I just built a 750mA LDO linear current regulator for driving Luxeons. I built this so I could treat my 5W luxeons a little more gingerly. I have a 5W with an exceptionally low Vf (6.24V @ 700mA). I chose a linear regulator because a LDO will give about the same efficiency as a step-down (or step-up) converter, but will be dead simple to build, and cheap.
I used Doug Owen's world torch circuit as the starting point (an LM334-based current regulator). The only difference is that I swapped the 2n3906 transistor for a Zetex FZT968 (sot223 format). To make soldering easier, I broke off the collector pin on the one side, and used the tab as the collector connection. The sense resistor (the coil of wire) is 0.08 ohms.
The PCB is a 1" circle. I made the pads by cutting channels in the copper cladding using a dremel and a cut-off wheel (takes a steady hand).
With full input voltage, at 750mA, the minimum dropout voltage is very near 0.1V. With reduced input voltage (out of regulation), at 400mA, the dropout voltage was down to around 0.06V (so practically in DD).
This circuit kicks the pants off using a resistor, and is quite cheap to build - about $4 in parts. The best part - it worked exactly as it should the first time!
I'll definitely be building a few more of these for the rest of my lights.
[ QUOTE ] evan9162 said:
I have a 5W with an exceptionally low Vf (6.24V @ 700mA). I chose a linear regulator because a LDO will give about the same efficiency as a step-down (or step-up) converter, but will be dead simple to build, and cheap.
[/ QUOTE ]
Okay, so a fella uses a 9V in batteries (since this can only do step down, and 123 is 1.8V basically discharged, and 1.8 V * 3 is 5.4V so, this guy goes out of regulation once the battery drops below 6.24V.
So 9V batt - 6.24V = 2.76 drop that the linear regulator has to do. 2.76V * 700mA = 1.932 Watts of heat made in the circuit. Input power is 9V * .7 = 6.3 W input power.
6.3 W - 1.932 W = 4.368 Watts delivered to the luxeon. Lets do a quick check, Lux V Vf is 6.24 * 700 mA = 4.368 W.
Okay we are on the right track. Now for efficiency, 4.368 Watts to the luxeon / 6.3 Watts input power = 69.3 % efficiency.
Sorry this is very far from a switching regulator. The recently announced regulators are +90 % efficient.
The voltage input range will be limited by how much power the PNP transistor dissipates (the FZT968 is limited to 3W). So it can only drop about 4.5V, so add 4.5V to the Vf of the luxeon to get the maximum input voltage. The LM334 its selv can take up to 40V input.
It'll be in regulation once the input voltage is about 0.1V above the Vf of the luxeon at 750mA.
Being a linear regulator, efficiency depends on how much voltage is being dropped by the regulator.
Ok, for instance, lets say 6 fresh NiMH batteries that start out at 8.23V under a 750mA load, and it's driving a 5W luxeon with a Vf of 6.8V at 750mA.
When the light is first clicked on, the regulator is dissipating 1.43W, while the luxeon, 5.1W. So it starts out at 82% efficiency. As the batteries drain, and their voltage drops, the efficiency only gets higher. When the batteries drop to 7.5V, it's at 90% efficiency.
Overall energy delivered to the luxeon from the batteries over the life of the batteries will be just over 90%.
So while this isn't a switching regulator, it's still hitting 90% overall efficiency.
[ QUOTE ] NewBie said:
[ QUOTE ] evan9162 said:
I have a 5W with an exceptionally low Vf (6.24V @ 700mA). I chose a linear regulator because a LDO will give about the same efficiency as a step-down (or step-up) converter, but will be dead simple to build, and cheap.
[/ QUOTE ]
Okay, so a fella uses a 9V in batteries (since this can only do step down, and 123 is 1.8V basically discharged, and 1.8 V * 3 is 5.4V so, this guy goes out of regulation once the battery drops below 6.24V.
So 9V batt - 6.24V = 2.76 drop that the linear regulator has to do. 2.76V * 700mA = 1.932 Watts of heat made in the circuit. Input power is 9V * .7 = 6.3 W input power.
6.3 W - 1.932 W = 4.368 Watts delivered to the luxeon. Lets do a quick check, Lux V Vf is 6.24 * 700 mA = 4.368 W.
Okay we are on the right track. Now for efficiency, 4.368 Watts to the luxeon / 6.3 Watts input power = 69.3 % efficiency.
Sorry this is very far from a switching regulator. The recently announced regulators are +90 % efficient.
Good job though!
[/ QUOTE ]
True, initial efficiency won't be so good. As the battery voltage drops, efficiency will improve.
You'll also get lower efficiency numbers for a lower Vf luxeon - so reducing the input voltage would be good in that case - though you can't really go lower than 3 123A cells, but if you were using AAs, that would be a good sign to use 5 cells instead of 6. So with that, we'd be dropping 1.2V instead of 2.7V, so power in the regulator will be reduced to 0.8W - which would put it at 4.4/5.2 = 85%. Though then, you drop out of regulation sooner, so it's a bit of a trade-off either way.
[ QUOTE ] evan9162 said:
The voltage input range will be limited by how much power the PNP transistor dissipates (the FZT968 is limited to 3W). So it can only drop about 4.5V, so add 4.5V to the Vf of the luxeon to get the maximum input voltage. The LM334 its selv can take up to 40V input.
It'll be in regulation once the input voltage is about 0.1V above the Vf of the luxeon at 750mA.
Being a linear regulator, efficiency depends on how much voltage is being dropped by the regulator.
Ok, for instance, lets say 6 fresh NiMH batteries that start out at 8.23V under a 750mA load, and it's driving a 5W luxeon with a Vf of 6.8V at 750mA.
When the light is first clicked on, the regulator is dissipating 1.43W, while the luxeon, 5.1W. So it starts out at 82% efficiency. As the batteries drain, and their voltage drops, the efficiency only gets higher. When the batteries drop to 7.5V, it's at 90% efficiency.
Overall energy delivered to the luxeon from the batteries over the life of the batteries will be just over 90%.
So while this isn't a switching regulator, it's still hitting 90% overall efficiency.
[/ QUOTE ]
Right, lets take your six cell case. NiMH discharged is about 0.9V. O.9V * 6 = 5.4 V is discharged. In the example you used, you used a 6.8Vf Lux V this time. You started at 8.23V. 1.5V of the available power is not used before you go out of regulation (don't forget to add dropout 8.23-6.8 and then +0.1 dropout. You do use 1.33V going from 8.23V then discharging to 6.9V (add the dropout) before you go out of regulation. So you are only in regulation for less than 50% of the battery power.
High Vf Lux V gives you more efficiency, but less runtime in regulation. Low Vf Lux V gives you alot less efficiency, more runtime in regulation.
[ QUOTE ] NewBie said:
Right, lets take your six cell case. NiMH discharged is about 0.9V. O.9V * 6 = 5.4 V is discharged. In the example you used, you used a 6.8Vf Lux V this time.
[/ QUOTE ]
I can't agree with this.
If you check, you'll find that 1.1 Volts, not .9 is the traditional end voltage for NiMH, and the difference in run time between the two is nearly trivial. NiMH cells hold 1.2 volts plus per for more than 90% of the discharge cycle meaing at the start we have 6.8/(1.25 X 6) or 90%, dropping to 1.2 per, 6.8/(1.2 X 6) or 94.4% at the 10% discharged point. It *averages* over 94%!
OTOH, how many switchers actually end up over say 85% oveall at *any* point on their V/I curves? Honest numbers, not marketing. Let alone averages that over the discharge cycle.
I still think that given a good match between source and load, LDO linear regulators are impossible to beat for efficiency and ease of construction and very hard to beat in terms of cost.
And in the end, we also need to consider that there are substancial losses elsewhere in the system (for instance from driving the LED too hard) that are far larger than this.
I don't recommend this circuit casually or because it's the only solution I know.
These were Energizer 2100s being discharged at 600mA into a constant current mode - notice that the time spent under 1.1V/cell amounts to 0.11/3.13 = 3.5% of the battery life - after 1V/cell, the pack voltage drops even faster. Discharging to 0.8V/cell doesn't gain much, and in multi-cell packs, you run the risk of reverse-charging one or more cell.
Other sets of rechargables I have show similar characteristics.
IMO the idea is to not think in terms of draining it to the nubs, but to consider that a 2000 mAh cell, not 2300 and simply plan to recharge at the point where the voltage starts to slide down.
Think of it as reserve, you don't ever use the last gallon in the gas tank, right? Well some do I guess.....
[ QUOTE ] Doug Owen said:
If you check, you'll find that 1.1 Volts, not .9 is the traditional end voltage for NiMH, and the difference in run time between the two is nearly trivial. NiMH cells hold 1.2 volts plus per for more than 90% of the discharge cycle meaing at the start we have 6.8/(1.25 X 6) or 90%, dropping to 1.2 per, 6.8/(1.2 X 6) or 94.4% at the 10% discharged point. It *averages* over 94%!
Doug Owen
[/ QUOTE ]
Looks like evan9162's NiMH cells hold 1.2V for 79.9 % of the discharge cycle.
[ QUOTE ] Doug Owen said:
OTOH, how many switchers actually end up over say 85% oveall at *any* point on their V/I curves? Honest numbers, not marketing. Let alone averages that over the discharge cycle.
Doug Owen
[/ QUOTE ]
I've built quite a few that do, *especially* buck switchers.
They get more efficient at the end, as Vin gets closer to Vout. I believe you are thinking of non-synchonous switchers, where that idea generally holds true.
I used all the same values, and simply replaced the 2N3906 transistor with a Zetex FZT968, and a different sense resistor.
I found some odd behavior in this circuit. It would go into regulation fine with a very low dropout voltage, but then, after increasing Vin a bit, the current would drop until Vin was increased about 0.4V higher, then back into regulation...observe: (note that this is for a 380mA version that I built for a 1W luxeon, but the behavior is the same for all of them)
After fiddling with the circuit, taking tons of measurements, then trying a few things out, I figured that the circuit was oscillating while the input voltage was in that range.
After replacing the 330 ohm resistor in the R-C network (between the base and collector of the transistor) with a 150 ohm resistor, stability has increased greatly - here is the result:
150 ohms still isn't optimal, so I'll do some fiddling sometime to find the optimal value - so in case someone wants to build circuits like these, they need to replace the 330 ohm resistor in the RC network with 150 ohms for better stability.
I found exactly the same thing at 350 mA using a Zetex PNP who's part number I can't find here. I ended up increasing the snubber cap as I recall (for the same reason, boost the negative feedback) more than I was happy with. I then went to a higer gain version (a 717 series part) and got good results with the same 390 ohm/.1 mfd snubber that seems to always work with the 3906 at lower currents.
Otherwise you get in trouble just as the transistor goes into (or comes out of) saturation. I suspect it's a phase shift issue rather than shift in gain.
There is also, of course, a smaller total AC signal available to work with than when we've got more headroom.
In extreme cases, you can even see it dim a bit. As you turn down the voltage, it gets dim, then bright, then dim again. My guess is this is as much due to loss of efficiency (note the transistor gets warmer?) as drop in average LED current.
Not to worry about it Al, you've been tossing so many good ideas around here for so long yer bound to forget one or two along the way...
I'm not exactly sure what Evan is using for his shunt, but I was using about two feet of wire wrap wire IIRC. By Ohm's law, 180 miliohms or so? That is I was running about 350 mA through the 1 Watt LS.
For the sense resistor, I used about 180 milliOhms for 380mA, just using some tiny, epoxy coated copper wire wrapped in a coil (I wrapped it around a spray can tube). I'm not sure what the wire gague is. I think it's about 10" of this stuff.
For the 750mA, it's about 80 milliohms of resistance.
With the 380mA version (the only one I've extensively tested), I get about 0.11V dropout right at the point of regulation. Vce is 0.037V for the FZT968 (hFe=75, Ib=5.1mA), and the sense resistor drops about 0.07V.
Thanks... that sounds pretty good to me.
I think a dropout of +0.1 volt over led voltage
is pretty darn good!
If you want to reduce oscillation try winding your
'sense' resistor forwards, then backwards, etc.
In other words, go around the core form once then bend the
wire carefully and wind in the other direction. You
still get two turns right next to each other, but every
other turn is going in the opposite direction. This
is the way some resistors are wound to reduce inductance.
BTW the only reason im mentioning this is because if
you replace the sense resistor with an inductor you
probably get an oscillator :-)
evan:
If you give me the dimensions i can tell you the approximate
inductance, but even a few microhenries or less can cause
oscillation. It's not a matter of how much inductance
is present, it's more about the circuit topology and
the switching speed of the components combined with
the inductance. With a small inductance, the only
difference is the circuit will oscillate at a higher
frequency than with a lower inductance.
Generally you dont want a linear circuit oscillating
if you can help it. Sometimes it doesnt matter but
if you can easily stop it you should.
Sometimes a circuit will oscillate about a given
operating point. In this case, it looks like it's
working normally until you look at the output wave,
which could be banging the LED with high current
pulses in order to achieve some average current level,
when it should be stable and at a nice, safe constant
current.
Newbie:
That's not a bad idea either, i'd try it that way
myself. It's a little easier to handle the wire.
If i did the estimate correctly you might end up with
as much as 0.7 micro henries with that coil.
Not a lot, but then oscillation doesnt require a lot :-)
If i get time tomorrow i'll plug an inductor in place
of the sense resistor and add 0.180 ohms series resistance
and see what frequency it oscillates at. Hopefully
the cap swamps most of the oscillations. I would guess
something in the 1MHz range wouldnt be surprising.
Perhaps you can hold it next to a radio, then a tv set,
and see if there is any disturbance in the appliances
sound/picture. Because it will most likely be a square
wave (if it is really oscillating) it will be sending
out harmonics way up there, so the radio or tv set should
pick up something, even if it's just a few lines in
the picture.
Newbie's solution sounds pretty good though if your
wire is insulated (like enamel). If it's not
insulated you'll have to use my suggestion and keep the
turns away from each other.
If you're getting the right current level and you dont
see any interference on the tv set then perhaps the
cap is swamping all the oscillations in which case
you wont even have to worry about it :-)
It's good to know it could happen though, and it may
break into oscillation at a given current level,
possibly high, low, or somewhere in between.
I've been bench-testing a version of this regulator on a breadboard - for the sense resistor, I have 3 0.47 ohm resistors in parallel for 0.156 ohms - giving a set current of 420mA. Removing one bumps the current down to 280mA.
For the RC network, I have a bunch of different capacitors, and a 1K pot. I wasn't able to obtain the same stability as my built circuits, even with the same 0.1uF, and adjusting the pot to obtain maximum current - the current dropped off gradually as the input voltage was increased.
Is it possible that the inductance from the coil is aiding in supressing some of the oscillation? Or could the construction difference (breadboard and wires vs. soldered PCB) account for the behavior difference?
evan, it starting to look like maybe the resistor
and cap would be better placed on the sense pin of
the ic chip. I took a quick look at this tonight,
and probably have more time tomorrow.
Anyway, when i put the cap across the "R" terminal
(on your schematic you posted in this thread)
and the 390 changed to 100 ohms in series with
the sense resistor, i reduced oscillations even
with higher value inductors.
Of course this requires a bench test...
The test would consist of the following...
1. remove the series 0.1uf cap and 390 ohm resistor.
2. place the 0.1uf cap across the "R" terminal
(on your schematic)
3. break the connection between the 'current sense'
resistor and the "R" terminal, and place a 100 ohm
resistor between the "R" terminal and the 'current
sense' resistor.
Test with a load and various voltage inputs.
This might actually work better then having the cap
and resistor where it is now, although it does need
actual bench testing. If you are able to try it
please let us know how well it works.
I'm not sure where to put the resistor - put it between R and GND? or R and Vin?
I did try the 100 ohm resistor between the R pin and the sense resistor - this caused the LED current to drop to a little less than half as before.
The 100 ohm resistor is restricting the base current of the transistor to 0.4mA, so is increasing the voltage drop of the transistor to about 0.9V. Meanwhile, the voltage drop of the sense resistor is reduced to 0.025V, so it's a bit of a wash in the dropout voltage department. Also, current slowly rises as the input voltage rises, since the base current is being reduced, and thus, the voltage drop of the sense resistor must increase to make up the 0.065V required between R and V- of the 334.
I've also found that (at least in my test circuit) replacing the whole RC network with a 2.5uF capacitor makes the circuit stable.
Yes, the 100 ohm resistor would go between the top
of the sense resistor
and the terminal marked "R" on your schematic.
You also need the capactor, from the "R" terminal to
ground.
Of course before you can connect the 100 ohm resistor
you need to break the existing connection between the
sense resistor and the "R" terminal. Instead
of a direct connection there you will now
have the 100 ohm resistor.
This is just something to try, as the simulation showed
less oscillation for larger inductance in the leads
of the sense resistor.
Come to think of it, i think i had another circuit that
required a small 'low pass filter' like this to be
inserted before the 'R' terminal of the LM334.
I'll have to look up that info next...
Ya gotta remember that there's current comin' out (or going into depending on what you know to be true...) the R lead. Basically it's the base current from the external PNP pass element. Expect a voltage drop across any series resistor between it and the shunt that will lower the total current.
750mA LDO current regulator for luxeons
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