1
\$\begingroup\$

Let's say I need to charge a 4m7 (4,700uF) capacitor to 5V, from a 1.5V battery with an internal resistance of 100 ohms. I'm trying to use a typical joule thief with something like soft-start, to limit the initial capacitor current.

I found a problem with unstable capacitor voltage, changing in different periods (about 10s). Sometimes the voltage drops to 2 volts, sometimes it rises to 3V.

Where is the problem? I can't find it.

schematic diagram

photo of hardware

Edit:

To make it clear, Q1 makes a pulsed voltage, after D1 is high voltage DC, which start to charge C1 via R3. When voltage on cathode of D1 is higher than 2.1V, Q2 opens and charges C1 directly. I added R3 to limit initial charging current to prevent attenuation of joule thief oscillation because of high initial load of capacitor C1.

After connecting the power supply, it starts nicely and I measure on C1 3.6V. After a few seconds, it slowly and continuously falls to 1.5V or 2V, and then rises to 3V again.

\$\endgroup\$
6
  • \$\begingroup\$ What's the point of Q2 and it's associated circuitry? \$\endgroup\$
    – TimWescott
    May 30, 2019 at 22:13
  • \$\begingroup\$ When on emitter of Q2 is more than 2.1V, bypass the R3. \$\endgroup\$
    – user208862
    May 30, 2019 at 22:37
  • \$\begingroup\$ Q1 is soft start and Q2 delayed rectifier.. They should call Joule Thief the Battery Killer. \$\endgroup\$ May 31, 2019 at 0:01
  • 1
    \$\begingroup\$ i.stack.imgur.com/AfE0h.png \$\endgroup\$ May 31, 2019 at 1:08
  • \$\begingroup\$ I have doubt, if it is a discharging of C1 back to boost converter, it become onloaded and shots off. \$\endgroup\$
    – user208862
    May 31, 2019 at 10:51

2 Answers 2

1
\$\begingroup\$

I think your circuit is oscillating, but I am not completely sure.

You will find that an inverting joule thief topology will work much better:

enter image description here

The zener is not yet conducting much, and that part of the circuit has to be done better -- as it is, it will not keep the supercap from being overcharged.

This is not your circuit exactly. The "transformer" is completely different. The 2N3904 makes a better joule thief transistor in this case because of it's better turn-off figure, and therefore less switching losses. Also making it a decent fit is the 200mA limit on the 2N3904 and the low power nature of this converter. It's always better to use a schottky diode for the catch diode (D3 above), as it is much more efficient, so the 1N5817 is much better than the 1N4148 and contributes to the success of this simulation.

At this point in the simulation (29 seconds), V2 is putting power out at -14.721mW, the battery internal resistance is wasting 9.63mW, D4 is dissipating at 2.9046mW, Q2=1.3675mW, Schottky D3=0.3021mW, R2 at 1K spends 0.2747mW, and C3 is only being charged at 11.533µW.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Hah, I opened this question, latter I canceled account. You are right. The inverted boost is much less sensitive to high output capacitance since it allows higher inductor voltage swing when cap is at 0V. \$\endgroup\$ Feb 22 at 13:42
  • \$\begingroup\$ @MichalPodmanický -- Glad to hear it answers your question. :-) \$\endgroup\$ Feb 22 at 22:27
1
\$\begingroup\$

It looks like the whole shebang is just not capable of supplying the initial current to the big cap. So when you start up, the output voltage of the oscillator- booster thing collapses, turning off Q2.The big cap then charges slowly via the 68R resistor. At some point Q2 turns on, and here we go again.

CR time of 4.7mF and 68R is around 320ms, from which I would expect things to happen a little quicker though. But I would guess that that is the basic mechanism, or something like it.

I'm wondering about Q2 - isn't that meant to limit initial current? try a cap from base to emitter so that it turns on slowly. You could also add an emitter resistor to make it a kind of slugged current source.

\$\endgroup\$
2
  • \$\begingroup\$ Point of Q2 is that bellow 2.1V the out cap is charged through 68R and above 2.1V the Q2 is switched. \$\endgroup\$ Feb 22 at 14:14
  • 1
    \$\begingroup\$ yes I see that. But I don't think it is limiting enough, so the supply gets pulled down. That's why I suggest the emitter resistor. That allows the OP to set a well defined max current and avoid that problem. \$\endgroup\$
    – danmcb
    Feb 22 at 14:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.