Circuit Surgery
Regular clinic by Ian Bell
Solving a circuit: iteration load lines and simulation
I
n the January issue of PE, a
couple of articles were featured which
mentioned load lines, and editor Matt
Pulzer suggested a Circuit Surgery article
looking at this topic.
A load line is a graphical means of
solving what a circuit will do without
having to perform detailed mathematical
analysis. In this article we will look at load
lines in the context of the various ways
to solve non-linear circuits – specifically
the diode-resistor circuit in Fig.1. This
is a straightforward circuit – a forwardbiased diode, that is a diode with current
flowing in the direction indicated by
the arrow shape of the diode symbol.
Assuming we know some basic circuit
theory, the voltage-current relationships
for a diode and a resistor, and the diode’s
characteristic parameter values, then we
might assume it would be easy to find
an equation for current in the circuit in
Fig.1. If there were two resistors instead
of a diode and a resistor it would be
easy with basic circuit theory; however,
it turns out not to be so simple to ‘solve’
the diode circuit.
Before going into details of the diode
circuit we will review the circuit theory
we need and solve a two-resistor circuit
as a point of comparison. Two of the key
principles of circuit analysis are known
as Kirchhoff’s laws (which date from
1854). One law states that, ‘the sum of
currents into a junction in a circuit is zero
(or total current in = total current out).’
This also means the current through a
I?
VD D
1 0 V
VD ?
D
+
1
V
–
R
1
Ω
VR ?
Fig.1. Diode-resistor circuit.
Simulation files
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.
58
set of components connected in series
is the same in each component (the
diode and resistor currents are equal
in the circuit in Fig.1). The other law
states that, ‘the sum of voltages round
any loop in a circuit is zero.’
Circuit laws and component
models
In the resistor characteristic equation
V = IR the R (the resistance) is referred
to in general terms as a ‘parameter’
(or model parameter). All component
characteristic equations will have one
or more parameters. We can analyse a
circuit algebraically using symbols for the
parameters (eg, R1, R2… for the resistors),
but if we need to know actual voltages and
currents, we have to obtain the specific
parameter values for every component.
Kirchhoff’s laws deal with either voltage
or current on their own. To fully analyse a
circuit, we need to know the relationships
between currents and voltages in the
Diode equation
circuit. These relationships are described
In order to include semiconductor
by the characteristic equations of each
components, such as the diode in Fig.1,
component. For our example, the
in circuit analysis we need a model;
characteristic equation for a resistor is
that is the characteristic equations for
V = IR, which many readers will recognise
general analysis, and actual parameter
as Ohm’s law. So, for example, for Fig.1,
values for specific calculations. We can
we can write: VR = IR1
use an understanding of semiconductor
physics to derive a characteristic equation
O h m ’s l a w s , a n d h e n c e t h e
for a diode – like Kirchhoff’s and Ohm’s
characteristic equation for the resistor,
laws, this is well established theory. For
was first obtained experimentally by
an idealised case we get the following,
Georg Ohm in the 1820s. It can also be
Solving a circuit: iteration load lines and simulation
relating the current through the diode
derived from the fundamental physics
(ID) to the voltage across it (VD):
of conductors, although historically
this happened much later. For a real
𝑉𝑉!
resistor, Ohm’s law is an approximation,
𝐼𝐼! = 𝐼𝐼" #𝑒𝑒𝑒𝑒𝑒𝑒 ' ) − 1,
𝑉𝑉#
the resistance may actually change
with applied voltage, and it will
This is known as the ‘Shockley ideal
probably change with temperature,
diode equation’. V T is called the
𝑉𝑉!
including that caused by self-heating
thermal
𝐼𝐼! = 𝐼𝐼"voltage.
𝑒𝑒𝑒𝑒𝑒𝑒 ' ) It is about 26mV at
𝑉𝑉# and is given by kT/e,
of the resistor. Furthermore, resistors
room temperature
generate electrical noise, and Ohm’s
where T is the junction temperature,
law only applies to the average current.
and e and k are physical constants, the
𝐼𝐼!
The mathematical way we choose to
charge
and Boltzmann’s
𝑉𝑉! =on
𝑉𝑉#the
𝑙𝑙𝑙𝑙 'electron
)
𝐼𝐼$
represent a component is referred to as
constant respectively.
The key parameter
the ‘model’ of that component.
for individual diodes is IS – the reverse
Most of the time, for resistive
saturation current – which can vary over
𝑉𝑉!!
− 𝑉𝑉%&
components that we use in circuits we
orders of
magnitude,
depending on the
𝐼𝐼 =
𝑅𝑅
can simply use V = IR in circuit analysis.
device and temperature,
for example
'
If choose to include other factors in our
10-14 to 10-10A for silicon diodes (the
calculations (and do so correctly) we get
default in LTspice is 10-14A).
𝑉𝑉!! − 𝐼𝐼𝑅𝑅& nature of the voltagea more accurate results at the expense
The exponential
𝐼𝐼 =
of more difficulty and complexity. For
current relationship
means that very little
𝑅𝑅'
manual calculation, too much complexity
current flows for low forward voltages,
may make the process error-prone, or
but it increases rapidly with applied
even intractable; and for computervoltage so 𝑉𝑉that
!! the diode ‘switches on’
𝐼𝐼 =
based analysis (circuit simulation), run
and conducts
𝑅𝑅& + 𝑅𝑅at' around 0.6 to 0.7V. For
times will increase with more complex
conduction at or close to these ‘on’ levels
models. As always in engineering, there
the exponential term is much bigger
are compromises – in this case between
than 1 and
𝑉𝑉!! we
− 𝑉𝑉!can simplify the diode
𝐼𝐼 = to:
accuracy and complexity.
equation
𝑅𝑅
Practical Electronics | May | 2021
𝑉𝑉!! − 𝑉𝑉# ln(𝐼𝐼⁄𝐼𝐼$ )
𝐼𝐼 =
𝑅𝑅
I?
it: iteration load lines and simulation
VD D
1 0 V
+
VR 1?
R
V
1
Ω
𝑉𝑉!
it: iteration load lines and simulation
–
𝐼𝐼! = 𝐼𝐼" #𝑒𝑒𝑒𝑒𝑒𝑒 ' ) − 1,
𝑉𝑉#
R 2
3 Ω
VR 2 ?
Fig.2. Two-resistors
𝑉𝑉! circuit.
𝐼𝐼! = 𝐼𝐼" #𝑒𝑒𝑒𝑒𝑒𝑒 ' ) − 1,
𝑉𝑉 𝑉𝑉!
𝐼𝐼! = 𝐼𝐼" 𝑒𝑒𝑒𝑒𝑒𝑒 '# )
𝑉𝑉#
If we know ID 𝑉𝑉and want to find VD we
!
can𝐼𝐼!rearrange
= 𝐼𝐼" 𝑒𝑒𝑒𝑒𝑒𝑒 ' this
) equation by taking
𝐼𝐼𝑉𝑉!#
natural
𝑉𝑉! = logarithms
𝑉𝑉# 𝑙𝑙𝑙𝑙 ' ) (ln – the inverse of
𝐼𝐼$ function) to produce:
the exponential
𝐼𝐼!
𝑉𝑉! = 𝑉𝑉# 𝑙𝑙𝑙𝑙 ' )
𝑉𝑉
!! − 𝑉𝑉𝐼𝐼%&
it:
iteration
load
lines
and
simulation
$
it: iteration load lines and
𝐼𝐼 = simulation
𝑅𝑅'
Two-resistor circuit
𝑉𝑉!!
−𝑉𝑉!𝑉𝑉%&
As
noted
above,
tackling the diode! )before
𝐼𝐼𝐼𝐼! =
''−𝑉𝑉𝐼𝐼𝑅𝑅
"" #𝑒𝑒𝑒𝑒𝑒𝑒
=𝐼𝐼 𝐼𝐼𝐼𝐼=
#𝑒𝑒𝑒𝑒𝑒𝑒
)&−
− 1,
1,
𝑉𝑉
!
𝑅𝑅
!! '𝑉𝑉
resistor
circuit
we
will
analyse a circuit
#
𝑉𝑉
#
𝐼𝐼 =
𝑅𝑅'
with two resistors
(Fig.2). After that we
will attempt to follow the same procedure
𝑉𝑉!! − 𝐼𝐼𝑅𝑅
𝑉𝑉!&
for𝐼𝐼 the
To analyse the
!)
𝐼𝐼 =
𝐼𝐼𝐼𝐼diode
𝑒𝑒𝑒𝑒𝑒𝑒
''𝑉𝑉circuit.
!
𝐼𝐼
=
𝑒𝑒𝑒𝑒𝑒𝑒
)can proceed as follows:
𝑉𝑉𝑅𝑅!!
! in""Fig.2
𝑉𝑉
circuit
' we
#
𝑉𝑉#
𝐼𝐼 =
𝑅𝑅& + 𝑅𝑅'
it: iteration load lines
and simulation
Current
in R1 is given by: I = VR1/R1
So VR1 = IR𝑉𝑉1!! 𝐼𝐼𝐼𝐼!
𝐼𝐼
=
𝑉𝑉! = 𝑉𝑉𝑉𝑉
𝑉𝑉# 𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙−''𝑉𝑉! ))
𝑉𝑉
!
# +
𝑅𝑅!!
𝑅𝑅𝐼𝐼!
𝐼𝐼 = & 𝑉𝑉 𝐼𝐼'$$
Current in R𝑅𝑅2 !(in terms of VR1 and VDD)
𝐼𝐼 = 𝐼𝐼" #𝑒𝑒𝑒𝑒𝑒𝑒 ' ) − 1,
is! given
by: 𝑉𝑉#
𝑉𝑉!!
− 𝑉𝑉𝑉𝑉%&
𝑉𝑉
!! −
!
𝑉𝑉
𝐼𝐼 =
= !! − 𝑉𝑉%&
𝐼𝐼𝐼𝐼𝑉𝑉
= − 𝑉𝑉𝑅𝑅#𝑅𝑅'ln(𝐼𝐼⁄𝐼𝐼$ )
!!
𝐼𝐼 =
𝑅𝑅'
𝑅𝑅 𝑉𝑉!
𝐼𝐼! = 𝐼𝐼" 𝑒𝑒𝑒𝑒𝑒𝑒
Eliminate
VR1'𝑉𝑉# )
𝑉𝑉!! 𝑉𝑉− 𝑉𝑉#−ln(𝐼𝐼
𝐼𝐼𝑅𝑅&⁄𝐼𝐼$ )
!! − 𝐼𝐼𝑅𝑅
𝐼𝐼 = 𝐼𝐼 = 𝑉𝑉!!
𝑉𝑉!&
𝐼𝐼 =
𝑅𝑅
𝑅𝑅
𝐼𝐼! = 𝐼𝐼" 𝑒𝑒𝑒𝑒𝑒𝑒
𝑅𝑅'' ' )
𝐼𝐼𝑉𝑉!#
𝑉𝑉! =
Solve
for𝑉𝑉#I 𝑙𝑙𝑙𝑙 ' 𝐼𝐼 )
𝑉𝑉$!
𝑉𝑉
𝑉𝑉!!
𝐼𝐼! 𝐼𝐼==𝐼𝐼" 𝑒𝑒𝑒𝑒𝑒𝑒
)
!!'
𝐼𝐼 =𝑉𝑉!!+−𝑅𝑅𝑉𝑉𝑉𝑉!#
𝐼𝐼 = 𝑅𝑅
𝑅𝑅&& + 𝑅𝑅''
𝑉𝑉!! 𝑅𝑅
− 𝑉𝑉%&
𝐼𝐼 =the component values, I = 10/(2kΩ
Using
𝑅𝑅'
𝑉𝑉
+ 3kΩ) =𝑉𝑉
2mA.
Using
V = IR we also get
!! −
!
𝑉𝑉
𝑉𝑉
!! −
!
𝑉𝑉
−V
𝑉𝑉!
𝐼𝐼
=
!!
VR1 =𝐼𝐼𝐼𝐼 =
4V
and
=
6V.
𝑅𝑅
R2
=
𝑅𝑅
𝑅𝑅
𝑉𝑉!! − 𝐼𝐼𝑅𝑅&
𝐼𝐼 =
Diode-resistor
circuit
𝑅𝑅'
Now,𝑉𝑉 let’s
attempt
to follow the same
⁄
−
𝑉𝑉
ln(𝐼𝐼
𝐼𝐼
#
⁄𝐼𝐼$$ ))
𝑉𝑉!!
!! − 𝑉𝑉
# ln(𝐼𝐼
𝐼𝐼𝐼𝐼 =
procedure
for
the
resistor-diode
circuit
=
𝑅𝑅
𝑅𝑅 current in a diode is given
in Fig.1. The
𝑉𝑉!!
by: I 𝐼𝐼==IS exp (VD / VT), so VD = VT ln (I
𝑅𝑅
+ 𝑅𝑅'current in R (in terms of
/ IS). Thus,& the
𝑉𝑉!
=V
𝐼𝐼 𝑒𝑒𝑒𝑒𝑒𝑒
''𝑉𝑉! ))
VD 𝐼𝐼𝐼𝐼and
) is:
!
𝑒𝑒𝑒𝑒𝑒𝑒
! = 𝐼𝐼"" DD
𝑉𝑉
𝑉𝑉##
𝑉𝑉!! − 𝑉𝑉!
𝐼𝐼 =
𝑅𝑅
𝑉𝑉
− 𝑉𝑉
!!
𝑉𝑉
!! − 𝑉𝑉!
!
𝐼𝐼𝐼𝐼 =
Eliminate
= V𝑅𝑅D
𝑅𝑅
𝑉𝑉!! − 𝑉𝑉# ln(𝐼𝐼⁄𝐼𝐼$ )
𝐼𝐼 =
𝑅𝑅
The next step would be to solve the
equation for I – to rearrange the equation
𝑉𝑉!
to make
but unfortunately,
𝐼𝐼! = 𝐼𝐼I" the
𝑒𝑒𝑒𝑒𝑒𝑒subject,
' )
𝑉𝑉# easily. Diode circuits
we can’t do this
produce transcendental equations which,
in general, are difficult or impossible
𝑉𝑉!! − 𝑉𝑉!
to solve.
𝐼𝐼 = A transcendental equation
𝑅𝑅
Practical Electronics | May | 2021
is one that involves a transcendental
function, which include the
exponentials/logarithms we have here.
A transcendental function is one which
cannot be expressed in terms of algebraic
operations – the name ‘transcendental’
refers to the idea that it ‘transcends’ or
goes beyond algebra. Actually, it is not
impossible to solve the diode-resistor
circuit equation, but it requires some
very advanced maths, specifically the
Lambert W-function, which most people
probably never encounter or use (see the
Lambert W-function Wikipedia page if
you want to know more).
Rule of thumb
If we cannot solve the circuit in Fig.1
by manipulating the circuit equations
algebraically, what can we do instead to
find the diode current? A simple answer,
which is commonly used in quick circuit
design calculations is to assume the
voltage across the diode (the forwardvoltage drop) is 0.7V. It is well known
that for typical operating currents in most
circuits the voltage across a forwardconducting standard silicon diode is
likely to be in the range 0.6 to 0.8V; so
0.7V is often taken as the rule-of-thumb
value to use. For other types of diodes,
which might have different voltages (eg,
LEDs) the forward-voltage drop is often
stated on the datasheet, so we can use
the same approach. In the resistor-diode
circuit of Fig.1, if the supply is not too
close to the expected diode voltage, the
expected range of forward-voltages does
not make much difference to the resistor
current and so we can get a reasonable
estimate just by assuming VD = 0.7V. If
we do this then we get VR1 = 9.3V and
I = V R1/I = 9.3/1000 = 9.3mA for the
circuit in Fig.1.
The 0.7V rule of thumb gives us an
approximate solution to the circuit, but
it may not be very accurate. What can
we do if we want to find a more accurate
value? Choosing 0.7V is basically an
informed guess about what is happening
in the circuit. Having made such a guess,
we can use the diode equation to find
the corresponding current and hence the
resistor voltage (VR1) with this current
flowing through it. We can then add
the resistor and diode voltages, which
should equal the supply voltage (if not,
Kirchhoff’s voltage law is not satisfied).
Guessing game
If our implied supply voltage is wrong,
we can make a new, hopefully better,
guess – for example, if the implied VDD
we calculated is too high, we can reduce
the assumed diode voltage a little (eg,
guess 0.68V instead of 0.7V) and see if
that improves things. We can keep on
guessing and refining our diode voltage
value, recalculating the current and
implied VDD, and guessing again, getting
closer and closer to the correct value.
Formally, this guessing process is referred
to as an iterative solution to the circuit.
Here is an example for Fig.1 in which
we will use IS = 1.0 × 10-14A and VT =
2.585mV (for a temperature of 27°C):
Guess: VD = 0.7V, so I = IS exp (VD / VT)
= 5.70mA (diode equation)
Thus: VDD = VR + VD = 6.400V – too low
– so increase VD guess to 0.725V
With: VD = 0.725V, ID = 4.986mA (from
the diode equation)
Thus: VR + VD = 15.711V – too high –
so decrease VD guess to 0.7125V
With: VD = 0.7125V, ID = 9.242mA (from
the diode equation)
Thus: VR + VD = 9.955 just too low – so
increase VDD guess to 0.7126V
With: VD = 0.7126V, ID = 9.278mA and
VR + VD = 9.990V
This final result implies the supply is
9.990V rather than 10V, a 0.1% error. We
could continue to add more significant
figures, but we have to stop at some point
when the result is ‘close enough’ – how
we decide what is close enough depends
on what we need in terms of accuracy.
Numerical methods
The guess and recalculate process we
have just used is a crude manual version
of what simulators such as LTspice are
doing when they solve a circuit. The
approach to finding the next guess is
more sophisticated than the ‘think of
a number’ approach implied above.
The mathematical techniques involved
are referred to as ‘numerical methods’,
or ‘numerical analysis’ and involve
numerical approximation rather than
symbolic manipulation. These have a
long history. For example, the NewtonRaphson method, which can be used by
circuit simulators, is derived from work
by Isaac Newton and Joseph Raphson
in the 17th century. Other numerical
methods are older still, going back to
the earliest mathematical writings on
Babylonian clay tablets.
As with our manual iteration, circuit
simulators must decide when to stop
a given calculation – when it is ‘close
enough’ – referred to as convergence in
simulator terminology. SPICE simulators
measure convergence in a couple of ways:
the difference between iterations and the
degree to which Kirchhoff’s current law
has been met. If the difference between
iterations is small enough it implies
the solution has been reached and new
steps will not get significantly closer. In
59
Fig.3.
I
LTspice
schematicload
Solving
Solving aa circuit:
circuit: iteration
iteration
load lines
lines and
and simulation
simulation
for circuit in
VD D
Fig.1.
L oad line
R
𝑉𝑉
𝑉𝑉!
𝐼𝐼𝐼𝐼!
= 𝐼𝐼 #𝑒𝑒𝑒𝑒𝑒𝑒 ' ! ) − 1,
! = 𝐼𝐼"" #𝑒𝑒𝑒𝑒𝑒𝑒 ' 𝑉𝑉 ) − 1,
𝑉𝑉##
our manual iteration above we checked
the implied supply voltage – effectively
checking Kirchhoff’s voltage law has been
met – SPICE’s checking of the current
law is similar – if the numerical values
of current throughout the circuit are in
compliance with the law sufficiently
closely then the solution is likely to be
good. SPICE uses a number of numerical
values to decide when convergence has
occurred, for example, abstol, the
absolute current error tolerance and
reltol, the relative error tolerance.
These values can be changed via user
options, for example in LTspice via the
SPICE tab of the Control Panel.
Simulation example
Fig.3 shows the circuit from Fig.1 drawn
as an LTspice schematic and includes
the diode model statement to set the IS
parameter to 1x10-14A. This is not strictly
required, as it is the same as the default,
but illustrates how this key parameter
can be set. We run an operating point
analysis in LTspice (.op directive) to
obtain the current and voltages. LTspice
simulates at a default temperature of
27°C so we should get similar results
to our manual calculation. The results,
which are shown below, are close to
our calculation.
--- Operating Point --V(diode): 0.712763
voltage
V(supply):10
voltage
I(D1):
0.00928724 device_current
I(R1):
-0.00928724 device_current
I(V1):
-0.00928724 device_current
Operating
point
𝑉𝑉
𝑉𝑉!
𝐼𝐼𝐼𝐼!
= 𝐼𝐼 𝑒𝑒𝑒𝑒𝑒𝑒 ' ! )
! = 𝐼𝐼"" 𝑒𝑒𝑒𝑒𝑒𝑒 '𝑉𝑉 )
𝑉𝑉##
VD D
V
𝐼𝐼𝐼𝐼!
!)
𝑉𝑉
=
𝑙𝑙𝑙𝑙
Fig.4.
𝑉𝑉!
= 𝑉𝑉
𝑉𝑉##line
𝑙𝑙𝑙𝑙 ''for
)a diode-resistor circuit.
!Load
𝐼𝐼𝐼𝐼$$
Loads lines
An alternative
𝑉𝑉
− 𝑉𝑉%&approach to iterative
𝑉𝑉!!
!! − 𝑉𝑉%&
𝐼𝐼𝐼𝐼 =
=
calculation,
𝑅𝑅
𝑅𝑅'' which does not use a
simulator, is to plot a load line graph.
This is a plot of the diode and resistor
characteristics
𝑉𝑉
!! −
& the same axes – the
𝑉𝑉!!
− 𝐼𝐼𝑅𝑅
𝐼𝐼𝑅𝑅on
&
𝐼𝐼𝐼𝐼 =
=where
point
the
two traces cross is the
𝑅𝑅
𝑅𝑅''
solution to the
circuit, also referred to
as the operating point. The term ‘load
line’ comes𝑉𝑉
the idea that we have an
!!
𝑉𝑉from
!!
=
active𝐼𝐼𝐼𝐼 component
= 𝑅𝑅 + 𝑅𝑅 (a diode or transistor)
𝑅𝑅&& + 𝑅𝑅''
controlling
the signal (output) to a load
(often a resistor). As we have seen,
circuits 𝑉𝑉
such−as
𝑉𝑉! that in Fig.1 cannot be
𝑉𝑉!!
!! − 𝑉𝑉!
𝐼𝐼𝐼𝐼 =
easily
using algebra, the load
=solved
𝑅𝑅
𝑅𝑅
line is a graphical
approach to solving
the circuit, which does not need a full
algebraic
⁄𝐼𝐼𝐼𝐼$ ))
𝑉𝑉!! −
−solution.
𝑉𝑉## ln(𝐼𝐼
ln(𝐼𝐼⁄
𝑉𝑉
𝑉𝑉
𝐼𝐼𝐼𝐼 =
For!!
a load
line $graph of the diode=
𝑅𝑅
𝑅𝑅 (Fig.1) we plot on the same
resistor circuit
graph (see Fig.4) the diode characteristic:
𝑉𝑉
𝑉𝑉!
𝐼𝐼𝐼𝐼!
= 𝐼𝐼 𝑒𝑒𝑒𝑒𝑒𝑒 ' ! )
! = 𝐼𝐼"" 𝑒𝑒𝑒𝑒𝑒𝑒 '𝑉𝑉 )
𝑉𝑉##
And the resistor current:
𝐼𝐼𝐼𝐼 =
=
− 𝑉𝑉!
𝑉𝑉
𝑉𝑉!!
!! − 𝑉𝑉!
𝑅𝑅
𝑅𝑅
The diode characteristic trace is simply
a plot of the diode’s current vs. voltage
relationship. The resistor current trace –
the load line – is a straight line because
a resistor has a linear current-voltage
Fig.5. Actual load line graph for the circuit in Fig.1.
60
D iode
ch aract eristic
relationship. The trace is the opposite
way round to plotting a resistor’s current
vs. voltage relationship on its own
because we are using the diode voltage
as the axis – increasing VD decreases
the resistor voltage (together they add
up to V DD ) and hence decreases the
resistor current. The load line is easy
to draw because it crosses the axes at
the I = 0 and VD = 0 points, which can
easily be found from the above equation
as VD = VDD and I = VDD / R respectively
(see Fig.4).
Plotting a load line manually on graph
paper specifically to solve a circuit like
Fig.1 is probably too much effort to
be worthwhile, but load line graphs
like Fig.4 are useful for visualising the
behaviour of circuits. We can see how
changing the resistor affects the operating
point and how different operating points
will change the effect of small variations
of diode voltage on diode current. Diode
load line graphs are often drawn with
some artistic license when used for
general discussion of circuit behaviour
– so that the shape of the diode curve
can be seen together with the full load
line (eg, Fig.4). To get a graph looking
like Fig.4 would require a supply of
around 0.8V and a resistor around 80Ω
– not really typical values.
Load lines in LTspice
Fig.5 shows the actual load line graph
for the circuit in Fig.1, obtained using
LTspice. It is not usual to plot a load line
using SPICE because, as we have seen, the
simulator can solve the circuit operating
point directly. However, Fig.5 nicely
illustrates the fact that the diode has an
almost constant voltage of around 0.7V
over a wide range of operating currents
when plotted on a voltage axis covering
the entire supply range (the diode curve
is almost a vertical line).
Fig.6 shows the same results plotted
over a small diode voltage range close
to 0.7V. Here we can see the diode
Fig.6. The load line results (Fig.5) zoomed to show the diode
characteristic shape.
Practical Electronics | May | 2021
characteristic curve
shape more clearly.
Note that the resistor
current is almost
constant over this range
(horizonal line). This
illustrates that with
typical supply voltages
of several times the
Fig.7. LTspice circuit used to plot load
diode forward voltage
line graphs (Fig.5 and Fig.6).
the resistor current will
not change much for different diode voltages. This brings us
back to the starting point of our discussion on this circuit where
we simply assumed the diode drop was 0.7V and consequently
obtained a current of 9.3mA. If we zoom in more and use the
cursor, we find that the point where the curves cross closely
matches the operating point simulation results given above.
A load line graph considers the component currents
independently over the entire supply range, so we can’t use
the circuit in Fig.3 directly to plot it. The LTspice schematic
in Fig.7 was used to obtain the graphs in Fig.5 and Fig.6. A
linear DC sweep analysis was used to step the voltage from
the V1 voltage source from 0 to 10V in 0.001V steps. This
voltage is applied directly across the diode so we can obtain
the characteristic current. The small steps are useful as we
are going to zoom in on the steep diode curve. It should be
noted that we get some unfeasibly large diode currents from
simulating the diode on its own over the 0 to 10V range – the
simulator is not limited by maximum power dissipation or
supply capabilities! In the actual circuit the maximum current
in the diode is limited by the resistor.
The resistor voltage is obtained by subtracting the currently
applied diode voltage (V(diode)) from the 10V supply
voltage (as V1 is swept) using a behavioural voltage source
(B1). Behavioural sources allow us to write equations for a
source’s output in terms of other circuit parameters and were
discussed in detail in the August 2020 Circuit Surgery.
Transistor load lines
To extend the load line theme a little, Fig.8 shows a load line
for the basic BJT common-emitter circuit in Fig.9. As with the
diode circuit, we simulate the transistor characteristics and the
load resistor separately to obtain the load line plot – the circuit
used is shown in Fig.10. We assume the transistor has been
biased with a base current of 40µA but have not specified how
this was done. Like the diode-resistor load line, the load line
Fig.8. Load line for the BJT circuit in Fig.9. The plot shows the
variation of collector current (IC) with collector-emitter voltage (VCE) at
various base currents (IB). The circuit’s operating point moves along
the load line as base current varies and is at the bias operating point
with no signal.
Practical Electronics | May | 2021
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for this circuit intersects the axes
at the supply voltage (5V) and the
current which would flow in the
R 1
IC
resistor if it was connected across
300Ω
the supply (5/300 = 16.7mA).
IBias
The operating point of the
4 0 µ A
circuit is the point where the
Q 1
VCE
2 N 2 2 2 2
load line intersects the transistor
output characteristic (IC vs. VCE)
curve corresponding with the
base bias. Fig.8 shows that for this
example the output voltage would
Fig.9. Basic BJT commonbe around 2.6V with no signal.
emitter circuit.
If the transistor was amplifying
a signal, then the base current would vary accordingly. The
load line provides an easy way to see how the output voltage
would vary – the operating point moves along the load line as
the base current changes. So, for example, if the input signal
caused the base current to vary from 30 to 50µA then the
output voltage would vary from about 2.1 to 3.2V. A similar
example was discussed for a 2N2700 MOSFET circuit in Mike
Tooley’s Kick Start article in the January 2021 issue.
VCC
5 V
Fig.10. LTspice schematic used to obtain the load line graph in Fig.8.
61
