[Syzygies]

Can a current mirror be used as a very low overhead trickle charger?

This is about as primitive a version as I can draw, that doubles the current through R. There are more sophisticated similar circuits, with more parts and/or higher overhead. This circuit won't have a dead flat charge curve, but we don't need one. With reasonable matching and adjustment, its charge curve should be good enough. The pattern is easily extended to charge multiple banks or increase the current multiplier.

I see a wide audience for Pimeta-class portable amps whose battery pack is recharged from the same stock iPod wall wart that one carries around already.

Unfortunately, a standard LM317 trickle charge circuit has 3-4 volts of overhead, and iPod wall warts put out 13V to 14V unloaded. It takes a bit of cleverness to charge a single 9V battery from one, e.g. a different resistor network, low dropout regulator, Schottky diode.

It takes more that a bit of cleverness to charge 8 NiMH AAA cells, whose charging voltage can reach 12V.

I've long thought that a charging circuit should operate beside the battery somehow, not in series, to get the overhead as low as possible. So I nearly fell out of bed when I reread this section of Art of Electronics last night.

Could this work? Should I experiment with it? If this works, the LM317 circuit starts to look like massive overkill, hunting squirrels with a bazooka.

Do the transistors need some diode protection? (In which case we lose some overhead!)

If this circuit works, then the next thing to explore would be adapting a fast charge circuit to operate beside the battery pack, reducing the voltage overhead by controlling the actual charge current using current mirrors.

Digi-Key carries the Zetex ZDS1009TA, a monolithic current mirror, for $1.25. It doesn't support current doubling like I've drawn above, but it could be used for similar circuits.

 

[Syzygies]

So everything I've built recently has worked, so I thought I'd skip breadboarding, and just build this:

Perhaps I accidentally applied too high a reverse voltage, but it's toast. Inert. Nice physical illustration of what I have in mind, but otherwise useless.

My idea here was that with a trimpot that covered the conceivable range needed, no need to match resistors, just tune the charger to the batteries and power supply in question.

Back to the breadboard, this circuit does work! However, transistors are all over the map if you don't match them, and the basic circuit is every bit as variable in its output as Art of Elec suggests. I think I need to measure my transistors and make predictions in Spice that I can confirm on a breadboard, before soldering again.

My breadboard circuit is successfully charging 8 AAA cells off the wimpier of my two iPod power supplies. They're not fully charged, but a standard LM317 charging circuit would have already thrown in the towel, so this is looking good...

 

[mirlo]

You can get around some of the transistor matching problem by using equal value emitter resistors in all the BJTs. Matching should be improved by roughly the number of times 26 mV divides into the voltage across the resistors. So if you put about 260 mV across the resistors, for example if you use 26 mA per leg use 10 Ohm resistors, you can expect maybe 10 times reduction in mismatch sensitivity. I attached a file to illustrate this witha multiply-by-three mirror that turns 26 mA into 78 mA.

I hope that makes sense. I realize I didn't explain it too clearly.

-- mirlo

 

[Syzygies]

Ok, my second try works great; I'm sure I fried my first board through a moment of stupidity.

This fully charged 8 AAA cells from a 13V iPod power supply, with a total charger voltage overhead of 1.35V. I'm sure it could go lower; I don't have a variable bench power supply for testing this.

The 4x "nominal" mirror is more like 10x. This took me by surprise, I had to solder a 1K resistor in series with the trimpot to bring it within range.

If one wants a "no brainer" design that doesn't require matching, then socket the 1K resistor I used. If you're lucky it can be a wire jumper. In any case, it can divide the voltage drop with the trimpot, and bring the trimpot into range.

Recall that the usual trickle charger voltage overhead is 3 to 4 volts. Even with low dropout regulators, the resistor alone accounts for 1.25V of overhead. If current mirror chargers can be tamed for general use, they represent a pretty significant improvement.

 

[Syzygies]

Quote:

Originally Posted by mirlo You can get around some of the transistor matching problem by using equal value emitter resistors in all the BJTs. Matching should be improved by roughly the number of times 26 mV divides into the voltage across the resistors. So if you put about 260 mV across the resistors, for example if you use 26 mA per leg use 10 Ohm resistors, you can expect maybe 10 times reduction in mismatch sensitivity. I attached a file to illustrate this witha multiply-by-three mirror that turns 26 mA into 78 mA.

Art of Electronics indicates something is true along these lines, but leaves the reader to work it out, and gives a schematic with huge resistor values. I first wanted "proof of concept", which I now have. I didn't want to give away any overhead on my prototypes. But, as they say, any design that depends this sensitively on the transistor parameters is a poor design. My next step was to figure out if a bit of resistance as you suggest would be worthwhile.

So you're a step ahead of me. Thanks!

Edit: Nice schematic! Your 10 ohm resistors give away 0.25V. I think we can spare this.

 

[Syzygies]

So I hooked up nine 1000 mAh NiMH AAA cells, to an iPod supply that can maintain 12.90V while this circuit delivers a 100mA trickle charge.

I got down to 0.40V overhead easily.

 

[mirlo]

Next step might be to add a current regulator to the input side of the current mirror, to make it insensitive to power supply variation (now the input current is too dependent on the power supply voltage)and some kind of voltage limiter to the output side.

What is the best end of charge voltage for NiMH cells? For Lithium Ion batteries, it is 4.2 to 4.25 volts ... I think I remember seeing somewhere that an appropriate end of charge for NiMH is 1.4 V per cell ... but I am not sure.

-- mirlo

 

[Syzygies]

Fast charging NiMH is subtle, and specialized controller chips like Tangent uses in his NiMH Battery Board offer various options: temperature, voltage dip. They also turn off the charging to more accurately measure the batteries periodically.

With trickle charging, I find that NiMH cells are too robust and inexpensive to need coddling. Perhaps what I do won't get 1000 charge cycles; I don't know and I'm not sure I care. What I find is that the voltage as measured while charging climbs to a terminal value 1.43 or 1.44 ... or 1.47 volts per cell, depending on the type, brand, etc., and stays there. Trickle charging is simpler and easier to predict than fast charging, and better for the battery.

People say that a C/10 to C/20 charge rate is trickle charging (where C is the nominal capacity in mAh), with C/10 too fast a rate to leave connected for days, and C/20 so slow as to perhaps be ineffective.

I read about current regulators as I went to sleep last night, you're right ahead of me! If the circuit stepped down to a lower current, say C/20, when the battery pack reached 1.40V per cell, that would be plenty of intelligence in the circuit. I'd like to keep the circuit small...

Let's not forget that a current mirror could be the basis for a fast charge circuit, again with 8 cells within reach of every iPod power supply. Perhaps mirlo's schematic with better current regulation is where we should quit for trickle charging, and the next step should be to adapt a fast charge circuit like Tangent uses.

 

[Syzygies]

So the 4x mirror pictured above as my "second try" has about a 5 mA reverse current, when a charging current isn't provided. I didn't notice this until now.

Surely I can prevent this with a Schottky diode, which I'll probably want as part of integrating this circuit into an amp, but that's another 0.35V of overhead.

Is it routine to get a bit of "backflow" from collector to emitter of PNP transistors, when they're reverse biased? Can I avoid this without a diode drop?

 

[Syzygies]

Quote:

Originally Posted by mirlo You can get around some of the transistor matching problem by using equal value emitter resistors in all the BJTs. Matching should be improved by roughly the number of times 26 mV divides into the voltage across the resistors. So if you put about 260 mV across the resistors, for example if you use 26 mA per leg use 10 Ohm resistors, you can expect maybe 10 times reduction in mismatch sensitivity. I attached a file to illustrate this witha multiply-by-three mirror that turns 26 mA into 78 mA.

I was able to derive this, it's a great exercise.

If one takes the Ebers-Moll equation for a transistor, and puts in an emitter resistor, one gets the recursive equation in I

I = I_S * exp[ ( V - I R ) / V_T ]

where V_T can be taken to be your 26 mV. Here, the exponential in the Ebers-Moll equation is our friend, because when we implicitly differentiate this equation, we can replace any instance of this exponential by I/I_S, simplifying the resulting equation.

Finding dI/dV, there's a factor of 1 / ( V_T + I R ) in the answer.

Finding dI/dI_S, there's also a factor of 1 / ( V_T + I R ) in the answer.

So in particular, your example of adding a resistor that drops the emitter voltage by 260 mV will decrease the sensitivities in the circuit by a factor of 11, as you say. I'm guessing that if overhead is tight, dropping the emitter voltage by 130 mV for a factor of 6 is plenty.

 

[SnoopyRocks]

Quote:

Originally Posted by Syzygies I was able to derive this, it's a great exercise.

You sound like a mathemtician or professor. The last time I heard something like this it was concerning the noise in an LNA. I found it a bit too painful to be a great exercise. That's a wonderful attitude to have.

Another way to describe what the resistor does is emitter degeneration. The resistor lowers the effective transconductance (gm=dIout/dVin: output current to input voltage) of the transistor. The effective transconductance (big Gm) is Gm=gm/(1+gm*R). When gm*R >>1, Gm~1/R and has very little dependence on the inherent transistor properties. Hence the matching improvement. This is what your analysis is telling you. You get it.

 

[guzzler]

Quote:

Originally Posted by SnoopyRocks You sound like a mathemtician or professor

He's both, or close

 

[mirlo]

So professor Syzygies, what is a syzygy?

;^)

--mirlo