Some questions about boost circuit/battery configs

Jabba

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Hi all,

I am new to this forum. Although I have been reading here a lot in the last time and found a lot of valuable information I still have some questions left.

What I plan to do is to equip my (homemade) LED-based lamps with conversion circuits to make them use up all the battery power. I want to use the same/similar circuit for all three lamps and will probably go with the ZLT+ design. The lamps use 3 AA cells which seems to make it a bit more difficult.
So before I go spending money for the parts I thought I ask the experts.
From what I understand it is

1. no good idea to run one white LED with the ZLT+ and 4,5 V
(i.e. 3 AA alkalines)?

2. a good idea to run 2 or more LEDs (Nichia 5mm ones) in
series with 3 AA alkalines ?

3. a good idea to run a 1W LS with a ZLT+ from 3 NiMHs?
The internal resistance of the batteries will reduce the
voltage to something close/below Vf. Currently, I run the
LS direct drive with a very small resistor.

4. The Zetex chip will run the batteries down to voltage of
roughly 0,8-0,9V, i.e. 0,3V per cell. Is this right? And if yes, doesn't
this very deep discharging damage the NiMH cells ???

Waiting for your comments. Thanks.

Jabba
 

pbarrette

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Hi Jabba,

I really like the simplicity of the Zetex circuit, but it does have some drawbacks:
It's a Constant Voltage circuit and LED's really should have a Constant Current source.
It's a boost circuit, so any Vin > LED-Vf is essentially a direct drive light.
There's no built-in way to stop the battery drain that you talk about in your #4.

On the other hand, if you want to use a 2AA configuration, then this could be a great circuit for you. Constant Voltage isn't really an issue if you don't care that the light from the LED will decrease as the battery is drained. A CC circuit would keep the LED at the same brightness until the battery dies, but with CV the light will taper off.

On the other hand, this circuit will drain a battery to it's absolute end. The 0.8V listed seems to be the required start-up voltage as opposed to the point where the circuit shuts down. If the battery can start the circuit going, it will drain down well below 0.8V.

So to answer your questions:
1) 3AA's is not a good idea for this circuit. 2AA's is.
2) This is doable. 2 LEDs in series would make Vin < Vf.
3) 3 NiMh's are not a good idea unless you are monitoring the light. 1 NiMh would be ok, but you should still be tracking the light.
4) Absolutely correct. If unattended, this circuit would likely do permanent damage to the batteries.

If you happen to be in the Wiesbaden area, I could always just give you a set of components for this circuit so you can play with it. I haven't quite figured out the postal system here though.

pb
 

mike101

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Hi pbarette,
I have some components to build the Zetex Circuit. As you stated the ZC was a CV circuit, so it might not be the best circuit for LED. I am thinking if someone can modify the circuit to CC using same components, then we can build a better circuit with least components.
Any idea?
 

Doug Owen

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If you're driving from 3 NiMH cells, I suggest you consider a LDO regulator. You'll not only get full usable charge, but it will not hurt the cells by over discharging as the current drops rapidly at about one volt per cell. Even if left on it doesn't seem to hurt them (I did this on purpose a dozen or so times to one set to see).

Cheap, simple, effective. Over 90% efficient in this configuration, something you'll never get with other schemes.

Doug Owen
 

pbarrette

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Hi Mike,

Well.. Ok.. I bluffed a little bit there.

Zetex claims that this IC is _already_ a constant current boost IC. The problem is that the IC is monitoring and stabilizing the current built up in the inductor rather than the current through the load (LED).

What this means is that if you feed the circuit a stable voltage then you will get a constant current at the load. But in reality, we are using batteries instead of a wired source. Batteries means that the voltage drops as the battery is discharged. As the voltage sags, the output current also drops. In a true, load measured CC setup, the IC would strive to maintain the current through the load regardless of the fluctuations in the input current.

So while the Z300 is not truely a CV regulator, it is also not truely a CC regulator. It is just easier for the purpose of battery powered, current load based applications (like LED flashlights) to treat it as a CV regulator since the voltage will still be boosted to maintain the Vf of the LED as the input voltage drops, but the current at the LED will drop.

I do think that someone (MrAl maybe?) has already posted a circuit modification that allows the use of this IC as a load measured CC driver, but it can't be done (to my knowledge) without more components than the reference circuit.

On the other hand, there are other, true CC boost IC's out there that have small component counts. I don't have any examples offhand since most of my boost circuit experience is with the Zetex IC, but I'll bet some searching through the forums would yield some good candidates.

Hope this helps,
pb
 

Jabba

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Re: Some questions about boost circuit/battery con

Hi pbarette,

thanks for your input.
I guess, I can work around the 3 AA issue by changing the wiring of the batteries from series to parallel giving me 1,5V or 1,2V, depending on the battery type. I think this should make the ZLT+ run quite effectively with the same amount of energy available. Since I currently do not really need to run the lamps from rechargable batteries (don't use them that often) I can also live with the ZLT+ draining the batteries to death.

Thanks for offering parts. Actually, I live rather close to Wiesbaden. Unfortunately our postal system wouldn't care much, whether if I'd lived next door to you or at the other end of the country. But since I have to get a good voltmeter for this project anyway, the cost for the parts isn't that much compared to it.
 

HarryN

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Re: Some questions about boost circuit/battery con

Just MHO, but you are making your life a lot more complex than it needs to be. As Doug Owen pointed out, the 3 x AA in series (regardless of chemistry) and either an LDO or resistor is a very nice match for your application.

You are talking about nominally a 1 volt drop and current in the range of 20 - 40 ma, which can easily be handled with a 50 ohm, 1/8 watt resistor for less than 10 cents in almost any package size you desire.

The most simplistic of the LDO variations on this is a Doug S post on the 14400 flashlight where he details his approach in a manner anyone can capture. (electronics section)

I won't say it cannot be done, but you will work very hard for a long time to make a setup that is more efficient as well.

When the cells are too dry to push this simple electronics package, they are pretty dead.
 

Jabba

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Re: Some questions about boost circuit/battery con

Hi pbarette,
I found two circuits by MrAl called ZLT-CCR and ZLT-CCR-2 which seem to enhance the ZLT+ in direction of a true CC device. I guess, that's what you mean. I will keep looking for some further possibilities, but since that's my first home-made circuit of that kind, I will probably stay with a really simple one and move the more sophisticated features to the second evolution.

Doug,
what parts did you use for your LDO circuit? I still like the idea of having a built-in protection for rechargables.

Regards,
 

pbarrette

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Re: Some questions about boost circuit/battery con

Hi Jabba,

MrAl and Andrewwynn show a very low component count LDO circuit here.

The total part count of andrew's modification is 4 plus the LED. The circuit consists of 2 transistors, 1 mosfet and 2 resistors. Andrew's design uses an SOT-23-6 package containing both of the 2 transistors to keep the part count low and deal with thermal differential problems in the circuit.

The ZLT-CCR-2 you found is the Zetex CC design I was thinking of. Basically adding a resistor divider network to fool the Isense pin by feeding it from both the battery and the inductor.

pb
 

andrewwynn

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Re: Some questions about boost circuit/battery con

small correction.. rather than two resistors i use a single Potentiomter.. you can wire it up either of two ways.. i haven't figured out which works better...

1) the 'high' Resistance goes through the whole pot.. and the wiper goes to the low resistance.

2) Wiper goes to Vcc.. and the two legs are the two resistances.

You can see pictures of the driver here: http://rouse.com/nano (drivers, electronics)

The circuit is very sensitive to thermal changes... when it's on the breadboard it doesn't do nearly as well as when it's inside the light.. it may get a bit warm inside there, but it stays very stable (temp)... it's amazing how much current swing you can get by blowing on the driver on the breadboard.

The LDO has 'built in protection' for my LiON batts.. based on the Vf of the LED.. once the batts are mostly dead.. they just output miliamps and then microamps 'til they just don't really output anything anymore.. I suppose it's conceivable that if you left it on for days it could hurt the battery.. but hours or overnight would probably just drain them to 2.5V or so.. just very sloooooowly at the end.


-awr
 

pbarrette

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Re: Some questions about boost circuit/battery con

Hi Andrew,

So is that .1omh SMT resistor just for measuring current? Because I clearly see 2 resistors there. 1 pot and 1 SMT.

The original MrAl LDO shows 3 resistors, all fixed.

pb
 

mike101

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Thank guys for the information. I think I would go for the ZLT-CCR-2 circuit because I have some Infinity Ultra and Solitaire as hosts. They are 1 cell lights.
Another question, if I use Lithium battery with the Zetex original design, how big the difference will be to compare with alkaline battery with the ZLT-CCR-2 (LUXI/LUXIII on 1 lithium AA/AAA)?
 

andrewwynn

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Re: Some questions about boost circuit/battery con

[ QUOTE ]
pbarrette said:

The total part count of andrew's modification is 4 plus the LED. The circuit consists of 2 transistors, 1 mosfet and 2 resistors. Andrew's design uses an SOT-23-6 package containing both of the 2 transistors to keep the part count low and deal with thermal differential problems in the circuit.
pb

[/ QUOTE ]

oops.. just a little confusion.. ... the part count is a little weird because of how you 'parse' the parts.

more clearly.. there are 4 components to my driver:

1) FET (power transistor)
2) dual NPN amplifier transistor
3) Sense resistor
4) Potentiometer

What you said was accurate, just misleading.. the 'two transistors' is in one package so it's one component.. and though pots are always labeled as a resistance 'RXX' on a schematic.. they are called a 'Pot' vs 'a resistor' since schematically they are two separate resistors in one package (just like the dualNPN).. so.. it's really 6 electrical elements in 4 component packages.

and yup.. the .1R is the Sense resistor

-awr
 

pbarrette

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Re: Some questions about boost circuit/battery con

Hi Andrew,

Ok.. I see the confusion now. I did mention that the total part count was 4, but I didn't mention that it's actually 3 resistors in 2 'parts'.

Though I consider a pot to be 1 component/part because you don't neccessarily need to use both of the variable terminals in the pot. Most of the time, I'm just using the wiper tap and one end so I'm just using 1 variable resistor.

In the case of your circuit, however, you are using the wiper and both ends to form two resistors where the resistance of one is directly proportional to the other.

In the end though.. One damn nice LDO in a small package with only 4 parts to buy.

pb
 

Jabba

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Re: Some questions about boost circuit/battery con

Hi HarryN,

I see your and Doug's point. It is a simple circuit. But I don't think it fits for every battery type. As far as I understood it, the LDO driver will run from the full battery voltage down to Vf and then drop out of regulation. For 3 NiMhs the start is 3,6V, for 3 Alkalines 4,5V. Luxeon states a typical Vf for the 1W LS of 3,42V (min=3,1V). Considering the voltage drop inside the NiMhs when drawing 350mA, this is pretty much the available voltage right from the start. So the LDO driver will stop regulating pretty soon.
For alkalines, running from 4,5V down to 3,42V looks much better. But on the other hand, I then don't need to protect the batteries and may want to run them to their very end. Alkalines should be usable downto at least 0,9V/cell (That's what I read elsewhere.) or 2,7V for 3 cells. So, when the light goes out there is still some energy left inside the batteries. How much energy this is and how this efficiency compares to using a 80%-efficient ZLT+ circuit on three alkalines in parallel (1,5V) I don't know. I guess, there is not that much difference.

I admit, that all this is very theoretical, and if someone tells me that his/her experience is clearly different, I will happily switch my opinion ;-)

Regards,
Jabba
 

pbarrette

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Re: Some questions about boost circuit/battery con

Hi Jabba,

With the LDO it's not that the circuit stops regulating, it's that the circuit no longer lights the LED since the Vf of the LED is greater than the input voltage.

Now, an LDO with 3 or more alkalines only makes sense (to me) if you are trying to push a high (>400mA), regulated current to the LED.

On the other hand, depending on the Vf bin of the LED and the specific batteries, NiMh's may work quite well. This is because NiMh's don't have the steep V drop under load that alkalines do and can generally maintain a much higher current draw. Also NiMh's maintain a fairly steady voltage until near the end of their charge.

So while 3xAA alkalines may start at 4.5v, under load that may decrease instantly to 4.1v and keep dropping until they are "dead" at 2.4-2.7v.

The 3xAA NiMh's on the other hand will likely start out at 3.9v, drop to 3.6v instantly under load, drop to 3.3v near the end of their charge, then finally be dead at 2.7-3v.

So if you have an LED (lux or otherwise) with a low Vf of ~3.3v, then an LDO with NiMh's would provide low battery protection, likely run the batteries through their whole charge, and probably be on-par or better with a boost circuit being fed 3v.

If you really wanted to get tricky with the alkalines, you could build both the LDO and the boost circuit and have a manual switch to switch between them. Start out with the LDO, then when the voltage drops too low for the LDO to function, switch over to the boost circuit.

I do recall a post hidden deep within the recesses of the forum where a schematic was developed that would do the switchover electronically. Some would say to just use a buck/boost circuit for that. The problem there though is that combination buck/boost IC's are notoriously inefficient compared to their dedicated counterparts. Thus the idea was tossed around to build 2 dedicated circuits and switch between them.

Hope this doesn't just confuse you more. /ubbthreads/images/graemlins/smile.gif

pb
 

evan9162

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Re: Some questions about boost circuit/battery con

[ QUOTE ]

With the LDO it's not that the circuit stops regulating, it's that the circuit no longer lights the LED since the Vf of the LED is greater than the input voltage.


[/ QUOTE ]

No, this is not true. If your LED has a Vf of 3.3V at 350mA, and you feed it an input of 3.1V, then more like 150mA will flow. The LED does not simply not light up if Vin < Vf at rated current.

The "turn-on" voltage for a luxeon is around 2.5V - that is the point at which it starts conducting current. It produces small amounts of light at that point.

When Vin ~ Vf with an LDO, the circuit stops regulating current, and the configuration is essentially direct drive.
 

pbarrette

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Re: Some questions about boost circuit/battery con

Ah..

Whoops.. I had totally forgotten about "turn-on" vs "Vf at rated current." I'm used to working with devices where Von is interchangeable with Vf.

Looking at it from that angle, a higher Vf lux would likely result in a higher Von and keep NiMh's safe.

Is that a workable solution?

pb
 
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