Joule Thief

SOME PERSONAL THOUGHTS TOWARDS UNDERSTANDING
THE JOULE THIEF CIRCUIT
JON KIRWAN
Abstract. This is a personal dialog regarding a steady-state analysis of a
Joule Thief, using a BJT as the switching element. Only algebra and ap-
proaches using averages are used here. No small signal, differential calculus
approaches are taken here, though the net results of such considerations may
be applied. The intent is to see how well the Joule Thief can be understood
quantitatively by more prosaic considerations.
1. Overview
The name “Joule Thief” is given to a particular kind of self-starting boost mode
switching power supply often used to operate white LEDs from a 1.5V battery
supply. It can even be used to operate a series chain of LEDs. It will continue to
operate until the battery’s energy is almost entirely consumed. Hence the name.
Date: November 29, 2013.
Figure 1. Canonical Joule Thief
1
THOUGHTS ON THE JOULE THIEF 2
In its simplest form, the Joule Thief replaces D
1
and C
1
shown in fig. 1 with
an LED. The difference is that fig. 1 allows a simpler specification of the load for
analysis. The output voltage must be higher and more positive than the input
supply voltage. Take special note of the winding arrangement of the transformer
(shown as L
1
and L
2
.) The usual method is to wind this transformer by hand
so that the inductance of L
1
and L
2
are taken as equal. It’s not necessary that
they be, though. D
1
may be a Schottky diode, as they tend to have lower forward
voltage drops and yield somewhat more efficient designs.
A base current via R
1
turns on Q
1
. Q
1
’s collector then approaches ground,
applying the battery voltage across the transformer primary, L
2
. This induces a
similar and aiding voltage (if L
1
is equal to L
2
and arranged as indicated) at L
1
,
adding to the EMF that drives current into Q
1
’s base. Roughly speaking, this
doubles the battery EMF and is why the Joule Thief works so well.
The rate of increase in the current in L
2
is determined by
V
L
2/L2 and rises along
a ramp towards I
peak
and is also Q
1
’s collector current. The magnitude of this
current represents the energy stored in L
2
’s magnetic field (according to eq. 7.1,
noted below.) So as the current rises linearly, the stored energy rises proportional
to its squared value. During this time, no energy is being transferred from L
2
to
the output capacitor, C
1
.
The base current of Q
1
remains roughly constant during this rise. At some
point determined by Q
1
’s beta, the magnitude of Q
1
’s collector current exceeds the
ability of the base current to sustain it. When that happens the collector current
cannot continue to rise. However, if the applied voltage across L
2
is maintained the
collector current must rise. So the voltage across L
2
declines rapidly towards zero.
But as the voltage across L
2
declines towards zero, so also does the aiding voltage
across L
1
. This reduction substantially reduces the base current that originally
supported the collector current in Q
1
, requiring a reduction in that collector current.
To reduce the collector current in Q
1
, the rate of change of current determined by
V
L
2/L2 must change sign and so V
L2
must itself change sign. This change of sign
constructs an opposing voltage in L
1
, further reducing the base current. Eventually,
this process turns Q
1
completely off.
(In fact, there is a danger of exceeding the modest V
EBO
reverse voltage, causing
it to zener. Especially in the case where the output voltage is substantially higher
than the battery voltage.)
During this voltage reversal phase, Q
1
remains off while the energy stored in L
2
is expended through D
1
and into charging C
1
. Eventually, the current falls towards
zero and reaches it and the magnetic field energy is exhausted. At this point, L
1
’s
opposing voltage cannot be supported and it no longer opposes the battery voltage.
Now the process described above can restart.
This paper focuses on the steady state case where the output voltage is roughly
stable and may be presumed higher than the input voltage. But initially upon
staring, C
1
will likely be uncharged and the input voltage supply will attempt to
charge it via L
2
and D
1
to a voltage slightly less than the input voltage magnitude.
During this moment, L
2
will act to oppose a sudden rush of current, so C
1
doesn’t
charge up instantly. This initial behavior isn’t examined here and the output voltage
is assumed to already be higher than the input voltage.
Of some interest in designing a Joule Thief is the value of L
2
, the operating
frequency, and the peak current that will be experienced by Q
1
. The input voltage,
THOUGHTS ON THE JOULE THIEF 3
Figure 2. Alternative Joule Thief
output voltage, and required output current compliance are required to be known
prior to the design process.
By the way, an alternative design for fig. 1 is found in fig. 2. The resistance used
for R
1
in fig. 1 is now divided in half between R
1
and R
2
in fig. 2. C
2
is added, as
well. The remainder of this discussion focuses on the earlier version in fig. 1 but
may also be applied to this alternative version of it.
Note that the discussion below may alternate between discussing energy and
power. Sometimes, it’s better to do the accounting in terms of energy and then
convert that to power for comparison (apples to apples, so to speak.) They are
related concepts but they are not the same thing. Just keep in mind that power is
energy per unit time and that the transfer of energy takes time.
2. Frequency and Time
In describing the above operation, two different phases were mentioned. The
first phase energizes L
2
with energy to be supplied later, by applying the battery
THOUGHTS ON THE JOULE THIEF 4
voltage across L
1
and Q
1
. The second phase delivers this stored energy, along with
additional energy from the battery source, to C
1
via D
1
.
Let t
on
and t
off
represent the time (in seconds) that Q
1
is on and off, respectively.
And let f represent the resulting number of such two-step cycles that operate per
second (expressed in Hz.)
(2.1) f =
1
t
on
+t
off
(2.2)
1
f
= t
on
+t
off
The total time required per cycle is the inverse of frequency of those cycles.
3. Preface on Efficiency
If the Joule Thief was 100% efficient, then the average value of its battery current
would be:
(3.1) I
in
= Iout
V
out
V
in
And since the L
2
causes the current to ramp up along a uniform slope to the
peak value and then back down along a different, but also uniform slope to 0, the
peak value itself would be just twice the average value. Or:
(3.2) I
peak
= 2 Iout
V
out
V
in
The fact is that a Joule Thief isn’t 100% efficient. So eq. 3.2 isn’t correct and
I
peak
will always be larger than indicated there.
4. Magnetic Flux in Inductors
Eq. 4.1 is the differential equation for an inductor.
(4.1) V = L
d I
d t
V is the voltage across the inductor, L is its inductance, and
dI
/dt is the change
in current versus the change in time. More rapidly changing currents imply larger
voltages.
For a Joule Thief, the value for V will be roughly fixed (but usually not the
same) during each phase. So V will take on two distinct values, V
on
and V
off
, for
the first and second phases.
If the inductor current rises from 0 to I
peak
in the first phase and returns to 0
in the second phase then the value for d I can be taken as exactly I
peak
for both
phases. Clearly, L is the same and can be taken as L
2
for both phases. Finally,
d t will be different for each phase but for design purposes can be taken as t
on
and
t
off
for the first and second phases, respectively.
Thus eq. 4.1 divides into two separate equations, one for each phase.
(4.2) V
on
= L
2
I
peak
t
on
(4.3) V
off
= L
2
I
peak
t
off
or,
(4.4) t
on
V
on
= L
2
I
peak
(4.5) t
off
V
off
= L
2
I
peak
THOUGHTS ON THE JOULE THIEF 5
Since L
2
and I
peak
are taken to be the same in both phases for reasons already
given, their product is also fixed and the same for both phases. So they may be
equated together:
(4.6) t
on
V
on
= t
off
V
off
I believe this is just another way of saying that the magnetic flux, measured in
volt-seconds or Webers, that is built up during the first phase must be dispensed
with in the second phase.
A simple rearrangement yields:
(4.7)
t
off
t
on
=
V
on
V
off
Eq. 4.7 will become useful, shortly.
5. Duty Cycle
The portion of each of the above described cycles or periods that is used in the
first phase to build energy in L
2
before turning Q
1
off is called the “duty cycle.”
A larger proportion of each cycle is required in this first phase for higher output
voltages than for lower ones. But it needs to be a value between 0 and 1 (0% to
100%.) (Actually, it has to be less than 1 because there must be time left over for
dispensing the stored energy, as well.)
The following two equations express this concept:
(5.1) D
on
=
t
on
t
on
+t
off
= t
on
f (5.2) D
off
=
t
off
t
on
+t
off
= t
off
f
The following relationships are also true:
(5.3) 1 = D
on
+D
off
(5.4) D
on
= 1 −D
off
(5.5) D
off
= 1 −D
on
Duty cycle shown above is in terms of t
on
and t
off
. But t
on
and t
off
aren’t
usually specified in the early design stages. What is probably better known earlier
is V
on
and V
off
. And eq. 4.7 suggests a way to specify the duty cycle.
Examine how the relationship in eq. 4.7 is applied to eq. 5.1 and eq. 5.2:
(5.6) D
on
=
t
on
t
on
+t
off
=
1
1 +
t
off
ton
=
1
1 +
Von
V
off
=
V
off
V
on
+V
off
and,
(5.7) D
off
=
t
off
t
on
+t
off
=
1
1 +
ton
t
off
=
1
1 +
V
off
Von
=
V
on
V
on
+V
off
or, in short,
(5.8) D
on
=
V
off
V
on
+V
off
= t
on
f (5.9) D
off
=
V
on
V
on
+V
off
= t
off
f
This is nice because V
on
and V
off
are usually easier to specify early in the design.
It’s time to examine them.
THOUGHTS ON THE JOULE THIEF 6
6. V
on
and V
off
When Q
1
is turned on, there is a slight voltage drop across V
CE
. This drop
decreases the applied voltage across L
2
during the first phase. (It also dissipates
power.) Since V
CE
goes from near 0 at the start of the t
on
period to its maximum
value of V
CEsat
at the end when I
C
= I
peak
, the effective value to use will be
somewhere between these two values. Because V
CE
tends to shift linearly with I
C
given a fixed I
B
, a good guess to use for duty cycle calculations is
V
CEsat
/2, which
should be subtracted from V
in
. (Keep in mind this only modifies the V
on
concept
as it applies to the t
on
period and not as it applies to the t
off
period.)
So a reasonable estimate of V
on
for duty cycle estimation is:
(6.1) V
on
= V
in
−
V
CEsat
2
When Q
1
is off, L
2
must develop sufficient voltage to overcome the diode drop
of D
1
and therefore must include a still higher effective output voltage. This drop
increases the voltage across L
2
during the second phase. (And again also dissipates
power.) Since V
D1
goes from its maximum value at the start of the t
off
period to
near 0 at the end when the magnetic field in L
2
is extinguished, the effective value
to use will be somewhere between these two values. Without better knowledge, a
good guess to use may be
V
maxD
1/2. But, in fact, that’s not the best value to use
because the voltage varies as the log of the current through it.
V
D1
=
1
t
off
_
t
off
0
V
D1
(t) dt
=
1
t
off
_
t
off
0
nkT
q
ln
I
t
I
s
dt
=
nkT
qt
off
_
t
off
0
ln
I
peak
(1 −
t
t
off
)
I
s
dt
=
nkT
qt
off
_
t
off
0
_
ln
I
peak
I
s
+ ln
_
1 −
t
t
off
__
dt
=
nkT
q
ln
I
peak
I
s
−
nkT
q
= V
maxD1
−nV
T
(6.2)
In short, there will be a small, constant subtraction from the peak voltage drop;
on the order of 30mV to 50mV for Schottky diodes. Given maximum voltage drops
on the order of 300mV for I
peak
= 150mA, it’s a modest adjustment. V
D1
increases
V
off
. (Keep in mind this only modifies the V
off
concept as it applies to the t
off
period and not as it applies to the t
on
period.)
A last point to keep in mind is that during the t
off
period, L
2
’s voltage is aided
by the battery voltage. So the value of V
off
is reduced by the battery voltage
(which increases t
off
.) The final equation looks like:
(6.3) V
off
= V
out
+V
maxD1
−nV
T
−V
in
Eq. 6.1 and eq. 6.3 may be applied to eq. 5.8 and eq. 5.9 in estimating the relative
portion of each phase of each cycle.
THOUGHTS ON THE JOULE THIEF 7
7. Using a Power Equation to Find I
peak
There are two ways of analyzing the larger picture. Accounting for energy sup-
plied and used in each two-phase cycle or else accounting for the power supplied
and used. These are nearly equivalent methods as the energy per cycle times the
frequency is power. Accounting for power will be used here. The sum of all the
power sources and dissipation sinks must be zero.
Inductors, such as L
2
, store energy in magnetic fields. The amount of work or
energy stored is based upon the current and the inductor value:
(7.1) W
L
=
1
2
I
2
peak
L
2
To convert this to power it must be multiplied by the frequency:
(7.2) P
L
=
1
2
I
2
peak
L
2
f
The battery source is supplying power during both phases. But eq. 7.2 already
accounts for the power supplied by the battery during the t
on
period. So only
the additional power supplied by the battery during the t
off
period needs to be
accounted for. Since the current declines linearly from I
peak
to 0 during this phase,
while the battery voltage remains constant, the power supplied by the battery
during this phase is:
(7.3) P
V
=
1
2
V
in
I
peak
t
off
f
The load power is dissipated all the time and is:
(7.4) P
load
= −V
out
I
out
The Schottky diode D
1
not only drops a varying voltage, but also dissipates
power during the t
off
period. The power dissipated is:
P
D1
= −
_
1
t
off
_
t
off
0
V
D1
(t)I
D1
dt
_
t
off
f
= −f
_
t
off
0
nkT
q
ln
_
I
t
I
s
_
I
t
dt
= −f
nkT
q
_
t
off
0
ln
_
I
peak
(1 −
t
t
off
)
I
s
_
I
peak
_
1 −
t
t
off
_
dt
= −f
nkT
q
I
peak
_
t
off
0
_
ln
I
peak
I
s
+ ln
_
1 −
t
t
off
__ _
1 −
t
t
off
_
dt
= −
1
2
I
peak
_
nkT
q
ln
I
peak
I
s
−
nkT
2q
_
t
off
f
= −
1
2
I
peak
_
V
maxD1
−
1
2
nV
T
_
t
off
f (7.5)
There is a very slight adjustment to V
maxD1
before applying it as one side of the
triangular area for the power computation, in short.
THOUGHTS ON THE JOULE THIEF 8
Q
1
’s V
CE
dissipation occurs duing the t
on
period and is:
P
Q1
= −
_
1
t
on
_
ton
0
V
CE
(t)I
C
(t) dt
_
t
on
f
= −f
_
ton
0
_
V
CEsat
t
t
on
__
I
peak
t
t
on
_
dt
= −fV
CEsat
I
peak
1
t
2
on
_
ton
0
t
2
dt
= −
1
3
V
CEsat
I
peak
t
on
f (7.6)
Q
1
base circuit losses during the t
on
period, including those dissipated by R
1
, will
depend on the assumed β value when calculating R
1
. But for reasonable choices,
dissipation will be less than 10% of Q
1
’s V
CE
dissipation. It’s easier to simply
multiply P
Q1
by such a factor than to perform a detailed calculation.
There are additional sources of loss. Resistance in L
1
will be insignificant com-
pared to R
1
and can probably be ignored. Resistance in L
2
may be important at
higher values of I
peak
. There are also losses during the short time that D
1
turns
off. These will be ignored here.
The following sums up the power sources and dissipation sinks and applies the
aforementioned additional 10% base circuit dissipation factor of 1.1 to P
Q1
. Eq. 5.8
and eq. 5.9 are applied below. Also, eq. 4.5 is applied to the first term to aid in the
solution.
0 = P
L
+P
V
+P
load
+P
D1
+ 1.1P
Q1
0 =
1
2
I
2
peak
L
2
f +
1
2
V
in
I
peak
t
off
f −V
out
I
out
−
1
2
I
peak
_
V
maxD1
−
1
2
nV
T
_
t
off
f −
1.1
3
V
CEsat
I
peak
t
on
f
2V
out
I
out
= I
2
peak
L
2
f +V
in
I
peak
t
off
f
−I
peak
_
V
maxD1
−
1
2
nV
T
_
t
off
f −
2.2
3
V
CEsat
I
peak
t
on
f
2V
out
I
out
= I
peak
t
off
V
off
f +V
in
I
peak
t
off
f
−I
peak
_
V
maxD1
−
1
2
nV
T
_
t
off
f −
2.2
3
V
CEsat
I
peak
t
on
f
2V
out
I
out
= I
peak
_
V
off
D
off
+V
in
D
off
−
_
V
maxD1
−
1
2
nV
T
_
D
off
−
2.2
3
V
CEsat
D
on
_
2V
out
I
out
= I
peak
__
V
off
+V
in
−V
maxD1
+
1
2
nV
T
_
D
off
−
2.2
3
V
CEsat
D
on
_
or,
(7.7) I
peak
=
2V
out
I
out
_
V
out
−
1
2
nV
T
_
D
off
−
2.2
3
V
CEsat
D
on
If
1
2
nV
T
is small compared to V
out
(almost certainly true) and if
2.2
3
V
CEsat
D
on
is similarly small compared to V
out
D
off
(probably also true), then the following
THOUGHTS ON THE JOULE THIEF 9
is good enough in most circumstances:
(7.8) I
peak
=
2I
out
D
off
=
2I
out
1 −D
on
Eq. 7.8 is often found in textbooks.
8. Efficiency Revisited
Now compare equations eq. 7.7 and eq. 7.8 with eq. 3.2. The efficiency can be
taken as the ratio and as derived from eq. 7.7 is:
(8.1) η =
_
V
out
−
1
2
nV
T
_
D
off
−
2.2
3
V
CEsat
D
on
V
in
Derived from eq. 7.8 (or by applying the same simplifications mentioned near
the end of the last section to eq. 8.1) is:
(8.2) η =
V
out
V
in
D
off
=
V
out
V
in
V
on
V
on
+V
off
9. Operating Frequency and Inductance
The total time per cycle should be,
(9.1) t
cyc
≥ t
on
+t
off
The frequency then clearly must be,
(9.2) f ≤
1
t
on
+t
off
Given I
peak
, though, eq. 4.4 and eq. 4.5 can be applied:
f ≤
1
L I
peak
Von
+
L I
peak
V
off
(9.3)
extracting out a common factor,
f ≤
1
L I
peak
1
1
Von
+
1
V
off
(9.4)
re-arranging the last factor,
f ≤
1
L I
peak
V
on
V
off
V
on
+V
off
(9.5)
The last factor in eq. 9.5 (or eq. 9.4) above is interesting, as it appears that V
on
and V
off
are treated like combining parallel resistors. In parallel resistance, each
conducts current separately that is proportional to the inverse of their value and
these currents add. In this case, each voltage separately determines time that is
proportional to the inverse of their value and these times add. Hence, the similarity.
Separating unknown terms from known terms in eq. 9.5 provides,
(9.6) f L ≤
1
I
peak
V
on
V
off
V
on
+V
off
The units of fL are Ωs. It’s not hard to see from eq. 9.6 that reducing the value
of L
2
will increase the operating frequency f.
THOUGHTS ON THE JOULE THIEF 10
In terms of D, applying either eq. 5.8 or eq. 5.9 creates two new forms from
eq. 9.6:
f L ≤
V
on
D
on
I
peak
f L ≤
V
off
D
off
I
peak
(9.7)
Which is used will probably depend on what’s handy.
10. Inductor Core Materials
As the above discussion shows, a Joule Thief does not require core saturation for
its operation. Core saturation would cause a sudden rise in Q
1
’s collector current
and if the core saturation is more controllable than Q
1
’s β, this would create a
more predictable frequency of operation. But for a hobbyist Joule Thief, an exact,
predictable frequency isn’t essential.
I like to imagine magnetic core materials as supplying magnetic spin domains
which act as magnetic short-circuits. They bridge over some physical space, in
effect. For example, if a core material has µ
e
= 125 then I imagine that if there
is a measurable magnetic path length of 12.5mm there is actually only a net of
0.1mm of vacuum gap within that physical length where energy can be stored.
In short, 12.4mm of the 12.5mm is short circuited by magnetic domains (where
energy is not and cannot be stored) leaving only 0.1mm of effective vacuum for
storing energy. Magnetic energy, as I imagine it, is stored (or transferred across
vast reaches via EM radiation) only in vacuum. (Though energy can certainly be
dissipated [distributed] into vibrational energy in particles of mass.)
It’s usually the case with a Joule Thief that a non-air-core toroid is used. It’s
convenient and the coupling between L
1
and L
2
is much better. There will be a
maximum flux density that can be supported – called B
max
– when using a toroid.
Ferrite cores will typically have a B
max
between 0.1T and 0.3T, while iron cores
range from perhaps 1T to 2T. The B
max
value in such inductor cores will limit the
choices otherwise allowed by eq. 9.6.
(It’s possible to use a nail or other short length of magnetic material to wrap
the transformer. This will improve the coupling of L
1
and L
2
but the very long
distance in air between the two ends will mean that the inductance is effectively
similar to an air core. I believe there is still the possibility of core saturation if
the flux density in volt-seconds per m
2
of cross section exceeds the core material’s
capacity. But the impact of that magnetic saturation will be less noticed because
the magnetic path length through air already dominates and saturating the short
length only extends this path length a relatively smaller amount than it would in
a toroid. It would be a soft change, in other words.)
When using something other than air for the core, the number of volt-seconds is
limited by the following relationship:
(10.1) V t ≤ B
max
A
c
N
In eq. 10.1, V t represents the number of volt-seconds. A
c
is the inductor core’s
magnetic window area (a cross-section) and N is the number of wire turns used to
make the inductor.
The volt-seconds are already known, though. As pointed out already in eq. 4.4
and eq. 4.5, the value may be computed by either examining t
on
and V
on
or else
examining t
off
and V
off
and, as eq. 4.6 shows, both products are equal to each
THOUGHTS ON THE JOULE THIEF 11
other. Either way, the right side of eq. 4.4 or eq. 4.5 can be inserted into eq. 10.1,
(10.2) I
peak
L ≤ B
max
A
c
N
Solving eq. 9.6 for L and substituting into eq. 10.2 and solving for f,
(10.3) f ≥
V
on
D
on
B
max
A
c
N
(10.4) f ≥
V
off
D
off
B
max
A
c
N
Eq. 10.3 and eq. 10.4 express limitations imposed when considering non-air-core
inductors and requires an operating frequency to be a certain value that is controlled
in part by the inverse of B
max
. For an inductor core with a given magnetic cross-
section, A
c
, and a specified number of turns, N, smaller B
max
values require higher
frequencies of operation in order to avoid saturation. (Eddy current losses become
increasingly important at higher frequencies.)
The equation for inductance is:
(10.5) L =
µ
0
µ
e
A
c
N
2
l
e
In Eq. 10.5, l
e
is the effective magnetic path length. µ
0
is the absolute perme-
ability of vacuum and in the SI system is the constant, π 4 ×10
−7
Hm
−1
. µ
e
is the
effective, relative permeability of the core material and is always multiplied by µ
0
in order to generate an absolute value.
Eq. 10.5 can be combined with eq. 10.2 to provide another useful relation:
(10.6)
l
e
N
≥
µ
0
µ
e
B
max
I
peak
Eq. 10.6 is set by a physical constant, the peak current, and the selected core
material; without any regard to shape or inductance, etc. That sets the required
minimum ratio between the magnetic path length and the number of windings used.
Of course, if core saturation is a desired result then by all means feel free to
violate this section’s equations in some designed way.
11. Final Caveats from the Author
It helps me to think about things, expressing what I’m thinking in quantitative
ways and to then compare the results of that with what LTspice, for example, tells
me about them. In doing so, I believe I came to better understand the way a Joule
Thief may and may not work.
Please keep in mind the fact that I’ve no training whatsoever in electronics. (I
really do mean an actual zero.) I am merely a part-time hobbyist. If the above
seems na¨ıve to you, that’s probably why.
If you have any helpful corrections or additions to make and feel able to offer
the time to help me gather them, I’d appreciate it very much.
A big weakness in mental concepts for me continues to be the application of
differential vector calculus to the magnetic and electric fields. They are some of
the simplest such applications, yet I still haven’t used them enough to really feel
comfortable there.
I am just beginning now to get an intuition that the A-field (not the B-field, but
the implications that an A-field must exist from the B-field) is important to under-
stand. And I hope to work harder at some point and see if that leads anywhere.
E-mail address: [email protected]