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 understoodquantitatively by more prosaic considerations.

1. Overview

The name “Joule Thief” is given to a particular kind of self-starting boost modeswitching power supply often used to operate white LEDs from a 1.5V batterysupply. It can even be used to operate a series chain of LEDs. It will continue tooperate 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 D1 and C1 shown in fig. 1 withan LED. The difference is that fig. 1 allows a simpler specification of the load foranalysis. The output voltage must be higher and more positive than the inputsupply voltage. Take special note of the winding arrangement of the transformer(shown as L1 and L2.) The usual method is to wind this transformer by handso that the inductance of L1 and L2 are taken as equal. It’s not necessary thatthey be, though. D1 may be a Schottky diode, as they tend to have lower forwardvoltage drops and yield somewhat more efficient designs.

A base current via R1 turns on Q1. Q1’s collector then approaches ground,applying the battery voltage across the transformer primary, L2. This induces asimilar and aiding voltage (if L1 is equal to L2 and arranged as indicated) at L1,adding to the EMF that drives current into Q1’s base. Roughly speaking, thisdoubles the battery EMF and is why the Joule Thief works so well.

The rate of increase in the current in L2 is determined by VL2/L2 and rises alonga ramp towards Ipeak and is also Q1’s collector current. The magnitude of thiscurrent represents the energy stored in L2’s magnetic field (according to eq. 7.1,noted below.) So as the current rises linearly, the stored energy rises proportionalto its squared value. During this time, no energy is being transferred from L2 tothe output capacitor, C1.

The base current of Q1 remains roughly constant during this rise. At somepoint determined by Q1’s beta, the magnitude of Q1’s collector current exceeds theability of the base current to sustain it. When that happens the collector currentcannot continue to rise. However, if the applied voltage across L2 is maintained thecollector current must rise. So the voltage across L2 declines rapidly towards zero.But as the voltage across L2 declines towards zero, so also does the aiding voltageacross L1. This reduction substantially reduces the base current that originallysupported the collector current inQ1, requiring a reduction in that collector current.To reduce the collector current in Q1, the rate of change of current determined byVL2/L2 must change sign and so VL2

must itself change sign. This change of signconstructs an opposing voltage in L1, further reducing the base current. Eventually,this process turns Q1 completely off.

(In fact, there is a danger of exceeding the modest VEBO reverse voltage, causingit to zener. Especially in the case where the output voltage is substantially higherthan the battery voltage.)

During this voltage reversal phase, Q1 remains off while the energy stored in L2

is expended through D1 and into charging C1. Eventually, the current falls towardszero and reaches it and the magnetic field energy is exhausted. At this point, L1’sopposing 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 uponstaring, C1 will likely be uncharged and the input voltage supply will attempt tocharge it via L2 and D1 to a voltage slightly less than the input voltage magnitude.During this moment, L2 will act to oppose a sudden rush of current, so C1 doesn’tcharge up instantly. This initial behavior isn’t examined here and the output voltageis assumed to already be higher than the input voltage.

Of some interest in designing a Joule Thief is the value of L2, the operatingfrequency, and the peak current that will be experienced by Q1. The input voltage,


Figure 2. Alternative Joule Thief

output voltage, and required output current compliance are required to be knownprior to the design process.

By the way, an alternative design for fig. 1 is found in fig. 2. The resistance usedfor R1 in fig. 1 is now divided in half between R1 and R2 in fig. 2. C2 is added, aswell. The remainder of this discussion focuses on the earlier version in fig. 1 butmay also be applied to this alternative version of it.

Note that the discussion below may alternate between discussing energy andpower. Sometimes, it’s better to do the accounting in terms of energy and thenconvert that to power for comparison (apples to apples, so to speak.) They arerelated concepts but they are not the same thing. Just keep in mind that power isenergy 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. Thefirst phase energizes L2 with energy to be supplied later, by applying the battery


voltage across L1 and Q1. The second phase delivers this stored energy, along withadditional energy from the battery source, to C1 via D1.

Let ton and toff represent the time (in seconds) thatQ1 is on and off, respectively.And let f represent the resulting number of such two-step cycles that operate persecond (expressed in Hz.)

(2.1) f =1

ton + toff(2.2)

1

f= ton + toff

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 currentwould be:

(3.1) Iin = IoutVoutVin

And since the L2 causes the current to ramp up along a uniform slope to thepeak value and then back down along a different, but also uniform slope to 0, thepeak value itself would be just twice the average value. Or:

(3.2) Ipeak = 2 IoutVoutVin

The fact is that a Joule Thief isn’t 100% efficient. So eq. 3.2 isn’t correct andIpeak 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 = Ld I

d t

V is the voltage across the inductor, L is its inductance, and dI/dt is the changein current versus the change in time. More rapidly changing currents imply largervoltages.

For a Joule Thief, the value for V will be roughly fixed (but usually not thesame) during each phase. So V will take on two distinct values, Von and Voff , forthe first and second phases.

If the inductor current rises from 0 to Ipeak in the first phase and returns to 0in the second phase then the value for d I can be taken as exactly Ipeak for bothphases. Clearly, L is the same and can be taken as L2 for both phases. Finally,d t will be different for each phase but for design purposes can be taken as ton andtoff for the first and second phases, respectively.

Thus eq. 4.1 divides into two separate equations, one for each phase.

(4.2) Von = L2Ipeakton

(4.3) Voff = L2Ipeaktoff

or,

(4.4) ton Von = L2 Ipeak (4.5) toff Voff = L2 Ipeak


Since L2 and Ipeak are taken to be the same in both phases for reasons alreadygiven, their product is also fixed and the same for both phases. So they may beequated together:

(4.6) ton Von = toff Voff

I believe this is just another way of saying that the magnetic flux, measured involt-seconds or Webers, that is built up during the first phase must be dispensedwith in the second phase.

A simple rearrangement yields:

(4.7)toffton

=VonVoff

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 thefirst phase to build energy in L2 before turning Q1 off is called the “duty cycle.”A larger proportion of each cycle is required in this first phase for higher outputvoltages than for lower ones. But it needs to be a value between 0 and 1 (0% to100%.) (Actually, it has to be less than 1 because there must be time left over fordispensing the stored energy, as well.)

The following two equations express this concept:

(5.1) Don =ton

ton + toff= ton f (5.2) Doff =

toffton + toff

= toff f

The following relationships are also true:

(5.3) 1 = Don +Doff (5.4) Don = 1 −Doff (5.5) Doff = 1 −Don

Duty cycle shown above is in terms of ton and toff . But ton and toff aren’tusually specified in the early design stages. What is probably better known earlieris Von and Voff . 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) Don =ton

ton + toff=

1

1 +toff

ton

=1

1 + Von

Voff

=Voff

Von + Voff

and,

(5.7) Doff =toff

ton + toff=

1

1 + tontoff

=1

1 +Voff

Von

=Von

Von + Voff

or, in short,

(5.8) Don =Voff

Von + Voff= ton f (5.9) Doff =

VonVon + Voff

= toff f

This is nice because Von and Voff are usually easier to specify early in the design.It’s time to examine them.


6. Von and Voff

When Q1 is turned on, there is a slight voltage drop across VCE . This dropdecreases the applied voltage across L2 during the first phase. (It also dissipatespower.) Since VCE goes from near 0 at the start of the ton period to its maximumvalue of VCEsat at the end when IC = Ipeak, the effective value to use will besomewhere between these two values. Because VCE tends to shift linearly with ICgiven a fixed IB , a good guess to use for duty cycle calculations is VCEsat/2, whichshould be subtracted from Vin. (Keep in mind this only modifies the Von conceptas it applies to the ton period and not as it applies to the toff period.)

So a reasonable estimate of Von for duty cycle estimation is:

(6.1) Von = Vin − VCEsat

2

When Q1 is off, L2 must develop sufficient voltage to overcome the diode dropof D1 and therefore must include a still higher effective output voltage. This dropincreases the voltage across L2 during the second phase. (And again also dissipatespower.) Since VD1

goes from its maximum value at the start of the toff period tonear 0 at the end when the magnetic field in L2 is extinguished, the effective valueto use will be somewhere between these two values. Without better knowledge, agood guess to use may be VmaxD1/2. But, in fact, that’s not the best value to usebecause the voltage varies as the log of the current through it.

VD1 =1

toff

∫ toff

0

VD1(t) dt

=1

toff

∫ toff

0

nkT

qlnItIs

dt

=nkT

qtoff

∫ toff

0

lnIpeak(1 − t

toff)

Isdt

=nkT

qtoff

∫ toff

0

[lnIpeakIs

+ ln

(1 − t

toff

)]dt

=nkT

qlnIpeakIs

− nkT

q

= VmaxD1− nVT(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 dropson the order of 300mV for Ipeak = 150mA, it’s a modest adjustment. VD1

increasesVoff . (Keep in mind this only modifies the Voff concept as it applies to the toffperiod and not as it applies to the ton period.)

A last point to keep in mind is that during the toff period, L2’s voltage is aidedby the battery voltage. So the value of Voff is reduced by the battery voltage(which increases toff .) The final equation looks like:

(6.3) Voff = Vout + VmaxD1− nVT − Vin

Eq. 6.1 and eq. 6.3 may be applied to eq. 5.8 and eq. 5.9 in estimating the relativeportion of each phase of each cycle.


7. Using a Power Equation to Find Ipeak

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 suppliedand used. These are nearly equivalent methods as the energy per cycle times thefrequency is power. Accounting for power will be used here. The sum of all thepower sources and dissipation sinks must be zero.

Inductors, such as L2, store energy in magnetic fields. The amount of work orenergy stored is based upon the current and the inductor value:

(7.1) WL =1

2I2peak L2

To convert this to power it must be multiplied by the frequency:

(7.2) PL =1

2I2peak L2 f

The battery source is supplying power during both phases. But eq. 7.2 alreadyaccounts for the power supplied by the battery during the ton period. So onlythe additional power supplied by the battery during the toff period needs to beaccounted for. Since the current declines linearly from Ipeak to 0 during this phase,while the battery voltage remains constant, the power supplied by the batteryduring this phase is:

(7.3) PV =1

2Vin Ipeak toff f

The load power is dissipated all the time and is:

(7.4) Pload = −VoutIout

The Schottky diode D1 not only drops a varying voltage, but also dissipatespower during the toff period. The power dissipated is:

PD1= −

[1

toff

∫ toff

0

VD1(t)ID1

dt

]tofff

= −f∫ toff

0

nkT

qln

(ItIs

)It dt

= −f nkTq

∫ toff

0

ln

(Ipeak(1 − t

toff)

Is

)Ipeak

(1 − t

toff

)dt

= −f nkTqIpeak

∫ toff

0

[lnIpeakIs

+ ln

(1 − t

toff

)](1 − t

toff

)dt

= −1

2Ipeak

(nkT

qlnIpeakIs

− nkT

2q

)tofff

= −1

2Ipeak

(VmaxD1

− 1

2nVT

)tofff(7.5)

There is a very slight adjustment to VmaxD1before applying it as one side of the

triangular area for the power computation, in short.


Q1’s VCE dissipation occurs duing the ton period and is:

PQ1= −

[1

ton

∫ ton

0

VCE(t)IC(t) dt

]tonf

= −f∫ ton

0

(VCEsat

t

ton

)(Ipeak

t

ton

)dt

= −fVCEsatIpeak1

t2on

∫ ton

0

t2 dt

= −1

3VCEsatIpeaktonf(7.6)

Q1 base circuit losses during the ton period, including those dissipated by R1, willdepend on the assumed β value when calculating R1. But for reasonable choices,dissipation will be less than 10% of Q1’s VCE dissipation. It’s easier to simplymultiply PQ1

by such a factor than to perform a detailed calculation.There are additional sources of loss. Resistance in L1 will be insignificant com-

pared to R1 and can probably be ignored. Resistance in L2 may be important athigher values of Ipeak. There are also losses during the short time that D1 turnsoff. These will be ignored here.

The following sums up the power sources and dissipation sinks and applies theaforementioned additional 10% base circuit dissipation factor of 1.1 to PQ1

. Eq. 5.8and eq. 5.9 are applied below. Also, eq. 4.5 is applied to the first term to aid in thesolution.

0 = PL + PV + Pload + PD1+ 1.1PQ1

0 =1

2I2peak L2 f +

1

2Vin Ipeak toff f − VoutIout

− 1

2Ipeak

(VmaxD1

− 1

2nVT

)tofff − 1.1

3VCEsatIpeaktonf

2VoutIout = I2peak L2 f + Vin Ipeak toff f

− Ipeak

(VmaxD1

− 1

2nVT

)tofff − 2.2

3VCEsatIpeaktonf

2VoutIout = Ipeak toff Voff f + Vin Ipeak toff f

− Ipeak

(VmaxD1

− 1

2nVT

)tofff − 2.2

3VCEsatIpeaktonf

2VoutIout = Ipeak

[Voff Doff + Vin Doff

−(VmaxD1

− 1

2nVT

)Doff − 2.2

3VCEsatDon

]2VoutIout = Ipeak

[(Voff + Vin − VmaxD1

+1

2nVT

)Doff − 2.2

3VCEsatDon

]or,

(7.7) Ipeak =2VoutIout(

Vout − 12nVT

)Doff − 2.2

3 VCEsatDon

If 12nVT is small compared to Vout (almost certainly true) and if 2.2

3 VCEsatDon

is similarly small compared to Vout Doff (probably also true), then the following


is good enough in most circumstances:

(7.8) Ipeak =2IoutDoff

=2Iout

1 −Don

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 betaken as the ratio and as derived from eq. 7.7 is:

(8.1) η =

(Vout − 1

2nVT)Doff − 2.2

3 VCEsatDon

Vin

Derived from eq. 7.8 (or by applying the same simplifications mentioned nearthe end of the last section to eq. 8.1) is:

(8.2) η =VoutVin

Doff =VoutVin

VonVon + Voff

9. Operating Frequency and Inductance

The total time per cycle should be,

(9.1) tcyc ≥ ton + toff

The frequency then clearly must be,

(9.2) f ≤ 1

ton + toff

Given Ipeak, though, eq. 4.4 and eq. 4.5 can be applied:

f ≤ 1L Ipeak

Von+

L Ipeak

Voff

(9.3)

extracting out a common factor,

f ≤ 1

L Ipeak

11

Von+ 1

Voff

(9.4)

re-arranging the last factor,

f ≤ 1

L Ipeak

Von VoffVon + Voff

(9.5)

The last factor in eq. 9.5 (or eq. 9.4) above is interesting, as it appears that Vonand Voff are treated like combining parallel resistors. In parallel resistance, eachconducts current separately that is proportional to the inverse of their value andthese currents add. In this case, each voltage separately determines time that isproportional 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

Ipeak

Von VoffVon + Voff

The units of fL are Ωs. It’s not hard to see from eq. 9.6 that reducing the valueof L2 will increase the operating frequency f .


In terms of D, applying either eq. 5.8 or eq. 5.9 creates two new forms fromeq. 9.6:

f L ≤ Von Don

Ipeakf L ≤ Voff Doff

Ipeak(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 forits operation. Core saturation would cause a sudden rise in Q1’s collector currentand if the core saturation is more controllable than Q1’s β, this would create amore 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 domainswhich act as magnetic short-circuits. They bridge over some physical space, ineffect. For example, if a core material has µe = 125 then I imagine that if thereis a measurable magnetic path length of 12.5mm there is actually only a net of0.1mm 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 (whereenergy is not and cannot be stored) leaving only 0.1mm of effective vacuum forstoring energy. Magnetic energy, as I imagine it, is stored (or transferred acrossvast reaches via EM radiation) only in vacuum. (Though energy can certainly bedissipated [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’sconvenient and the coupling between L1 and L2 is much better. There will be amaximum flux density that can be supported – called Bmax – when using a toroid.Ferrite cores will typically have a Bmax between 0.1T and 0.3T , while iron coresrange from perhaps 1T to 2T . The Bmax value in such inductor cores will limit thechoices otherwise allowed by eq. 9.6.

(It’s possible to use a nail or other short length of magnetic material to wrapthe transformer. This will improve the coupling of L1 and L2 but the very longdistance in air between the two ends will mean that the inductance is effectivelysimilar to an air core. I believe there is still the possibility of core saturation ifthe flux density in volt-seconds per m2 of cross section exceeds the core material’scapacity. But the impact of that magnetic saturation will be less noticed becausethe magnetic path length through air already dominates and saturating the shortlength only extends this path length a relatively smaller amount than it would ina toroid. It would be a soft change, in other words.)

When using something other than air for the core, the number of volt-seconds islimited by the following relationship:

(10.1) V t ≤ Bmax Ac N

In eq. 10.1, V t represents the number of volt-seconds. Ac is the inductor core’smagnetic window area (a cross-section) and N is the number of wire turns used tomake the inductor.

The volt-seconds are already known, though. As pointed out already in eq. 4.4and eq. 4.5, the value may be computed by either examining ton and Von or elseexamining toff and Voff and, as eq. 4.6 shows, both products are equal to each


other. Either way, the right side of eq. 4.4 or eq. 4.5 can be inserted into eq. 10.1,

(10.2) Ipeak L ≤ Bmax Ac N

Solving eq. 9.6 for L and substituting into eq. 10.2 and solving for f,

(10.3) f ≥ Von Don

Bmax Ac N(10.4) f ≥ Voff Doff

Bmax Ac N

Eq. 10.3 and eq. 10.4 express limitations imposed when considering non-air-coreinductors and requires an operating frequency to be a certain value that is controlledin part by the inverse of Bmax. For an inductor core with a given magnetic cross-section, Ac, and a specified number of turns, N , smaller Bmax values require higherfrequencies of operation in order to avoid saturation. (Eddy current losses becomeincreasingly important at higher frequencies.)

The equation for inductance is:

(10.5) L =µ0 µe Ac N

2

leIn Eq. 10.5, le is the effective magnetic path length. µ0 is the absolute perme-

ability of vacuum and in the SI system is the constant, π 4 × 10−7 H m−1. µe is theeffective, relative permeability of the core material and is always multiplied by µ0

in order to generate an absolute value.Eq. 10.5 can be combined with eq. 10.2 to provide another useful relation:

(10.6)leN

≥ µ0 µe

BmaxIpeak

Eq. 10.6 is set by a physical constant, the peak current, and the selected corematerial; without any regard to shape or inductance, etc. That sets the requiredminimum 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 toviolate 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 quantitativeways and to then compare the results of that with what LTspice, for example, tellsme about them. In doing so, I believe I came to better understand the way a JouleThief may and may not work.

Please keep in mind the fact that I’ve no training whatsoever in electronics. (Ireally do mean an actual zero.) I am merely a part-time hobbyist. If the aboveseems naıve to you, that’s probably why.

If you have any helpful corrections or additions to make and feel able to offerthe 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 ofdifferential vector calculus to the magnetic and electric fields. They are some ofthe simplest such applications, yet I still haven’t used them enough to really feelcomfortable there.

I am just beginning now to get an intuition that the A-field (not the B-field, butthe 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]

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