2
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Note: The question part is the fixing of the negative current at the transistor base. It looks like there's a problem with negative current at the LED(s) -- this is not expected to be answered in this question, but will be a separate question, if needed. Finally, I asked for comments, because I want to know what others think about this circuit, but again, it is not really part of this question. (If you feel like fixing the LED problem or commenting, please do, but that will have nothing to do with acceptance of a particular answer). I just said all of that to try to avoid a "too broad" disqualification.


I found this interesting alternative Joule Thief, which the creator named the "Super Efficient Joule Thief". I thought up a name for this. Would a good name for this be the Half-Hartley Joule Thief? A comment in the link calls it a modified Colpitts, but I think it's closer to a Hartley.

This particular Joule Thief alternative is interesting because it is a resonant type I haven't seen before. It is also interesting to me because the inductors are not coupled, making it easier to throw together with a few scrounged inductors, meaning that you don't have to wind your own traditional Joule Thief coupled inductor.

Here is the original schematic from the designer (he uses 4 of the same kind of 82uH axial inductor, probably not terribly efficient.):

re-spin from source, with original schematic and BOM parts list

In the pictorial schematic above, the diodes are backwards and make a short-circuit, as astutely observed by one commenter.

Though I just thought of something... Maybe they are implementing a buck converter arrangement, the flywheel action.

The schematic circuit diagram above has the diodes in seemingly correctly, but in the video, that schematic seems wrong, so be wary.

In my first schematic, I removed the diodes because they didn't seem to do anything:

First Schematic

Then I modified the circuit a bit, mainly combining the three serial inductors into one, but also selecting real inductors.

Almost all of the Joule Thief alternatives have reverse base current, so is there a way we can solve this? Or how I can know how much is too much?

The schematic of my version of the "Super Efficient Joule Thief" (attribution included):

schematic

Here is a picture of the resonant behavior (Isn't it beautiful! Except for the hiccough, of course!):

Resonant Behavior

Some traces showing reverse base current, and also voltage traces on the base and on the collector of N1 (per request in comments):

voltage and current traces from 0 to 50us

Here are some calculated values taken from 0ms to 8ms:

Ib(N1) :

  • 48.496µA AVG
  • 107.24µA RMS

V(N1b)*Ib(N1) :

  • 1.0388mW AVG
  • 8.3102µJ integral

0.5*(ABS(V(N1b)*Ib(N1))-V(N1b)*Ib(N1))

  • 23.751µW AVG
  • 190.01nJ integral

Actually, this reverse base current does not seem so bad, but I often have problems with reverse current at the base. I've put my time in. Would you please tell me how to solve this?

Note: I have not yet built this circuit because of the multiple problems -- the base problem, as well as the LED reverse current problem.

Here is the LTspice XVII source code for your convenience:

Version 4
SHEET 1 1420 864
WIRE 64 -48 -48 -48
WIRE 96 -48 64 -48
WIRE 208 -48 176 -48
WIRE 288 -48 208 -48
WIRE 672 -48 288 -48
WIRE 704 -48 672 -48
WIRE 672 -16 672 -48
WIRE 288 0 288 -48
WIRE -48 16 -48 -48
WIRE -48 16 -80 16
WIRE -48 80 -48 16
WIRE 64 80 64 -48
WIRE 96 80 64 80
WIRE 208 80 208 -48
WIRE 208 80 160 80
WIRE 672 96 672 64
WIRE 672 96 464 96
WIRE 704 96 672 96
WIRE 672 160 672 96
WIRE 464 208 464 96
WIRE 112 256 80 256
WIRE 288 256 288 80
WIRE 288 256 192 256
WIRE 368 256 288 256
WIRE 400 256 368 256
WIRE -48 352 -48 160
WIRE 80 352 80 256
WIRE 80 352 -48 352
WIRE 464 352 464 304
WIRE 464 352 80 352
WIRE 672 352 672 224
WIRE 672 352 464 352
WIRE -48 368 -48 352
FLAG -48 368 0
FLAG -80 16 V1
FLAG 704 96 D1
FLAG 368 256 N1b
FLAG 704 -48 L2
SYMBOL voltage -48 64 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
WINDOW 3 18 92 Left 2
SYMATTR Value 1.5
SYMATTR InstName V1
SYMBOL res 272 -16 R0
SYMATTR InstName R1
SYMATTR Value 1K
SYMATTR SpiceLine tol=1 pwr=0.1
SYMBOL cap 96 96 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName C1
SYMATTR Value 100nF
SYMBOL ind 80 -32 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 5 56 VBottom 2
SYMATTR InstName L1
SYMATTR Value 100µH
SYMATTR SpiceLine Ipk=5 Rser=0.028 Rpar=0 Cpar=0 mfg="Coiltronics" pn="CTX100-5-52"
SYMBOL ind 656 -32 R0
SYMATTR InstName L2
SYMATTR Value 270µH
SYMATTR SpiceLine Ipk=5.2 Rser=0.0688 Rpar=25434 Cpar=0 mfg="Coilcraft" pn="PCV-2-274-05"
SYMBOL LED 688 160 M0
WINDOW 3 -61 76 Left 2
SYMATTR Value LXK2-PW14
SYMATTR InstName D1
SYMBOL npn 400 208 R0
WINDOW 0 62 21 Left 2
WINDOW 3 22 48 Left 2
SYMATTR InstName N1
SYMATTR Value 2SCR553P
SYMBOL res 208 240 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 22K
TEXT 120 336 Left 2 !.tran 0 16ms 8ms startup
TEXT -616 184 Left 2 ;orig source before mods:\nSuper Efficient Joule Thief DIY \nby GEORGE CHANIOTAKIS in leds\non "instructables" web site\nhttps://www.instructables.com/id\n/Super-Efficient-Joule-Thief-DIY/
TEXT -680 -24 Left 2 !.meas iD1 AVG I(D1)*1000\n.meas iV1 AVG -I(V1)*1000\n.meas Pin AVG -V(V1)*I(V1)*1000\n.meas Pout AVG V(D1)*I(D1)*1000\n.meas Eff param 100*Pout/Pin
TEXT -712 -24 Right 2 ;92.38 mA\n214.95 mA\n322.4  mW\n285.0  mW\n88.4062 %
TEXT 408 24 Left 6 ;88%
TEXT 592 280 Left 5 ;92 mA
TEXT -1184 -56 Left 2 ;Ib(N1)\n=========\n48.496µA AVG   \n107.24µA RMS\n \nV(N1b)*Ib(N1)\n===========\n1.0388mW AVG\n8.3102µJ integral\n \n0.5*(ABS(V(N1b)*I(R1))-V(N1b)*I(R1))\n=========================\n15.079µW AVG\n120.63nJ integral
RECTANGLE Normal -240 128 -880 -64 2
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13
  • 3
    \$\begingroup\$ The original circuit in the linked article schematic at least dissipates also the last joule from the battery. It's a short circuit. You have removed the short. \$\endgroup\$
    – user136077
    Commented Jun 14, 2022 at 7:04
  • 3
    \$\begingroup\$ What's the concern about reverse base current? Some (transiently) is helpful (or required, even) to turn off the transistor; this is normal operation. If it's about avalanche breakdown, can you show a circuit and waveform demonstrating it? \$\endgroup\$ Commented Jun 14, 2022 at 8:07
  • 3
    \$\begingroup\$ On a separate note, I ran through a bunch of JT circuits, years ago when it was first a bit of a fad; I determined that most are trying to be a blocking oscillator: en.wikipedia.org/wiki/Blocking_oscillator and therefore the always-on bias, and AC losses, due to the lone resistor driving the base, are the main killer. Using an RC network enforces BO behavior, sharpening the switching waveform and greatly improving efficiency. This is somewhat beside the matter about your resonant example, of course. \$\endgroup\$ Commented Jun 14, 2022 at 8:10
  • 2
    \$\begingroup\$ Microservices, @TimWilliams is speaking truth. An RC network improves efficiency of the commonly found joule thief circuit. A lot. It can also be designed to greatly reduce the likelihood of avalanching the base-emitter junction. \$\endgroup\$
    – jonk
    Commented Jun 14, 2022 at 9:32
  • 2
    \$\begingroup\$ The original circuit also shows the red and black wire colors from the battery backwards. I use a QX5252F 4-pins IC plus one little inductor from a cheap solar garden light to make a Joule Thief. \$\endgroup\$
    – Audioguru
    Commented Jun 14, 2022 at 15:39

3 Answers 3

4
+50
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The question part is the fixing of the negative current at the transistor base.

My simulation shows a "normal" negative voltage.

Has anybody tried out this circuit in a laboratory? What are the "true" waveforms?

Don't know if this can help, but I see no "abnormal" base negative current voltage.

Efficiency is 91 % with this transistor (interactively changed some components).

Tried others BJT (2N2222, but efficiency is "lower" ?).

enter image description here

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11
  • 1
    \$\begingroup\$ microcap v12 spectrum-soft.com/download/mc12cd.zip \$\endgroup\$
    – Antonio51
    Commented Jun 15, 2022 at 14:33
  • 1
    \$\begingroup\$ Your base voltage drops to ONLY -2.0V but -6V is the spec'd limit so what is the problem? \$\endgroup\$
    – Audioguru
    Commented Jun 15, 2022 at 14:59
  • 1
    \$\begingroup\$ I thought any amount of negative base current was a problem because it would damage the b-e junction and the beta of the transistor. That's not true at all! \$\endgroup\$ Commented Jun 16, 2022 at 0:35
  • 1
    \$\begingroup\$ @MicroservicesOnDDD How can current flow backwards unless the -5v spec is exceeded. I don't understand. The current clearly is related to the slope of the voltage so it's a current that flows into/out of a capacitance. This has nothing to do with the sign of the voltage on the base, only with how fast the voltage is moving about. To decrease this current, you need to slow things down. That's all. Base capacitance doesn't change so much with negative bias to make the negative bias responsible for this. You'd have this phenomenon even for 0.1V positive spikes on the base. \$\endgroup\$ Commented Jun 16, 2022 at 0:42
  • 1
    \$\begingroup\$ The base-emitter of a transistor is a diode. It conducts current when it is forward-biased WITH VOLTAGE that is 0.6V or more. When it is reverse-biased then it does not conduct any current until THE VOLTAGE is 5V or more that causes an avalanche breakdown current that damages it. The charging of tiny stray capacitance is another different current. \$\endgroup\$
    – Audioguru
    Commented Jun 16, 2022 at 18:17
3
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How to reduce it? It’s low enough. No problem there.

It’s really simple. Reverse base voltages aren’t forbidden. There’s a limit on them - usually small compared to the collector-emitter forward voltage limit. Look in the transistor datasheet to see what’s the maximum allowed reverse base voltage – it’ll usually be >4V and the small voltages you apply in reverse are lower than that. So no problem.

You only damage the transistor if the reverse base voltage goes beyond the maximum allowed.

In your particular simulation, the base current you observe is proportional to the slope of the base voltage, and seems to flow just the same whether base is positive or negative. So it's a current that flows into/out of base capacitance, and is not related to junction conduction at all.

It looks like the base capacitance in the transistor model you're using is super high, or maybe the transistor just has a huge base capacitance. Select one with a lower capacitance then. At least now you know what to select. But low base capacitance won't help with overall efficiency if the current gain is low at the collector current you operate in. So lowering base capacitance can't be continued when the gain starts going down too quickly. It likely requires experimentation with real parts.

But first of all, slow down. The current flowing into the base is not somehow "bad" because it's negative voltage. It's high because the base voltage moves too fast. If you make the negative base excursion 10x slower, the current will drop 10x as well.

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7
  • \$\begingroup\$ So the spikes in current that I'm seeing are interactions of the circuit with the capacitance of the junction, but since the voltage spec. of the junction is not being exceeded, it's okay? The second circuit I showed has a negative spike of 180mA, but the voltage trace (not shown) never even goes negative. So I've been looking at the wrong thing? \$\endgroup\$ Commented Jun 15, 2022 at 18:09
  • \$\begingroup\$ I thought voltage was the pressure, and the avalanche was the damage (indicating that I should be looking at the current)?! \$\endgroup\$ Commented Jun 15, 2022 at 18:38
  • 1
    \$\begingroup\$ In SPICE, you can plot formulas. Just plot the difference between the base and emitter voltage on that transistor. As long as the difference is less than about 5V (or whatever limit the datasheet provides), the B-E reverse bias is not a problem. If you confirm this, then the currents are caused by something else. \$\endgroup\$ Commented Jun 15, 2022 at 18:47
  • 1
    \$\begingroup\$ The capacitance is tiny. Try adding some capacitance - say 50pF - to the base and see how the current changes. That will tell you whether capacitance had anything to do with it. And besides: LTSpice does not simulate reverse B-E breakdown as far as I know. To see reverse breakdown on a transistor's B-E diode, you'd need to have a physical circuit assembled. So your problem is not breakdown from -1.8V on the base. The current waveform looks like it's related to some capacitance, but it seems way too big. Looks like the transistor model you're using is bonkers somehow. \$\endgroup\$ Commented Jun 16, 2022 at 0:29
  • 1
    \$\begingroup\$ Slow down. The circuit is oscillating 10x too fast. Micro power and fast are mutually exclusive. Parasitic capacitances are a big part of it. \$\endgroup\$ Commented Jun 26, 2022 at 16:04
1
\$\begingroup\$

Your concern over base current is overly cautious: it depends on the voltage.

In normal operation, at turn-on, some excess positive current is needed, to overcome capacitance. Conversely, some excess negative current is needed at turn-off, to remove stored charge. (This is a period during which base voltage seems to hang in place, at about a diode drop: it's the same phenomenon as PN diode reverse recovery.)

This is a purely dynamic effect. When switching slower than, say 10µs, the stored charge is recombined internally (more or less, in effect: consumed by base current), and no negative current will be sensed externally as it goes from forward bias to reverse.

Statically, nothing interesting happens between 0 and Vebo. Some leakage current flows through the reverse-biased junction.

When Vebo is exceeded, avalanche (breakdown) current begins to flow. This is the mode of current flow which causes cumulative damage. Also due to the relatively high voltage drop, it can quickly overheat the transistor, causing thermal failure. The higher voltage means higher energy on internal constitutents ("hot carrier" electrons/holes), which can force charges into surface oxide, or knock atoms out of place. The usual effect is reduced hFE, particularly at low currents; hFE can be restored somewhat by annealing at ~soldering temperatures.

There are other aspects to avalanche breakdown as well (most dramatically, C-E punchthrough as in avalanche switches), but this is going beyond scope, and I don't think these effects apply to E-B or E-C avalanche.

See:

BJT In Reverse Avalanche Mode

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4
  • \$\begingroup\$ So does a diode reverse to the B-E junction solve the problem? \$\endgroup\$ Commented Jun 20, 2022 at 22:49
  • 1
    \$\begingroup\$ @MicroservicesOnDDD See Antonio51's answer: it's not a problem in this circuit at all. In others, a diode can do, yes. \$\endgroup\$ Commented Jun 20, 2022 at 23:39
  • 2
    \$\begingroup\$ That 2SCR553P is a big-area transistor with over 100pf base-emitter capacitance. Substitute a dinky transistor like 2N3904, and that reverse-base current turns into a tiny little pip... not a problem. \$\endgroup\$
    – glen_geek
    Commented Jun 25, 2022 at 13:15
  • \$\begingroup\$ @glen_geek -- I'm specifically using that transistor because it is the most efficient that I have found in the LTSpice Library. Tried them all using LTSpice "AKO" feature (AKO = A Kind Of -- a type of aliasing feature). \$\endgroup\$ Commented Jun 5, 2023 at 20:52

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