Circuit Surgery
Regular clinic by Ian Bell
Strain gauge circuit revisited
I
n the November 2019 issue we
looked at strain gauges and differential signals in an article inspired by
a post on the EEWeb forum from Scott
Siler, who wrote: ‘I’m an aerospace engineer working on a side project privately
to gather data from a strain gauge using
an NI DAQ (a data acquisition device
from National Instruments). The output is
0-36mV DC so I need a gain of 3 to better
utilise the DAQ. My EE experience is limited to a couple of classes I took during
my bachelor’s degree. I have an AD8628
op amp and built a basic, non-inverting
negative feedback circuit, as shown in
the attached picture (see Fig.1). The rail
voltage is 0-3V DC. It does not seem to
have any gain, but instead the output
voltage is actually lower than the input.’
We received an email from John Ellis,
who though that I had not diagnosed the
issues with the Scott’s circuit in sufficient
detail. John wrote: ‘I read Ian Bell’s article
on strain gauges and differential amplifiers
with interest. I thought, however, that the
article might have provided a little more
information about the problem faced by
Scott Siler, which may have been more
help to your readers in designing their
own sensor circuits.’
John also provided his own analysis
of the circuit, deriving the asymmetrical
gain formula for the circuit in Fig.1
(which shows that it is not a good
differential amplifier) and also pointing
out possible problems with the commonmode input voltage (which may be the
main reason it did not work). We will
look at these aspects of op amp circuit
design in this article.
In response to John’s criticism, there are
a few reasons why we do not necessarily
look at the specific details of problems
posted on the forum, but cover a topic
in more general terms, as we did with
the November 2019 article. First, the
full details are not always posted on
the discussion thread – this was the
case here to some extent – we do not
have a complete schematic of Scott’s
circuit (although we can guess) and we
do not know all the details of the load
cell used. Second, specific problems are
often solved by forum contributors long
before an article can be published, so it
makes sense for Circuit Surgery to take
a broader look at the topic. Third, there
are often several themes that can be
developed from the issues discussed in
the forum thread, and often there will not
be space to address them all in a single
article. Thus, we often have multi-part
articles and sometimes return to a topic
on request – which is exactly what we
are doing here.
For the benefit of readers who do not
have the November 2019 article to hand,
we will briefly recap strain gauges, load
cells and differential and common-mode
signals to provide background and context
for discussion of the op amp circuits.
Strain Gauges
Strain is a measure of the deformation
(change in size or shape) of an object. A
strain gauge is a sensor whose electrical
resistance changes with deformation.
It typically comprises a thin, flexible,
insulating foil that supports a long
conductive strip, typically in a zig-zag
V2
–
V1
+
U1
A differential signal is carried on
two wires other than ground, so we
can observe the voltage on each wire
individually (eg, V1 and V2). The actual
signal is equal to the difference in the
R1
Strain gauge
R3
Vout
R2
R4
Vout
Fig.1. Scott’s circuit, as discussed in this
article and Circuit Surgery, November 2019.
44
Differential and common-mode
signals
Vexcite
Rf
10kΩ
Ri
5kΩ
pattern (see Fig.2) with contacts to
connect the device to a measurement
circuit. Strain gauges are often built into
larger devices, called ‘load cells’, in which
strain gauges are attached to a specifically
designed metal body that deforms when
force is applied to the device. Load cells
have many industrial applications in
force and weight measurement.
Strain gauges are typically used in
Wheatstone bridge circuits, as shown in
Fig.3. The bridge comprises two potential
dividers in parallel, with the output
voltage being the difference between the
two divider voltages. The differential
voltage from a bridge does not have
the offset associated with a simple
potential divider, so it can be amplified
without the offset causing the amplifier to
saturate. The circuit in Fig.3 has a single
strain gauge with three fixed resistors, a
variety of possible physical measurement
scenarios use multiple stain gauges.
The bridge can contain one, two, three
or four strain gauges depending on the
setup used. In some situations additional
resistors may be used (eg, between the
excitation voltage and bridge) to fine tune
the behaviour of the circuit.
Fig.2. A typical strain gauge cemented to the
substrate under measurement.
Fig.3. A common strain gauge bridge
circuit – in this example R1 is the strain
gauge, but other arrangements are used.
Practical Electronics | March | 2020
Strain Gauge Circuit Revisited
= common-mode
𝐴𝐴! 𝑉𝑉! − 𝐴𝐴! 𝑉𝑉!
!"#the
to 𝑉𝑉as
gain, Acm. The
common-mode gain is equal to the
difference between the two individual
gains, so in the ideal case, where the two
𝐴𝐴! + 𝐴𝐴the
! value of (A2 − A1) is 𝑉𝑉! + 𝑉𝑉!
gains are equal,
𝑉𝑉 =
𝑉𝑉! − 𝑉𝑉! + 𝐴𝐴! − 𝐴𝐴!
2
zero!"#and the 2common-mode
input has
no effect on the output.
Strain
Circuit
Revisited
We
canGauge
rewrite
the above
equation as:
𝐴𝐴! 𝑉𝑉! − 𝑉𝑉! + 𝐴𝐴!"
𝑉𝑉!"# =
𝑉𝑉
!"# = 𝐴𝐴! 𝑉𝑉! − 𝐴𝐴! 𝑉𝑉!
Fig.4. Example differential signal: 2V peak, 1kHz sinewave with 1.5V DC common
mode. The upper traces (red, green) are the individual voltages. The lower trace
(magenta) is the differential signal.
𝑉𝑉! + 𝑉𝑉!
2
Where Ad is the differential voltage gain
and is Acm, the common-mode gain.
𝐴𝐴!
𝐴𝐴! +the
𝐴𝐴! influence
The smaller
of common- 𝑉𝑉! + 𝑉𝑉!
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
𝑉𝑉!"# = = 20log!" 𝑉𝑉! − 𝑉𝑉dB
! + 𝐴𝐴! − 𝐴𝐴!
𝐴𝐴
mode signals on
amplifier,
2 the differential
2
!"
the better the amplifier. The ability of an
op amp to reject common-mode signals is
expressed as the ratio of the differential
𝑉𝑉! is
+ 𝑉𝑉
𝑅𝑅!
and common-mode
gains; this
called
!
𝑉𝑉− = 𝑉𝑉𝐴𝐴!! 𝑉𝑉! − 𝑉𝑉! + 𝐴𝐴!"
the!"#
‘common-mode
rejection
2ration’
𝑅𝑅!
(CMRR), which is often expressed in
decibels as follows:
input). One way of looking at what the
voltages between the two wires, each
differential amplifier does is to think of
measured with respect to ground.
each input as having a separate gain.
So if the two voltages on the wires are V1
The output is then made up from the
and V2, then the differential signal is (V1
non-inverting input signal times its gain
– V2). However, in order to fully describe
𝐴𝐴!
(A1) minus the inverting input signal
a differential signal we need to state two
𝑅𝑅! 20log
dB
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
!"
1 + = 𝑉𝑉
Strain
Circuit
Revisited
!
things – the differential signal itself,
times
its Gauge
gain (A
).
We
can
write
this
as
𝐴𝐴
!"
2
𝑅𝑅!
and the voltage they have in common –
an equation as follows:
The CMMR of the AD8628 used in Scott’s
called the ‘common-mode voltage’. The
𝑉𝑉!"# = 𝐴𝐴! 𝑉𝑉! − 𝐴𝐴! 𝑉𝑉!
circuit is very high at 130dB (the commoncommon-mode voltage is the average of
mode𝑅𝑅gain
is more
the voltage on the two wires (V1 + V2)/2.
!
𝑅𝑅!
𝑅𝑅! than three-million-times
− =than
𝑉𝑉
𝑉𝑉!"#
𝑉𝑉! −
𝑉𝑉!
smaller
An ‘ordinary’ signal carried on a single
If A1 = A2 = Ad then this equation becomes
𝑅𝑅! !1 +the
𝑅𝑅! differential
𝑅𝑅! gain).
wire is described simply by its voltage
the ideal case of Vout = Ad(V1 − V2), which
𝐴𝐴! + 𝐴𝐴Ideally,
+ 𝑉𝑉!amp amplifiers
with respect to ground. Such signals may
we noted above.
the gains for the 𝑉𝑉!Op
!
𝑉𝑉 =
𝑉𝑉! − 𝑉𝑉! + 𝐴𝐴! − 𝐴𝐴!
be referred to as ‘single ended’; where it
two!"#
inputs are2equal; however,
this is not
It2is useful to consider Scott’s circuit in
is necessary to make a distinction from
the case for real differential amplifiers.
terms of𝑅𝑅the
general differential amplifier
! ! + 2𝑅𝑅!
+ 𝑅𝑅discussed
𝑉𝑉!
differential signals in a discussion.
With a little algebraic manipulation we
properties
above. However, first
𝐴𝐴!1 =
𝑅𝑅! 2𝑅𝑅
Fig.4 shows an example differential
can
rearrange
the above
equation so
we will briefly! describe the most basic
Strain
Gauge Circuit
Revisited
signal – this is a 1kHz sinewave with a
that it includes the term (V1 𝑉𝑉−! +
and commonly used op amp amplifier
V2𝑉𝑉)!plus
𝑉𝑉!"# = 𝐴𝐴! 𝑉𝑉! − 𝑉𝑉! + 𝐴𝐴!"
peak voltage of 2V and common-mode
circuits – the inverting or non-inverting
some
other
terms.
It
is
useful
to
do
this
2
Strain Gauge Circuit Revisited
component which is 1.5V DC. The upper
amplifier configurations,
because we can then see an ‘ideal part’
𝑅𝑅!
𝑅𝑅! as shown in
𝑉𝑉!"# = 𝐴𝐴! 𝑉𝑉! − 𝐴𝐴! 𝑉𝑉!
𝑉𝑉!"# In
= both
1 +cases𝑉𝑉!negative
−
𝑉𝑉feedback is
plot shows the individual signals (V1
Fig.5.
– the differential gain multiplying (V1 −
𝑅𝑅!
𝑅𝑅! !
applied via a pair
of resistors,
which act
and V2) and allows the 1.5V commonV2), and an ‘error part’ – the rest of the
𝑉𝑉 = 𝐴𝐴! 𝑉𝑉! − 𝐴𝐴! 𝑉𝑉!
𝐴𝐴!
as a potential divider feeding a fraction of
mode signal to be observed.!"#The lower
equation. We get
dB
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 20log!"
the output voltage back to the inverting
plot shows the differential signal as a
𝐴𝐴!"
𝐴𝐴! + 𝐴𝐴!
𝑉𝑉! + 𝑉𝑉!
input. The gain of these circuits is related
single waveform.
𝑉𝑉! − 𝑉𝑉! + 𝐴𝐴! − 𝐴𝐴!
𝑉𝑉!"# =
𝑅𝑅! + 2𝑅𝑅!
2
2 the ratio
to
of the resistor values, which
𝐴𝐴! =
𝐴𝐴! + 𝐴𝐴!
𝑉𝑉! + 𝑉𝑉!
2𝑅𝑅!
sets the proportion
of the output fed back
Differential amplifiers 𝑉𝑉!"# =
𝑉𝑉! − 𝑉𝑉! + 𝐴𝐴! − 𝐴𝐴!
2
2
by the potential divider. The amplifier as
A differential voltage amplifier is a circuit
𝑅𝑅!
−
𝑉𝑉
a whole is either inverting (negative gain)
in which
which has two inputs and amplifies
+ A )/2
𝑅𝑅! !we can see that (A
𝑉𝑉 1+ 𝑉𝑉! 2
or non-inverting (positive gain) depending
the voltage difference between them –
corresponds
ideal! equation,
𝑉𝑉!"# = 𝐴𝐴! with
𝑉𝑉! − A
𝑉𝑉!d in+the
𝐴𝐴!"
2 The
on whether the input signal is routed
thus, it amplifies the differential signal
and is the average of the two gains.
𝑉𝑉! + 𝑉𝑉!
to the inverting or non-inverting input.
at its input. If its inputs are
contains (V1 + V2)/2, which
V!2− 𝑉𝑉!other
𝑉𝑉!"#V=
𝐴𝐴! 𝑉𝑉
+ 𝐴𝐴term
1 and
!"
2
Scott’s circuit (Fig.1) is similar to the
and its differential voltage gain is Ad,
is the common-mode
input signal (the
𝑅𝑅!
1 + of 𝑉𝑉
!
circuits
in Fig.5 – the feedback arrangement
average
the
input
voltages),
which
then its output is Ad(V1 – V2). Its output
𝑅𝑅!
𝐴𝐴!
is the same, but it has two inputs instead
is 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
multiplied
by !"
a gain which
we refer
may be differential (in which case it
= 20log
dB
𝐴𝐴!"
is referred to as a ‘fully differential’
𝐴𝐴!
Vin
Rf
Rf
amplifier), or single ended.
A standard
+
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
= 20log!"
dB
V
𝐴𝐴
out
!"
op amp is a differential amplifier with
U1
𝑅𝑅!
𝑅𝑅!
Ri
Ri
Vin
𝑉𝑉!"# = 1 + – 𝑉𝑉! −
𝑉𝑉! –
a single-ended output. However, op
V2
–
Vout
𝑅𝑅!
𝑅𝑅
𝑅𝑅!
!
U1
Vout
amps are very commonly used to build
U1
−
𝑉𝑉!
𝑅𝑅!
+
V1
+
amplifiers circuits which have both
Rf
𝑅𝑅! – these
Ri
single-ended inputs and outputs
−
𝑉𝑉!
Ri
𝑅𝑅
Rf
!
are the well known inverting and
non𝑅𝑅! + 2𝑅𝑅!
Inverting
Non-inverting
inverting configurations (see Fig.5 and
𝐴𝐴! =
A
AC = 1 + R f / R i
𝑅𝑅C!= –R2𝑅𝑅
f/R
!i
later discussion).
1+
𝑉𝑉!
𝑅𝑅!
An ideal differential amplifier only
𝑅𝑅! V
Fig.5. Basic op amp amplifiers: inverting (left) and
Fig.6. The standard differential op
amplifies the difference between
1
1+
𝑉𝑉
𝑅𝑅! !
non-inverting (right).
amp amplifier.
(non-inverting input) and V2 (inverting
Practical Electronics | March | 2020
𝑉𝑉!"# = 1 +
𝑉𝑉!"# = 1 +
𝑅𝑅!
𝑅𝑅!
𝑉𝑉! −
𝑉𝑉
𝑅𝑅!
𝑅𝑅! !
𝑅𝑅!
𝑅𝑅!
𝑉𝑉 −
𝑉𝑉
𝑅𝑅! !
𝑅𝑅! !
45
poor one. The differential gain is not
the value required by Scott (3), but this
could be fixed with different resistors.
This incorrect differential gain value is
unlikely to be the reason that the circuit
did not perform as Scott hoped.
The circuit configuration in Fig.1 may
be a bad differential amplifier, but it is
not a useless circuit in other contexts.
Superposition
A common use for this circuit is to
Having
identified
this
relationship
Strain
Circuit Revisited
Strain Gauge
Gauge
simultaneously amplify a signal and shift
between
Scott’sCircuit
circuitRevisited
and the basic op
its DC level. For example, consider a
amp configurations, we can use some
signal with range ±0.5V, which needs to
circuit theory called the ‘superposition
𝑉𝑉
= 𝐴𝐴
− 𝐴𝐴! 𝑉𝑉
!
!
𝑉𝑉!"#
𝐴𝐴!! 𝑉𝑉
𝑉𝑉
𝑉𝑉formula
!"# = to
! − 𝐴𝐴!
!
be input to an ADC with a 0-5V input.
theorem’
obtain
a
for
its
gain.
Strain Gauge Circuit Revisited
A gain of 5 is required to map the ±0.5V
The superposition theorem states that for
range to 0-5V, but it is also necessary to
a linear circuit we can find a voltage or
shift the DC level from 0V to 2.5V. The
current
interest
by𝑉𝑉taking all the sources
𝑉𝑉!"# =of 𝐴𝐴
𝑉𝑉
−
𝐴𝐴
!
!
!
!
𝐴𝐴
𝑉𝑉! +
!
𝐴𝐴!! +
+ 𝐴𝐴
𝐴𝐴all
+ 𝑉𝑉
𝑉𝑉!! configuration in Fig.1 powered from
! the other sources to 𝑉𝑉!
circuit
in 𝑉𝑉
turn,
setting
=
𝑉𝑉
−
𝑉𝑉
+
𝐴𝐴
−
𝐴𝐴
𝑉𝑉!"#
𝑉𝑉!! − 𝑉𝑉!! + 𝐴𝐴!! − 𝐴𝐴!!
!"# =
2
2
2
2
a ±5V or greater supply can handle the
zero, working out the relevant circuit
required signals. If we connect the signal
values with just that one source active,
the non-inverting input (V1 in Fig.1)
repeating this
for𝐴𝐴all the sources, and then 𝑉𝑉 to
𝐴𝐴! +
!
! + 𝑉𝑉!
𝑉𝑉!"# =the results from
𝑉𝑉! − 𝑉𝑉the
𝐴𝐴! − 𝐴𝐴!
adding
individual
with a resistor ratio satisfying 1 + Rf/Ri
! +
2
2
𝑉𝑉
𝑉𝑉!! +
+ 𝑉𝑉
𝑉𝑉!!
sources
value.
= 5, that is Rf/Ri =4, will get the required
𝑉𝑉
−
!"# =
! 𝑉𝑉
! +
!"
𝑉𝑉!"#
=to𝐴𝐴
𝐴𝐴get
𝑉𝑉!!the
− 𝑉𝑉
𝑉𝑉required
+ 𝐴𝐴
𝐴𝐴!"
!
!
2sources
For Scott’s circuit we have two2
gain of 5 with respect to this input. The
– the inputs V1 and V2. We don’t have to
gain with respect to the inverting input
will then be −4 (−Rf/Ri). A DC input at the
worry about the supplies – they
are
fixed
𝑉𝑉! + 𝑉𝑉!
= 𝐴𝐴! 𝑉𝑉that
𝐴𝐴!"
DC𝑉𝑉!"#
voltages
do
not
carry signals,
V2 input in Fig.1 will be amplified by −4,
! − 𝑉𝑉
! +
𝐴𝐴
2
!
𝐴𝐴do
! not change the
and
(at least
ideally)
so if we connect this input to a DC level of
dB
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
= 20log
!"
!" 𝐴𝐴
𝐴𝐴!"
!"as long as the op
output voltage or gain,
2.5/−4 = −0.625V (for example, obtained
amp is operating normally. The output
via a potential across the negative supply
due to V2 with V1 grounded
rails) this will produce the required +2.5V
is:
𝐴𝐴!
dB
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 20log!"
DC output when the signal is 0V.
𝐴𝐴!"
𝑅𝑅
𝑅𝑅!!
−
𝑉𝑉
!
− 𝑅𝑅 𝑉𝑉!
Alternative circuits
𝑅𝑅!!
As we discussed in the previous article
there is a standard op amp circuit for
The output due to V1 with V2 grounded is:
𝑅𝑅!
a ‘proper’ differential amplifier (see
−
𝑉𝑉!
𝑅𝑅! 𝑅𝑅
𝑅𝑅!!
Fig.6). Comparing this with Fig.1 we
1
1+
+ 𝑅𝑅 𝑉𝑉
𝑉𝑉!!
see that it is essentially the same circuit
𝑅𝑅!!
with a potential divider at the V1 input.
The divider attenuates the V1 input to
Adding these together we get:
𝑅𝑅!
compensate for the higher gain via the
1+
𝑉𝑉
𝑅𝑅! ! 𝑅𝑅
𝑅𝑅
inverting input in Fig.1. The differential
𝑅𝑅!!
𝑅𝑅!!
𝑉𝑉
= 1+
𝑉𝑉
𝑉𝑉
𝑉𝑉!"#
𝑉𝑉! −
−
𝑉𝑉!
!"# = 1 +
gain of the circuit in Fig.6 is simply Rf/
𝑅𝑅!! !
𝑅𝑅!! !
Ri – note that the two pairs of resistors
We can compare this equation to the nonhave the same values. As discussed in
𝑅𝑅! amplifier
𝑅𝑅! with separate
ideal
differential
the previous article, this circuit only
𝑉𝑉!"# = 1 +
𝑉𝑉! −
𝑉𝑉
𝑅𝑅
𝑅𝑅! !
gains (A1𝑅𝑅
achieves high CMRR (limited by the op
A!2!), as described
above;
𝑅𝑅and
+ 2𝑅𝑅
2𝑅𝑅
!! +
!
𝐴𝐴
=
𝐴𝐴!!finding
=
amp) if the pairs of Rf and Ri resistors
and
the
differential gain as the
2𝑅𝑅
!!
average of A1 and A2, we get:
both have very accurately matched values
– so it is difficult to achieve very high
𝑅𝑅! + 2𝑅𝑅!
CMRR with discrete resistors. Another
𝐴𝐴! =
issue with this circuit is its relatively low
2𝑅𝑅!
input impedance, which means that it
may load signal sources such a sensors,
A bad circuit?
potentially leading to measurement
With Scott’s resistor values the differential
errors. As discussed in the previous
gain is: (5 + (2 × 10)) / (2 × 5) = 2.5.
article a solution to this is to use an
Similarly, we can find the commoninstrumentation amplifier circuit or IC.
mode gain from A1 – A2 which is simply
unity. Thus the CMMR is just 2.5 (about
8dB) and is more than a million times
Common-mode input
worse than that of the op amp itself!
As indicated above, the poor performance
Using a high performance op amp does
of Scott’s circuit as a differential amplifier
not guarantee a high performance circuit.
is probably not the reason that it failed to
Previous discussion has dismissed
work. The gain of 2.5 would still provide
Scott’s circuit as not being a differential
a reasonable output (not ‘lower than
amplifier, but this is not strictly true –
the input’). The reason is more likely to
it is a differential amplifier, just a very
be related to the common-mode input
of one. In fact, Scott’s circuit is like a
combination of the two basic amplifier
configurations. If we ground the V1 input to
Scott’s circuit we get an inverting amplifier
with V2 as its input. If we ground the V2
input we get a non-inverting amplifier
with V1 as its input.
46
applied to the circuit in Scott’s system.
Assuming the load cell Scott used is
similar to Fig.3 and that all four bridge
resistors have approximately the same
value (as is typical) then the signal from
the load cell will be a small differential
voltage (Scott states 0 to 36mV) with a
common-mode voltage of half the load cell
excitation voltage (due to the almost-equal
resistances in the potential dividers). Scott
stated that his load cell is operated on
12V, but did not provide further details.
If we assume the simplest case – the
load cell is connected to a 12V supply
relative to ground – then its commonmode output would be 6V. If we just
consider the gains calculated earlier we
might expect the output of Scott’s circuit
to be 2.5 times the differential signal
(about 0 to 90mV), with a common-mode
output of one times the 6V input – that
is 6V. There is a problem here – Scott
states that the AD8628 op amp circuit
(Fig.1) is operating from a 3V supply
(and its maximum supply is 5V). If our
assumptions about the load cell power are
correct then the common-mode voltage
into the circuit in Fig.1 would be well
beyond the supply voltage. Typically, op
amps do not work under such conditions.
Op amps have a characteristic called
‘common-mode input range’ which
specifies the extremes of commonmode input at which they will work
correctly. Some op amps (including the
AD8628) can cope with common-mode
signals equal to the supply voltage –
often referred to as a ‘rail-to-rail input’
capability. We need to check what the
datasheet says about the AD8628’s
common-mode input range and its
behaviour with overvoltage inputs.
The AD8628
The AD8628 is made by Analog Devices
and described as a zero-drift, singlesupply, rail-to-rail input/output op amp.
It is not an ordinary op amp in that
it achieves its high precision through
a combination of auto-zeroing and
chopping techniques (patented by Analog
Devices). Analog Devices state that this
allows the op amp to maintain its low
offset voltage over a wide temperature
range and over its operating lifetime,
and that it has low noise compared with
other auto-zero amplifiers. It is available
in a variety of packages – the common
8-lead SOIC and MSOP forms, and the
less usual (for op amps) 5-lead TSOT and
SOT-23 packages (as used by Scott). See
Fig.7 for the 5-lead pin-out. Strain gauge
amplifiers are listed among the suggested
applications for the device.
Auto-zeroing and chopping amplifiers
are designed to deal with the difficulty
of high-precision amplification of lowfrequency signals (typically a few hertz
Practical Electronics | March | 2020
OUT 1
V– 2
this occurred in Scott’s system. Another
potential issue with common-mode
overvoltage, which applies when it
occurs temporarily, is that some autozero amplifiers take a long time to recover
when the overload is removed. The data
sheet for the AD8628 states that it has a
much shorter recovery time than typical
auto-zero amplifiers. This problem is not
relevant in Scott’s case as the presumed
overload is permanent.
5 V+
AD8628
Top view
(not to scale)
+IN 3
4 –IN
Fig.7. AD8628 pinout for
5-lead TSOT and SOT-23
packages.
Simulation
Fig.8. LTspice schematic for
simulating the circuit in Fig.1.
and below). Depending on the application,
these amplifiers may also be required to
handle much higher frequencies, adding
to the difficulty of creating suitable circuit
designs. Low-frequency signals typically
occur in certain sensing implications,
such as strain gauges, where the measured
signal changes slowly. The problem is the
low-frequency noise and offsets which
are inherent in electronic amplifiers
and which render ordinary op amps
inadequate for these tasks. The solution
makes use of switching circuits and, as
indicated by the dual approach taken by
the AD8628, there are two basic ways to
implement these circuits – auto-zero and
chopping. We discussed these concepts
in depth in Circuit Surgery in the June
and July 2018 issues.
Overvoltage
The ‘rail-to-rail input’ part of the AD8628
description indicates that it can handle
input signals close to the supply, but the
datasheet goes into more detail. It states
that if either input exceeds either supply
rail by more than 0.3V, large currents
may flow through the ESD (electrostatic
discharge) protection diodes in the
amplifier. These diodes are connected
between the inputs and each supply rail
and are normally reverse-biased. However,
they can become forward-biased if the
input voltage exceeds the supply voltage.
If excessive current flows as a result of
this the device can sustain permanent
damage. In circuits where overvoltage
may occur the datasheet recommends
series resistors at the inputs to limit the
current to 5mA. We do not know enough
about Scott’s system to be certain about
the level of current that might have been
delivered via the strain gauge – although
Scott did not mention anything about his
op amp being destroyed.
As well as damage when currents are
not limited, common-mode overvoltage
can cause strange behaviours in some
differential amplifiers, specifically
the output can suddenly jump in the
opposite direction to the supply rail –
a phenomenon known as output phase
reversal. With the AD8628, limiting
the overvoltage input current to 5mA
should prevent this. Again, we do not
have enough information to know if
LTspice had a model for the AD8628, so
we can try a simulation. The schematic is
shown in Fig.8. Here we have set up the
input with specific common-mode and
differential signals – 1.5V common mode,
and 40mV peak-to-peak differential, are
obtained with the values shown on the
schematic, but these are easily changed.
This approach is used as it provides direct
control of the input signal and because we
do not know the exact circuit of Scott’s
load cell. We start with the commonmode level at half the op amp supply
voltage so that the circuit will operate
correctly and we can confirm the gains
calculated earlier. The sources have some
output impedance (R1 and R2 at 200Ω)
as the load cell would have non-zero
output impedance, but values used are
an arbitrary choice as we do not know
the load cell details.
The results of the simulation are
shown in Fig.9. We see that the input
signal (v(1) – v(2)) is slightly lower
than the source signal (v(source1) –
v(source2)) due to loading of the source
by the amplifier circuit. The differential
input amplitude is about 38.3mV peakto-peak and the output is about 96.7mV
peak-to-peak, confirming the expected
differential gain of 2.5 calculated above.
The output is centred on 1.5V – equal to
the common-mode input, confirming the
common-mode gain of 1. Experimenting
with the common-mode voltage indicated
correct operation with 0.05V and 2.95V
common-mode inputs, confirming the railto-rail operation. At 6V common mode
the output is basically saturated at the
3V supply, with the signal showing at a
level of tens of microvolts. We cannot be
sure this simulation exactly replicates
the situation in Scott’s circuit, but if his
load cell common-mode output was 6V
then that is the most likely reason why
his circuit did not work.
Simulation files
Fig.9. Results from simulation in Fig.8 of the circuit in Fig.8.
Practical Electronics | March | 2020
Most, but not every month, LTSpice
is used to support descriptions and
analysis in Circuit Surgery.
The examples and files are available
for download from the PE website.
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