DC operating conditions (in particular,
the collector current, IC)
2) Calculate small signal model parameters: gm, rπ, re
3) Eliminate DC sources: replace voltage sources with
shorts and current sources with open circuits
4) Replace BJT with equivalent small-signal model.
Choose most convenient depending on surrounding
circuitry
5) Analyze
re =
1
gm
Voltage gain with small signal model
ic
+
RC
vc
ic
eliminate
DC sources
and apply
T-model
+
vbe
re
vbe
-
vc  ic RC  
Find the gain using a small
signal model:
ie

RC  ie RC
vbe  ie re
Voltage gain 
re 

gm

vc  ie RC
R

 C
vbe
ie re
re
1
gm
v
so volt agegain c   g m RC
vbe
gm =
IC
VT
How to build a Real Common
emitter amplifier
• Why bother with 2 voltage
supplies?
• Use a voltage divider R2/R1 to
provide base-emitter voltage to
correctly bias the transistor.
DC condition: the voltage divider
VBE  VB  VE  0.6V
• The voltage divider
VB  I E RE  0.6V
should provide sufficient
R1
voltage to place the
now VB  VBB
R1  R2
transistor in active mode
(base-emitter forward
R1
I E RE  0.6V
so

biased):
R R
V
1
• Current through resistors
should be >10 times base
current for stability
2
BB
Vbb
IC
 10
R1  R2

Amplifier specifications
• What other parameters of an amplifier do
we care about?
– Voltage gain
– Dynamic range
– Frequency response (bandwidth)
– Input impedance
– output impedance
Voltage Gain
• Voltage gain
• Use small signal model (short Voltage sources and capacitors)
ground
vb  ie (re  RE )
vc  ic RC  ie RC since  1
αie
voltage gain
vout vc
RC
 
vin vb
re  RE
usually re<<RE
vc
RC

vb
RE
• Voltage gain is only defined by resistors RC and RE
ground
Frequency response (Bandwidth)
• Normally interested in providing a
small, AC signal to the base
• Use capacitors to remove ("block")
any low frequency (DC) component
("capacitively couple the signal to the
base") which could affect the bias
condition
• C1 forms a high-pass filter with R1in
parallel with R2 (Assuming the AC
impedance into the base is large).
• Cut off frequency ω0=1/RC, so to
remove frequencies <fmin:
C1 
1
 RR 
2f min  1 2 
 R1  R2 
Frequency response (Bandwidth)
• Also worthwhile to place a capacitor
on the output
• C2 forms a high pass filter with RL.
• Cut off frequency ω0=1/RC, so to
remove frequencies <fmin:
C2 
1
2f min RL
Dynamic Range
• Maximum voltage output = Vbb
• Minimum = 0
• Beyond this the signal becomes 'clipped' or distorted
• To get the maximum possible voltage swing, both positive and negative,
set VC=0.5 VBB
• Maximum 'dynamic range'
VC
Input impedance
• Consider the circuit without the voltage
divider resistors. What's the small signal (AC)
input impedance at the base, rb?
vb
ib
v b = (re + Re )ie
i
ib = e
b +1
vb
rb =
= (b + 1)(re + Re )
ib
b + 1 » b; re << Re
rb » bRe
rb =
rOUT
rb
• Including voltage divider resistors in parallel
• Input signal sees a total input impedance
rIN= R1 // R2 // rb
RB
Output impedance
rOUT
vo io RC
 
 RC
io
io
ROUT
rb
• If RL=10kΩ and we want a low frequency cutoff of 20Hz, What is C2?
• If VBB=15V and IC=2mA what is the output impedance?
DC condition
R1
I E RE  0.6V

R1  R2
VBB
Vbb
IC
 10
R1  R2

Frequency response
C1 
1
 R1 R2 

2f min 
 R1  R2 
C2 
1
2f min RL
Impedance
rb » bRe
rIN » R1 || rb || R2
rOUT
vo io RC
 
 RC
io
io
Gain/Dynamic range
VC  0.5VBB
vc
RC

vb
RE
Impedances
• Why do we care about the input and output impedance?
• Simplest "black box" amplifier model:
ROUT
VIN
RIN
AVIN
VOUT
• The amplifier measures voltage across RIN, then generates a voltage
which is larger by a factor A
• This voltage generator, in series with the output resistance ROUT, is
connected to the output port.
• A should be a constant (i.e. gain is linear)
Impedances
• Attach an input - a source voltage VS plus source impedance RS
RS
VS
ROUT
VIN
RIN
AVIN
• Note the voltage divider RS + RIN.
• VIN=VS(RIN/(RIN+RS)
• We want VIN = VS regardless of source impedance
• So want RIN to be large.
• The ideal amplifier has an infinite input impedance
VOUT
Impedances
• Attach a load - an output circuit with a resistance RL
RS
VS
ROUT
VIN
RIN
AVIN
• Note the voltage divider ROUT + RL.
• VOUT=AVIN(RL/(RL+ROUT)
• Want VOUT=AVIN regardless of load
• We want ROUT to be small.
• The ideal amplifier has zero output impedance
VOUT
RL
Operational Amplifier
• Integrated circuit containing ~20 transistors
Operational Amplifier
• An op amp is a high voltage gain amplifier with high input
impedance, low output impedance, and differential inputs.
• Positive input at the non-inverting input produces positive output,
positive input at the inverting input produces negative output.
• Can model any amplifier as a "black-box" with a parallel input
impedance Rin, and a voltage source with gain Av in series with an
output impedance Rout.
Ideal op-amp
• Place a source and a load on the model
RS
+
vout
RL
-
• Infinite internal resistance Rin (so vin=vs).
• Zero output resistance Rout (so vout=Avvin).
• "A" very large
• No saturation
• iin=0; no current flow into op-amp
So the equivalent circuit of an
ideal op-amp looks like this:
Schematics
 An amplifier will not work without a power supply. And a more complete
diagram looks like the figure below, which also indicates the standard pin
configuration.
Function
Pin
2
Inverting input
3
Non-inverting input
4
V- supply
6
Output
7
V+ supply
Measuring Impedances
RS
VS
ROUT
VIN
RIN
AVIN
VOUT
RL
• Assuming you can only vary RL and RS, how would you measure the
input and output impedances of the amplifier?
Measuring Impedances
RS
VS
ROUT
VIN
RIN
AVIN
VOUT
RL
• With the black box model, it is simple to measure the input and
output impedances of an amplifier
• To measure the input impedance, vary RS until the output
voltage has dropped to half ; then RS=RIN= input impedance
•To measure the output impedance, vary RL until the output
voltage has dropped to half ; then RL=ROUT= output impedance
Cascaded Amplifiers
• Easiest way to increase amplification is to link amplifiers together
ROUT1
VIN1
RIN1
A1VIN1
ROUT2
VOUT1
VIN2
RIN2
A2
VIN2
• Ideal amplifiers; VOUT2=A1A2VIN1
• In reality, take account of voltage divider action due to input and
output impedances
VOUT2