9V by a ni-cd battery with USB charging


Poking even once in all the drawers looking for a 9V battery charged for a multimeter that has shut in time of need (how could it be otherwise?), I thought it would be nice to have a box of these batteries. I have a lot of small rechargeable batteries around, usually with a capacity of 100..500 mA / h. But there is not much to 9V, it consumes the tester? 5..10 mA with such power fatigue any step up to have a decent real efficiency.

thinking, I decided to abandon the PWM controller and stabilization with protections, any voltage between 8V and 9,5V I'm okay to power the multimeter, in the worst case scenario I will see the battery light on. After rummaging through the net to find inspiration for good circuits I had the lighting a self-oscillating push-pull with a small number of turns of the windings on a small ferrite ring.

I love the push-pull circuits! Good, solo 3 necessary and sufficient components. Furthermore, in the schemes single cycle, the ferrite rings do not work, you need a gap in the nucleus, without which the nucleus itself will immediately become saturated.

The pattern is visible here,

In the manufacturing process, I could not resist experimenting with different rings and spirals, in these times of "house arrest" due to the epidemic Time it is the only thing that is missing, the conclusion is that everything works, the difference is only in the operating frequency and efficiency.

The number of turns is not critical, it is only important to maintain the ratio of the windings I and II in the region of 2,4 – 2,6 (for an input voltage of 3,7 V and output 9,0 V). The circuit will act as a volt and a half, the output voltage is set by the ratio of turns of the windings.

Since the BE junction of each transistor acts as a rectifier, it is necessary to check in the manual the parameter of the maximum reverse voltage for this transition – not all low-power transistors will. I took the first transistor I had in the BC328 drawers (I bought in industrial quantities to average between fairs 1 e 3 cents per piece), I got a maximum base emitter voltage of 5 V – no problem, if the transition is interrupted, the rupture is reversed in the Zener diode and everything works as it should.

It is better to choose a ferrite ring of maximum size and permeability allowed: the converter frequency will be lower, the ferrite losses will be smaller, the efficiency will be greater, the interference will decrease considerably. In when it is still an oscillator. For the experiment, I built the transformer on a ring 10 x 7 x 3 mm (circa) when the size is minimal it is difficult to be precise, I wrapped 2 x 10 turns on the primary and 2 x 25 revolutions per secondary.

Access: it works, the conversion frequency is about a half MHz. I entered the circuit in the case of tin now exhausted stack, I turned on so the tester I lit not far from the analyzer spectrum and I detected a carrier with a low level of conversion to the frequency and two harmonics still lower than another was not measurable covered by background noise. There was no signal in the next room.

So the devil is not so terrible as it is described, but it is better to take a larger ring and lower the operating frequency. The circuit starts to work as expected only when the output is at least some kind of load, at least 0,5 mA. If not, the converter does not consume almost anything. But there is a peculiarity.

If a transformer is realized on a small ring with a small number of turns and an operating frequency of several hundred kilohertz, the circuit may ignite even without load, work independently, apparently due to microscopic leakage current (if the circuit does not work, the input voltage is less than the fall at the junction of the emitter transistors) and initiated by the charge of the output capacitance current.

This is another reason to choose a larger ring, with a working frequency 150 kHz, I have not encountered any problems. A clear ignition when the load is connected and off off when disconnected.

The points that indicate the beginning of the windings are not in vain on the diagram. It is easier to wind the two primary windings and the two secondary windings with double wire, then connect the beginning of one with the other end (middle point) and here it is. But there is a probability of 50% that the circuit does not start from the first start.

I'm probably been unlucky but I, the probability in my case was much higher! If there is a persistent output voltage, half a volt below the input terminals of one of the windings must be exchanged will start to operate.

The input capacitor, and in particular the output capacitor, It must be taken with minimal loss. Although you can leave 0,1 microfarad ceramic inlet and outlet, in all circuits, the capacitors are already at supply I preferred not to risk it and put the electrolytic.

In my circuit, the ripple with a load of 6 mA was about 2 mV, I do not know if it was just a ripple, and not some other interference: these small voltages are not visible on my oscilloscope antediluvian. In this period I have this at home, my laboratory equipped with instruments is valid for the off-limits now too far away from my home and therefore unattainable without risk to my health and other.

The circuit obviously does not tolerate an output short circuit, which it has been verified in practice by a random short. A couple of BC328 souls fallen in battle have flown to their paradise. The diagram shows four other components, battery, two transistors also PNP, I had already started to use those because changing! In a last resistance, this circuit recharges the battery with a minimum number of components.

This circuit is a current mirror, they are found in large quantities within the analog integrated circuits, It is useful because it does make the transistor of the variable resistance function with a very reduced number of components.

Current mirrors (current mirror)

This circuit converts the current applied to its input in a current absorbed, that is in a precise relationship with the first.

In the current mirror circuit principle, illustrated in the version with PNP transistors in figure, he transistor TR1 it is connected as a diode (base and collector short-circuited) and it is disposed across the base-emitter junction of TR2

An input reference current

It is applied to the diode connected as BJT (TR1), determining a base-emitter voltage (VBE1) adequate to IB1.

The base-emitter voltage (VBE2) of the transistor TR2 is forced to equal VBE1.

Relative to the node N you have:

IR =IC1 + IB1 + IB2 .

Assuming the two identical transistors, gives


it follows that

IB1 = IB2 e IC1 = IC2 = I2,

and bearing in mind the relationship in the linear operating zone between the base and collector currents

IB = IC/b

you have:

In the case of b >> 2, you have:

Since the variation of VBE with the temperature it is generally small compared to VCC, I2 it is substantially constant and depends only on the values ​​chosen for VCC ed R.

One of the current mirror constructed with NPN transistor works in exactly the same way, but we must keep in mind that the lines of current are reversed and that its output current absorbs and does not behave as a current generator.

The most peculiar characteristics of the current mirror are its simplicity and the fact that it requires only one base-emitter voltage drop; in this way it is possible to use for other purposes the remaining available voltage. Since we want to reload it with a USB 5V from the output circuit seems optimal, with the only fall of 0.7V have sufficient voltage to recharge my battery with a nominal 3.6V with a constant current

I2 = (5-0,7) / 470 = 9mA

around so as to charge it the night and if I urgently I can always use the meter while charging. The charging circuit when it is not connected to the USB jack simply isolates itself from the circuit does not absorb any current.

I did not realize a handout for this application, I realized all on a matrix board cutout of 22mm x 28mm with external battery to it, the dimensions of the battery in height are similar to the dimensions of the inside of the battery shell to be used. A generous layer of tape to isolate the whole and a colatina of hot glue make the whole stable and insensitive to vibrations or shocks.

The USB cable that I stole a muse no longer functioning does not cause too much hassle. My intent was not to change the meter, for the future I will see whether to insert a micro USB plug directly into basettina SMD to eliminate even the cable.

I intend to redo all in SMD in the near future to easily replicate the basettina and have more available space by using smaller components with the exception of the coil and battery that can not be shrunk. For the current situation I'm happy to have a working unit and functional.

Amilcare Greetings

6 replies
  1. Paulo Dias
    Paulo Dias says:

    Hi! (now with voltges corrected…)

    Just a doubt on the output voltage…

    Isn’t it: V(battery) – V(diode) – V(secondary)?

    That is: 3.7v – 0.7v – (-3.7 x 2.5) = 6.25v ?

    Thanks a lot for the article!

    Regards, Paulo

  2. Paulo Dias
    Paulo Dias says:


    Just a doubt on the output voltage…

    Isn’t it: V(battery) – V(diode) – V(secondary)?

    That is: 3.7v – 0.7v – (-10v) = 5v ?

    Thanks a lot for the article!

    Regards, Paulo

  3. Amilcare
    Amilcare says:

    No I did not want to put a micro USB socket but at the moment there is not much and I opted for a cable recovered from an old wheel mause. If you look closely at the photos of the multimeter you can see the cable well.

  4. theremino
    theremino says:

    This time I really have to say, damn how beautiful this scheme is!
    And the idea of ​​the battery for testers is also excellent.
    I voted for 10 but if it were possible I would have also given 12.

    I have a question though, how can you make the connector for charging?
    You managed to make it fit in the bottom wall?


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