Constant current for 16 white LEDs?

snipped-for-privacy@att.net wrote in news: snipped-for-privacy@m73g2000cwd.googlegroups.com:

Google "LED driver IC".

--
Jim Yanik
jyanik
at
kua.net

PWM

Hi Bill, this might be a bit too crude for your liking, however, it may be worth your time to look..

A boost converter should work, but you must make sure the output voltage does not fall below the total forward voltage, unless you want a PWM brightness control. You can tie all the LEDs in series, which will require

48 VDC, and make a boost converter that puts out at least 50 VDC. The tricky part is to regulate the current without burning up too much power in a series dropping resistor. It is better to use inductance to maintain a constant current.

You could use a 5 volt to 15 VDC DC-DC converter, and add a current regulator, to drive 4 strings of four in series.

I am working on a design to drive up to 12 high power white LEDs* (3 watts each, at 700 mA, 4.5 Volts), from a 12 VDC SLA battery. I worked out some boost circuits using the Tina simulator, and it looks promising. I will be building an actual prototype soon. The same basic design could be used for your application. I can send you a schematic you may use, if you have a copy of Tina. You can get a free demo, or purchase a copy for about $30, from

or get a free limited version at

(See the CREE website for Cree XLamp3_7090.pdf)

Paul

Well, the problem with the boost converter is the ramp current won't increase past a few hundred milliamps before the transistor comes out of saturation. I have tried several inductors and transistors and ended up with the same problem.

I am experimenting with a large 25mH inductor with low resistance of

200 milliohms or less and a power MOSFET (IRF620). I am driving the gate with a (0 to 8 volt) squarewave at about a 800Hz rate. It works well at a 1/4 watt power level, but if I try and increase the ramp time by lowering the frequency (from a generator) to obtain more inductor current, the transistor comes out of saturation and starts dropping a large portion of the supply voltage and the efficiency goes way down. The supply voltage is 4 volts.

I must be missing something fundamental. As I understand it, inductor current will increase with time and stored energy is 1/2 LI^2. So, to increase output power, the time should be extended so that more current flows into the inductor. But that idea only seems to work within a small range.

-Bill

The boost converter I simulated (and also breadboarded) uses a 555 timer to drive the gate of a MOSFET. I used a 5 uH inductor at about 100 kHz, and also a 50 uH inductor at about 20 kHz. These inductors are about 1" long and 1/2" diameter. I plan to use a toroidal inductor about 3/4" dia and

1/2" thick. These inductors have DCR in the uOhm range, and can handle current up to several amperes. The overall volumetric efficiency increases with frequency up to a point, where switching losses and other effects become important. My circuit essentially pulls one side of the inductor to ground, allowing the current to build up to several amperes (but below saturation), and then opens the connection to allow the current to charge a fairly large filter capacitor. The output voltage ratchets up over a period of time until it reaches a maximum, which depends on the load and other factors. I added a sampling resistor in series with the LED load (simulated with a zener and a resistor), and drive an NPN transistor with its collector to the RESET line of the 555. When the voltage is high enough to make the zener conduct (or the LEDs light), current will increase until the transistor shuts down the 555 until more voltage is needed. Remember that you will need to store a lot more current in the inductor than what you want to get out. For your 4.5 VDC to 48 VDC at 20 mA, assuming a 50% duty cycle, the peak inductor current may need to be as high as 40 times the output current, or close to 1 ampere. For my circuit, I was looking at peak currents up to 50 amps, so I needed a MOSFET with a very low Rds-On. I will be breadboarding a new version soon. A rough schematic is as follows:

L1 D2 R1

+12VDC----+----+-UUU--+--->|---+--/\\/\\/---> To LEDs | | | | | D1 - --D-- | Q1-C LEDs return | To Q1-B

Short answer is PWM.

Long answer is to pulse 3 sets (5 to 6 each) of them at 25% to 33% duty cycle, adjusted with battery voltage.

Have you looked on the Elektor UK site? ISTR a chip that does just that.

...

Yes, I have a similar circuit but the problem is getting enough inductor current at low voltage. If I raise the supply voltage to 9 volts, I can get 60 volts out into 2700 ohms, or 1.3 watts which is enough for this application. Input current is about 190mA, so the efficiency is about 1.3/1.7 = 76%

But it won't work at low voltage of 4 volts, and that's the problem. Why does Vce Sat increase as supply voltage decreases?

-Bill

Bill, Your MOSFET's the problem -- 4V is not enough to turn an IRF620 fully on. If you look at Fig. 1 on page 3 of the data sheet, you'll see that it's not even good for 200mA at Vgs=4.5v.

I'm sure I have a suitable FET I could spot you, or you could substitute an appropriate bipolar transistor.

Best regards, James

I don't think the problem is the FET since I am driving the gate with a

0-8 volt square wave from an external generator. The on-resistance is 180 milliohms. But the same problem occurs with a bi-polar transistor driven with 100mA of base current. The inductor current never ramps high enough before the voltage drop on the whatever transistor used starts climbing and overheating the device. It won't stay saturated with a low Vce drop using a 4 volt supply. But it will work at higher voltages.

The question is, how do i get 1.5 amps or more into an inductor of low resistance (200 milliohms or less) using a 3 to 4 volt supply without overheating the transistor?

Something funny is going on with the inductor at low voltage, but I don't know what it is. I have tried several values from 3 to 25 mH inductors.

-Bill

If the on-resistance of the switch is low, it shouldn't get very hot.

Are you trying to make a discontinuous conduciton boost mode switcher? Charge time with no resistance should be t = i*l/v with v=battery voltage. Given resistance in the switch and inductor, the charge time will be longer. For discharge, it is the same equation, but V is the voltage across the inductor, which should be the voltage that the diode clamp the output less the battery voltage. With just a diode string out the output, I would expect the discharge to be faster than the equation.

While on the surface, a good design would seem to one where you just build up current in the inductor then dump it, I think in practice a simple voltage output boost converter will be more useful since you can LDO the ouput voltage to power the gate drive of the power fet. You could add current limiting, so long as the resultant output voltage is sufficient to drive the LDO.

I wondered that too, but took Bill's earlier response to Paul Schoen to mean Bill's got some sort of current-limiting or duty cycle control is his actual circuit.

Bill, your setup is perfectly ordinary and should work. Something is wrong.

Driving with a signal generator you'd better be sure the inductor fully discharges during each discharge pulse, otherwise current will accumulate from one cycle to the next--reaching very high levels--until something breaks (or screams).

The inductor-charging time needed increases as supply voltage falls.

Vce(sat) does _not_ increase as Vcc falls; it increases with higher collector current (or Vds/drain current). Therefore, either you're not driving it hard enough, it's oscillating, or you're passing a lot more current than you think.

Maybe you're hitting continuous mode. What's your drive's duty-cycle? Frequency?

Maybe your inductor is saturating. Substitute a power resistor, and see if your transistor can drive it. Or, leave the inductor in and watch the drain current with a 'scope and see if it spikes after ramping smoothly. That's very possible--your inductance is huge.

Maybe you've underestimated the peak inductor current needed to get ~60v @ 20mA from 4V ? You'll only be discharging at most 1/15th of the time, so you'll have to discharge at 15 x 20mA (300mA) to get the current you want in continuous mode, possibly much more in discontinuous mode.

In this situation I'd prefer a few series LED strings in parallel -- it improves reliability (redundant LED chains, reduces peak currents, eases switching stress).

Hey, speaking of switching stress, please tell us you're using a suitable fast-recovery rectifier -- NOT a 1n4001 ;-)

Best, James Arthur

Actually, I just noticed he said he was using a square wave. In a boost converter, discharge will take less time than charge, thus there is a need to change the duty cycle.

I did a boost converter control scheme that used a current sense circuit to limit the charge cycle, and a comparator to detect the end of the discharge cycle. The end of the inductor that flies high can be used to detect the end of discharge when it falls lower than the battery voltage. It is simple, though there are plenty of gotchas I could go into if he picks such a scheme.

Now in the case of driving leds, I think you could build a controller strictly based on current sensing. I haven't done this, but here is the scheme. Pick a peak current limit to charge the inductor. When the peak current limit is reached, let the inductor fly. The end of the discharge cycle would be detected by a lower current limit being reached. Say the upper limit was 20ma, and the lower limit was 10ma. Then you would generate a triangle wave of current in the inductor between 10ma to 20ma.

All that said, I'm more comfortable using a voltage limit on the boost, with some trick to limit the current in the leds. Discontinuous boost converters are stable beasts. I still like using the boost output to feed a LDO which in turn controlls the gate drive of the power fet.

Yes, so he's not in continuous mode. Either he's saturating his inductor, or there's something he's not told us.

[snip]

Remember, the inductor discharges only briefly; average LED current = (average inductor discharge current) * (inductor discharge duty cycle). Here, ignoring losses, he'll need about 15x, or 300mA average discharge current for 20mA @ 60v in the LED chain. That is, 300mA peak inductor current is needed in continuous mode (0% ripple current), or twice that if the inductor discharge time exactly discharges the inductor (100% ripple current).

No need for LDOs--a current-mode boost works great.

Best regards, James Arthur

I think it's running continous mode at best performance. The scope waveform is a series of clean pulses (20 volts peak into 560 ohms). If I let the inductor fully discharge where the waveform falls to the supply voltage (3 step waveform) the output power is too low. It seems to work better if the inductor doesn't completely discharge.

Yes. T=LI/E, so as E goes down, T goes up to keep L and I constant. But for the same power level, I must also increase as E falls.

The supply current reads 220mA at 4 volts, or 880 milliwatts.

Cycle time is 390uS. 20 volt output pulse width where the inductor discharges is about 60uS.

The inductor core measures 1.5 inch diameter by 1/2 inch high and is wound with around 100 turns of 18 gauge enamel copper wire, resistance is less than 200 milliohms. The inductance measures somewhere between

20 and 30 millihenrys depending on how you measure it. I resonated the inductor with a 1uF cap in series and parallel and got different results. But it's a big inductor, around 20-30 millihenrys, probabably not saturating at 700mA.

Well, I figure 16 LEDs at 3 volts each and 20mA should be about 1 watt total. So, with a 4 volt input, I need 250mA or a little more to cover losses. In continuous mode, the duty cycle is about 3:1 so the peak inductor current should be about 700mA. The thing is running at 230mA (DC input) and I get 20 volts into 560 ohms which is a little less than

1 watt, but in the ballpark, but I can't get any more power out. That's the problem No matter what I tweek, (frequency, duty cycle, etc.) the output power won't go past 1 watt with a 4 volt input.

Yes, I'm using a 1N4001 because the frequency is less than 3Khz, so there should be plenty of time to turn a slow diode on and off.

-Bill

The LDO is to safely limit the VGS drive, should he go bootstrapped power fet rather than bipolar. I think this was 4V design, but I don't recall the final operating voltage. That is what is nice about the circuit that self times (i.e. sets the current based on a peak level). I've also designed switches where you build timers based on the supply voltage and output voltage, i.e. something like an analog computer. But the self timed operation I described works well, at least in how I implemented it for a voltage output boost.

The self timed circuit is also independent to a some degree of the inductor value and ESR. It just has lots of gotchas to look out for. Start up, things like that.

[snip]

Ah, gotcha.

I've designed "self-timed" converters similar to what you describe, but can't comment. Here's one, however, you might enjoy:

Best, James Arthur

[snip]

1/390uS = 2,560 Hz.

[snip] [snip: 1.5"x0.5" inductor, 100t of #18, d.c.r. = 0.2 ohms]

You left out an important detail--the core material--but we'll assume for now it's a hardy beast and not saturating.

[snip]

Hey, that doesn't sound like squarewave drive! Squarewave means 50% duty. And no fair changing the load--560 ohms is not 16 LEDs in series.

3:1 drive duty (~75%) notwithstanding, the max duty cycle of your converter must equal the ratio of inductor charge time to discharge time, which is set by the input and output voltages--you don't get to choose! You can choose lower duty cycles, but must then run even higher switching currents to compensate.

Your 2.6KHz frequency is much too high for this new Vin/Vout combo unless you're using continuous mode, and if you try to drive it in continuous mode with a signal generator, the inductor current's going to run away. That's your problem.

To charge a 30mH inductor from zero up to 700mA from 4V will take: (30mH*0.7A)/4V = 5.3 milliseconds.

Discharging from 700mA to 0 at 20 volts will take (30mH*0.7A)/(20V - Vcc) = 1.3 milliseconds. Frequency of operation = 152 hertz.

Duty cycle = 5.3 milliseconds / (5.3 mS + 1.3mS) = 80% (on), 20% (off).

In practice, "on" time will have to be longer to make up for losses.

You can switch faster than 152 Hz and attain the necesary inductor current using a continuous mode converter, but you *must* have feedback to keep the inductor current from running away, and preferably a current-mode converter (cycle-by-cycle current limiting on the big FET) for stability.

[snip]

You might be surprised, but I think a 1N4001 may be too slow for even this. Let me check the recovery time... Okay, John Fields measured

20uS, and Helmut Sennewald reports 30uS, so I suppose you slide THIS time ;-)

Best regards, James Arthur