Post on 04-Jun-2018
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DC TRANSISTORS BIASING
in active regionaF
AMPLIFIERS WITHTRANSISTORS
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Remark: input signal - low
enough for the amplifier to workin a linear region around the OP
Op-ampamplifier
revisited
VTC
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transistor utilization as amplifier (CS, CE)
in active region (aF), the transistor operates around the dcoperating point (OP)
Necessity for dc transistor biasing
VPS dc supply
VI set the OP: (VO,IO)
vi input voltage
(to be amplified)vo output voltage
(amplified voltage)
superposition of the
variable signal over the
dc regime
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Operation of the amplifier (CS, CE)
Who is responsible for the gain ?
VPS
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Voltage transfer characteristicvO(vI), inverting amplifier
Small signal:operation of theamplifier in the
narrow linear regionaround the OP
Maximum swingof the input signal:
often determined based on thelinearity considerations
slope voltage gain
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DC biasing setting the OP
Operation of the transistor as amplifier:
the transistor biased as close as possible to the middle of the aF
the instantaneous (mobile) operating point - in the active region
the input variable signal kept small (linear region around OP)
OP:
stabile and predictibile
independent of the transistor parameters
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MOSFET biasing
2)( ThGSD VVI =
simple
the current in the OP, ID,depends on the transistor
parameters, and VTh
cannot assure the stability ofthe quiescent point.
PS
GG
GGS V
RR
RV
21
2
+=
DDPSDS IRVV =
DR
1st
variant 3 resistors,single supply
),(OP DDS IV
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V2.154.26.7
4.2
21
2 =+
=+
= PSGG
GGS V
RR
RV
( ) A808.02.1500)( 22 === ThGSD VVI
V67.208.01.295 === DDPSDS IRVV
Example 1 RG1=7.6M; RG2=2.4M;RD=29.1K; VPS=5V
VTh =0.8V; =500A/V2. OP ?
V4.08.02.1 === ThGSDSsat VVV
FDSsatDS aVV inisrtransistothe>
( ) ( ) V7.22/4.052/ =+=+ DSsatPS VV V67.2=DSVThe transistor is biased in themiddle of its active region
A)80,V67.2(Q
DR
Resize the circuit to have the
transistor in the middle of its activeregion for ID=120A
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MOSFET biasing
DSGGGS IRVV =
VGSdepends also on the drain currentID
ID , RSID, VGS , ID the circuitwithstands to the variation tendency of ID
negative feedbackdue toRS ensure the OP stability for variation of
certain parameters
increases the complexity of computationalrelations
2)( ThGSD VVI =
2nd
variant 4 resistors, single supply - cont.
OPTIONAL
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Example 2
RG1=3M; RG2=1M;RD=3K; RS=1K; VPS=20V
VTh =2V; =0,5mA/V2.
? What is the OP ?
SDGGGS RIVV =2)( ThGSD VVI =
ID
2-8ID
+9=0; ID
in mA
V6,14)13(35,120
)(
=+=
=+= SDDPSDS RRIVVID1=6,65mA and
ID2=1,35mA
ID1 is not suitable;results VGS
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Example 3
MOSFET: VTh
=2V; =0,25mA/V2; VPS
=20V
? Choose the resistances to obtainID=1mA in the OP.
2)( ThGSD VVI =
V4
25,0
12 =+=+=
DThGS
IVV
VDSsat=VGS-VTh=2V
T- active region VDS(2V; 20V).Q : We chose VDS=9V
DSDPSDS IRRVV )( += K111
920
=
=
=+D
DSPS
SD I
VV
RR
VGG
OPTIONAL
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Example 3 cont.
MOSFET: VTh
=2V; =0,25mA/V2
? Choose the resistances to obtainID=1mA in the OP
RD,RS also sets the gain. For now we
can consider VS=4V acrossRS :
K41
4===
D
S
SI
VR
K7411 ==DR
V844 =+=+= SGSGG VVV
== K200;K300 21 GG RR
OPTIONAL
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MOSFET biasing
Usual in integrated circuits: biasing with current sources
ID independent of the transistor parameters
GGGSDPSDS VVIRVV += GSDPSDS VIRVV +=Voltage across the current source: VGG-VGS
3rd
variant current source, single and differential supply
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BJT biasing, usual variant in discrete circuits
oppositely to MOSFET, for BJT appears:
- base currentIB different from zero
- through collector and emitter do not flow exactly thesame current
CBBCEIIIII
1)1( +=+=+=
EC II One can approximate
BC II =
precise calculation:make use ofIB
approximate calculation:neglectingIBcompared to the current throughthe resistive voltage divider in thebase of the transistor
),(OP CCE IV
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Approximate calculation
IBmuch smaller than the
current flowing through thebase divider
PSBB
B
BB VRR
R
V 21
2
+=
E
BEBB
EC R
VV
II
=
)( ECCPS
EECCPSCE
RRIV
RIRIVV
+
=
RE
is very important for setting
and stabilizing the OP, through anegative feedback mechanism
IC; IE; VRE; VBE; IC
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Precise calculation
Thevenin theorem: VBB, RB
IC=IE+IB IE
IE insensitive to variations:
)1( +>>
BE
RR
BE
RR 10>
RB1,RB2 small values for the
independence of OP on
RB1, RB2 high values for the high
input resistance
IEinsensitive to temperature
variations due to VBE
V1.0>>BBV
aVBEvariation of 0,1V can be neglected visa vis the VBB=35V
)( ECCPSEECCPSCE
RRIV
RIRIVV
+
=
)1/( ++=
BE
BEBBE
RRVVI
IE=(+1)IB
EEBEBBBB IRVIRV ++=
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VPS=15V;RB1=10k;RB2=4.7k;
RE=1.5k; RC=1.8k; =150
Approximate calculation
Exact calculation
IC
= ?
VCE =?
VC = ?
VE = ?
IC = 2.73mA
VCE = 6V
VC = 10.1V
VE = 4.1V
IC = ? IC = 2.7mA
Example 4
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Example 5
Values of the resistances
so that T is biased in aF@ IC=2mA ?
VPS=12V, =100
Usually wechoose:
V4123
1
3
1===
AlBBVV
=
=
= k65.12
7.012)3/1(
E
BEBBE
I
VVR
PSPS
BB
BBB VV
RRRV
31
21
2 =+
= 21 2 BB RR =
mA84.1)1100/(7.1465.1
7.04
)1/(=
++
=
++
=
BE
BEBBE
RR
VVIVerification:
RB2=22K;RB1=44K
B
E
RR 10>
EBB
BB RRR
RR10
21
21