Download capacitively coupling, single supply, current sources

Capacitively coupled
amplifier
vI(t) =VI+vi(t)
We want to amplify only vi(t)
Role of R3?
 R2 
vo (t ) = vi (t )1 + 
 R1 
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Op-amp amplifiers operated from a
single power supply
5.1K
51K
How can be vi(t) amplified in the case of a unipolar supply?
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Solution: VTC translation
+VPS
+VPS
VBIAS
VO=+VPS/2
• obtaining the
biasing voltage
• equivalence in
steady-state regime
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VPS
 R2 
 R2 
 R2 
vO = (vi + VBIAS )1 +  = vi 1 +  + VBIAS 1 + 
 R1 
 R1 
 R1 
The value of dc gain is to high
How can it be lowered to unity?
 R 
 R 
R
vO = vi 1 + 2  + VBIAS 1 + 2  − VBIAS 2
R1
 R1 
 R1 
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Complete circuit
Equivalent circuit in
steady-state regime
10K
VBIAS=6Vdc
VPS
10K
vi
5.1K
51K

R
v O = (v i + V BIAS ) 1 + 2
R1


R 
v O = v i  1 + 2  + V BIAS
R1 


R
 − V BIAS 2
R1

VBIAS=6Vdc
Equivalent dc circuit ?
Equivalent ac circuit ?
What is the solution for an inverting op-amp amplifier with single supply?
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Op-amp current sources
vI
iO =
R
the current doesn’t depend on RL
adjustable current if R is replaced with a
resistance in series with a potentiometer
the current of the source can be modified by
modifying vI – voltage controlled current source
none of RL terminals can be connected to the ground,
therefore we have a floating load
? What if we have to return the load to ground ?
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Current source with grounded load
Howland source
NF and PF
NF - dominant
R3
R
K =
vo ,OA =
vo ,OA
R3 + R4
R+R
−
R1 || R L
R || R L
K =
=
R1 || R L + R R || R L + R
+
Because R||RL < R,K − > K+,
+
−
v
=
v
rezults NF,
+
v I − v + v o, AO − v
+
iO = i1 + i 2 =
R1
R2
R3
v =v =
v0 ,OA
R3 + R4
+
vI
i0 =
R
−
v0,OA = 2iO RL
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the resistors – must be very well paired in order to have a
perfect (ideal) current source (the output resistance tends to
infinite)
If
R1 R3
=
R2 R4
vI
iO = −
R1
Practical solution: op amp + transistor
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Op-amp current follower
The current source doesn’t generate power
The power in RL comes from the op-amp power supplies
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Current Sources Using Op Amp and T
Floating load
current
sink
vI
iE =
R
β vI
vI
iO =
≈
β +1 R R
RL iO < VCC − v I − VCEsat
Better linearity
Series-series NF (voltage-current): the small signal output resistance
of the current source (seen by RL) is approximately a times larger than
the one in the absence of the NF, (without the op amp).
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Current Sources Using Op Amp and T
Load returned
to ground
Current
source
β
vI vI
iO =
≈
β +11+ 2 R R
1
β1
Adjustable current source:
• modifying vI - voltage-controlled current source
• a potentiometer in series with R.
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