The properties of operational amplifiers
Operational amplifier applications
Electronics – Operational amplifiers
Prof. Márta Rencz, Gergely Nagy
BME DED
November 19, 2012
Non-ideal OPAMPs
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
The operational amplifier
The properties of ordinary amplifiers are determined by their
structure.
The properties of operational amplifiers (OPAMP) are
determined by the negative feedback realized by the
circuit surrounding the OPAMP.
The first stage is a differential
amplifier: the OPAMP has two
inputs and their potential
difference is amplified:
Vout = Ad · (V+ − V− )
+
_
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Negative feedback I.
Negative feedback: a portion of the output signal is fed back and
is subtracted from the input.
The input of the amplifier:
Vd = Vin − β · Vout
In steady state:
Vout = A · Vd = A · (Vin − β · Vout )
The properties of operational amplifiers
Operational amplifier applications
Negative feedback II.
The gain of the circuit:
Av =
Vout
A
=
Vin
1+β·A
As β · A 1:
Av ≈
1
β
The gain is determined by the feedback.
Non-ideal OPAMPs
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Gain vs. frequency
If the feedback circuit contains resistive elements only:
fC · Av = fC0 · A = fT = GBW
The product of the gain and the bandwidth of a negative
feedback amplifier is equal to the transit frequency (or GBW).
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Ideal OPAMP
Properties of an ideal OPAMP
Ad → ∞
rd → ∞
rout → 0
where rd is the differential input resistance and rout is the
output resistance of the amplifier.
A non-ideal, cheap OPAMP (µA741) has the following properties:
A = 2 · 105
rd = 2 MΩ
rout = 75 Ω
In most applications ordinary OPAMPs can be treated as ideal.
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Basic principles of calculation
As the OPAMP’s voltage gain is a very large value (∼ ∞), the
following approximations can be applied:
1
the two inputs are assumed to be at the same potential
V+ = V−
2
the input current is zero.
With these approximations the circuit is linear, thus it can be
calculated using the linear methods.
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Non-inverting amplifier
+
The KCL for the ϕ node:
_
Vin
Vout − Vin
=
R1
R2
Av =
Vout
R2
=1+
Vin
R1
A different way of calculation is to find the β of the feedback
circuit. This is a voltage divider:
β=
Av ≈
R1
R1 + R2
1
R2
=1+
β
R1
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Inverting amplifier
The potential of the inverting
input (ϕ) of the OPAMP is
0 V, as the two inputs have to be
at the same potential.
_
+
The KCL for the ϕ node:
Vin Vout
+
=0
R1
R2
Av =
Vout
R2
=−
Vin
R1
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Voltage follower
Although the voltage gain is 1, this is a
very useful circuit as
+
the input generator has a very
little load (due to the large input
resistance),
_
the output resistance is very small,
thus almost the entire output
voltage drops on the load.
rin → ∞
As β = 1, the gain is:
Av =
Vout
=1
Vin
rout → 0
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Summing amplifier
_
+
Using the equation describing the gain of the inverting amplifier
and the theorem of superposition:
Vout = −
R
R
R
V1 +
V2 + . . . +
Vn
R1
R2
Rn
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Subtracting/Differential amplifier I.
The theorem of
superposition is used:
if V1 = 0 we get an inverting
2
amplifier: Vout1 = − R
R1 V2
_
+
_
+
if V2 = 0 we get a non-inverting
amplifier:
+
_
2
When the V2 = 0: ϕ = R1R+R
V1
2
R2
The gain of the non-inverting amplifier: Vout
ϕ = 1 + R1
2
Thus the relationship between V1 and Vout : Vout2 = R
R1 V 1
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Subtracting/Differential amplifier II.
The results from the previous
slide:
_
Vout1 = −
+
Vout2 =
R2
V2
R1
R2
V1
R1
The overall result can be gained by summing the two equations
above:
Vout =
R2
(V1 − V2 )
R1
The properties of operational amplifiers
Operational amplifier applications
Integrator
Integrators can be defined with the following formula:
Z
vout (t) = K · vin dt + vout (t = 0)
The KCL for ϕ:
_
+
vin
dvout
+C
R
dt
Thus the transfer function of this circuit is:
vout
1
=−
RC
Zt
vin (t)dt + vout (0)
0
Non-ideal OPAMPs
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Digital voltmeter I.
This is an example of the application of an integrator.
_
+
_
+
1 At t1 the voltage to be measured is switched to the input
and is integrated for a constant Tf period of time.
2 At t2 the input is switched to a negative reference voltage
(Vf ): the capacitor is discharged.
3 The time it takes to discharge the capacitor (Tx ) is
measured: t3 − t2 .
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Digital voltmeter II.
_
_
+
+
The charge stored in the capacitor:
Q=
Vf
Vx
· Tf =
· Tx
R
R
Thus:
Vx = Vf ·
Tx
Tf
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Instrumentation amplifier I.
+
_
+
_
_
+
This circuit is realized in ICs, the value of the resistors is set
very precisely.
This is a differential amplifier.
Its gain is set by an external resistor to value between 1
and 1000 usually.
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Instrumentation amplifier II.
OPAMP 3—R2 —R3 realize a
differential amplifier.
+
_
+
Vout =
R3
(V1 − V2 )
R2
_
The current of RG :
_
+
V+ − V−
RG
flows through the
IG =
As the input current of the OPAMP is zero, IG
R1 resistors, thus:
ϕ1 − ϕ2 = I · (RG + 2 · R1 )
Thus
Vout
R3
=
R2
2 · R1
1+
(V+ − V− )
RG
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Logarithmic amplifier
_
+
Works for positive input voltages.
The current of the resistor and the diode are equal.
The diode’s voltage is the negative of the output voltage.
Thus:
Vin
= I0 e−Vout /mVT − 1
R
Vout = −m · VT · ln
Vin
I0 · R
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
Exponential amplifier
_
+
The current of the resistor and the diode are equal.
Thus:
−
Vout
= I0 e−Vin /mVT − 1
R
Vout ≈ −I0 · R · eVin /mVT
The properties of operational amplifiers
Operational amplifier applications
Precision rectifier
+
_
Vout = Vin
Vout = 0
when Vin > 0
otherwise
Non-ideal OPAMPs
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
The parameters of non-ideal OPAMPs I.
The transfer characteristic curve of a non-ideal OPAMP:
The input voltage has to be non-zero (V0 ) to have a zero
output – this is the offset voltage.
The speed at which the output voltage changes is limited –
slew rate (SR):
dVout
SR = max
dt
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
The parameters of non-ideal OPAMPs II.
The output voltage of an OPAMP is also limited:
the supply voltage is the absolute maximum the output can
reach,
but most OPAMPs stop below the supply voltage and can
not pull their output entire up to it.
OPAMPs that are able to reach the supply votage at their outputs
are called rail-to-rail OPAMPs.
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
The parameters of non-ideal OPAMPs III.
Common-mode gain:
If the potentials of the inputs of an ideal OPAMP are equal
then the output voltage is zero.
In the case of non-ideal OPAMPs, the output voltage is a
function of the common voltage at the inputs. This is the
common-mode gain.
If V− = Vc − V2d and V+ = Vc + V2d (where Vc is the common
voltage at the inputs and Vd is the differential input):
differential voltage gain:
Ad =
Vout
Vd
common-mode voltage gain:
Ac =
Vout
Vc
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
The parameters of non-ideal OPAMPs IV.
The better the OPAMP, the bigger the differential gain and
the smaller the common-mode gain:
Ad Ac
Thus a quality factor of an OPAMP is the ratio of two gains,
which is called the Common-mode rejection ratio
(CMRR):
Ad
CM RR =
Ac
CMRR is around 104 for common OPAMPs.
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
The parameters of non-ideal OPAMPs V.
The model of the input of a non-ideal OPAMP:
_
+
Common-mode input
resistance: rc (GΩ)
Differential input
resistance: rd (MΩ)
Operating point input
currents: Ib+ , Ib−
The properties of operational amplifiers
Operational amplifier applications
The parameters of non-ideal OPAMPs VI.
Ad
CMRR
GBW
rd
rc
Ib
V0
SR
rout
µA741 (BJT)
2 · 105
90 dB
1 MHz
2 MΩ
109 Ω
80 nA
1 mV
0.5 V /µs
75 Ω
TL081 (JFET)
2 · 105
86 dB
4 MHz
1012 Ω
1014 Ω
20 nA
3 mV
16 V /µs
100 Ω
Non-ideal OPAMPs
The properties of operational amplifiers
Operational amplifier applications
Non-ideal OPAMPs
The parameters of non-ideal OPAMPs VII.
The above circuits all assumed to contain ideal OPAMPs.
Due to the effect of input resistances the basic circuits need
to be modified for higher precision.
It is important that the load resistance on the two inputs
be equal:
_
+
The properties of operational amplifiers
Packaging of OPAMPs
Operational amplifier applications
Non-ideal OPAMPs