Blue Shark with custom resistor?

[MrNaz]

I am about to order 4 Blue Shark drivers, is it possible to have them ordered with 0.09R resistors? I would like to aim for an output of about 1.2A. If not, can I order them with no resistors and then add my own?

I will be driving 4 series connected XR-E LEDs from a 7.2V LiIon battery pack. I will be providing good heat sinking for both the driver and the LEDs.

Is the Blue Shark built to take this current? I notice that on your web site 1A is the max option that is listed, but I read in a thread that the old Shark would take a 0.05R resistor and output 2A. Is this also true of the Blue Shark? Can it handle this sort of load?

[dat2zip]

I am about to order 4 Blue Shark drivers, is it possible to have them ordered with 0.09R resistors? I would like to aim for an output of about 1.2A. If not, can I order them with no resistors and then add my own?

I will be driving 4 series connected XR-E LEDs from a 7.2V LiIon battery pack. I will be providing good heat sinking for both the driver and the LEDs.

Is the Blue Shark built to take this current? I notice that on your web site 1A is the max option that is listed, but I read in a thread that the old Shark would take a 0.05R resistor and output 2A. Is this also true of the Blue Shark? Can it handle this sort of load?

Adding your own resistor??? The resistor is a SMT 0603 resistor and at this size configuration there are not a lot of values to choose from. You can solder a 2nd SNT resistor on top of the 0.1 ohm resistor to lower the value to up the max output.

The actual limiting factor is not the output current, but, the input current under worst case situation. This occurs at the lowest operating battery voltage and the Blue Shark should be kept under 3A. With a 7.2V battery pack which appears to be two li-ion batteries then the low voltage or near dead condition is around 3V/cell or 6V. With 4 LEDs driven at 1A each the Vf will be around 3.5V or so. That's 14V or so for the output. 6V and 14V is just over a 2:1 ratio. Taking into consideration of converter efficiency (worst case) puts you in the 770mA output range.

Of course good heatsinking and keeping the heatsink cool will allow you to achieve numbers higher than this, but, this is the ballpark napkin calculation.

Your actual configuration and testing will be the only way to determine if it will work or not since there are a lot of variables here.

Wayne

[MrNaz]

Adding your own resistor??? The resistor is a SMT 0603 resistor and at this size configuration there are not a lot of values to choose from. You can solder a 2nd SNT resistor on top of the 0.1 ohm resistor to lower the value to up the max output.

I've decided that leaving the standard resistor on there and taking off the trimpot is the way I should go. 1A is enough for my application, and I don't need the trimpot, so that should be fine. This is what I've ordered from The Shoppe.

The actual limiting factor is not the output current, but, the input current under worst case situation. This occurs at the lowest operating battery voltage and the Blue Shark should be kept under 3A. With a 7.2V battery pack which appears to be two li-ion batteries then the low voltage or near dead condition is around 3V/cell or 6V. With 4 LEDs driven at 1A each the Vf will be around 3.5V or so. That's 14V or so for the output. 6V and 14V is just over a 2:1 ratio. Taking into consideration of converter efficiency (worst case) puts you in the 770mA output range.

Er, do I understand right? 770mA is the current being drawn from the battery at low voltage? Surely not, surely the current draw from the battery has to be higher than the delivered current? I calculate:

(14.5V/6V) = 2.4 boost factor at worst case.
(2.4 * 1A) / 85% = 2.85A drawn from the battery when the battery is low.

Is that correct?
Also, how would I determine how much heat the driver is dissipating?

Of course good heatsinking and keeping the heatsink cool will allow you to achieve numbers higher than this, but, this is the ballpark napkin calculation.

Heat is a non-issue in this application. The driver will be mounted directly to the torch body using the Mag D shark sinks (I'm sure you've seen my order from the Shoppe by now ), which is a single large piece of aluminium with heat fins, and the light is a dive light so it will be used underwater. Here's the application:
http://www.mrnaz.com/?s=publish-gallery&album=57

[dat2zip]

I've decided that leaving the standard resistor on there and taking off the trimpot is the way I should go. 1A is enough for my application, and I don't need the trimpot, so that should be fine. This is what I've ordered from The Shoppe.

Er, do I understand right? 770mA is the current being drawn from the battery at low voltage? Surely not, surely the current draw from the battery has to be higher than the delivered current? I calculate:

(14.5V/6V) = 2.4 boost factor at worst case.
(2.4 * 1A) / 85% = 2.85A drawn from the battery when the battery is low.

Is that correct?
Also, how would I determine how much heat the driver is dissipating?

Heat is a non-issue in this application. The driver will be mounted directly to the torch body using the Mag D shark sinks (I'm sure you've seen my order from the Shoppe by now ), which is a single large piece of aluminium with heat fins, and the light is a dive light so it will be used underwater. Here's the application:
http://www.mrnaz.com/?s=publish-gallery&album=57

the calculation I was showing for worst case shows that driving more than 770mA might put you at risk of pushing the driver to hard thermally and that may cause problems. Under the conditions of the calculations and the stock sense resistor set for 1A may be at risk of thermal shutdown.

Power dissipated in the converter board is dependant on converter board efficiency. At any given time converter board efficiency runs from 60-65% efficiency. For worst case assume in ther 60-70% efficiency.

Let us pick 70% for example. That means 30% is absorbed in the converter board.

Power in the LEDs (load) is Vf * LED_Current or 14.5V * 1A or 14.5W

30% of 14.5W will be the power lossed in the converter board or 4.35W. Total input power will be the sum of the two or 14.5W + 4.35W = 18.85W.

Power = Voltage * Current.

or more precisely

Input_Power = Input_Battery_Voltage * Input_Battery_Current

If we now solve for battery current substituting in Power and voltage we have.
18.85W = 6V * Current
3.14A = Current

Input current computes to 3.14A under these conditions.

This is just a guess. Only real measurements will give you more accurate numbers.

Also, remember that how the Shark is attached and where the heat sink for the converter board is mounted will have a major impact on converter board efficiency. As the converter heats up efficiency goes down. It is very non-linear and rises almost exponentially as the converter heats up.

Wayne

[MrNaz]

the calculation I was showing for worst case shows that driving more than 770mA might put you at risk of pushing the driver to hard thermally and that may cause problems. Under the conditions of the calculations and the stock sense resistor set for 1A may be at risk of thermal shutdown.

Aah, so that was a napkin calculation of the theoretical max safe current that can be delivered to the LEDs. That's fine and everything, but I didn't carve my flashlight out of a single lump of metal because I wanted to drive my LEDs at 770mA

Also, remember that how the Shark is attached and where the heat sink for the converter board is mounted will have a major impact on converter board efficiency. As the converter heats up efficiency goes down. It is very non-linear and rises almost exponentially as the converter heats up.

If I have excellent heat sinking (and I really mean excellent, not merely "pretty damn good"), could I safely push 3.2A - 3.5A into the driver on the input side? I really do want to get in the area of 1.4A into each of my LEDs.

I'm driving a 4S array of XR-E LEDs using a 2S set of Li-Ions. So I need about 14V out of as low as 6V. Have I perhaps selected the incorrect driver for the purpose?

Otherwise, I could wire up the LEDs in 2S2P array and then push 2.8A at them. Could the shark do that from a 2S set of Li-Ions? Could it be configured to push current that high on the output side?

From this discussion, it seems that I may have selected the wrong driver for my application...

[dat2zip]

MrNaz,

Maybe the wrong driver. I think I did not stress strong enough that boost drivers are more difficult in power management. It is more efficient for a buck driver than it is for a boost. It is a well established fact in the design community that step downs are more efficient and do not tax components as hard as a step up converter. Notice the Shark Buck has a 3Amp output rating and uses the same driver IC as the Blue Shark which can on a good day muster 1A to the output. That's if the wind is blowing in the right direction and the sun and moon are in perfect ying yang alignment.

If you had any choice at all in the matter always side on the step down converter and only use a step up when there are no other solutions.

Always run LEDs in series is always the primary choice above all else.

You might be better to use two Drivers vs one ans split the drive requirement among two drivers cutting the power/heat factor in 1/2.

I think an old wise man once said:
"When in doubt buck"

Wayne

[MrNaz]

If you had any choice at all in the matter always side on the step down converter and only use a step up when there are no other solutions.

Always run LEDs in series is always the primary choice above all else.

You might be better to use two Drivers vs one ans split the drive requirement among two drivers cutting the power/heat factor in 1/2.

I think an old wise man once said:
"When in doubt buck"

Yes, I suspected this all along, but I guess I needed someone to explicitly say it to me.

I like the idea of using two drivers, but I'm not entirely sure how to do that. I have 2S Li-Ion battery pack and 4 LEDs. Splitting the LEDs into 2 means that as soon as the Vbatt falls below Vf of each parallel arm of the LED split, the power goes into DD, and I'd rather avoid DD.

I have even considered the idea of using 4 buck drives, but then I have Vbatt being stepped down by a 2:1 factor, which is also a pretty big step down. Can buck drivers step 8.4V down to 3.7V efficiently?

Ideally, I would like to use the batteries to power the LEDs in 2S2P configuration, but that would require a driver that could both boost and buck over the cycle of the batteries. This would be ideal, as there would not be a large difference between Vbatt and Vf. Does such a thing exist?

[dat2zip]

I believe you can wire the two Blue Sharks in parallel and set each one for 1/2 the output current. One can be the master and control the other one as a slave. Multiple slaves can be hooked to one master. This works even with the Remora. The one with the Remora would be the master and the others would be slaves.

There should be a thread in the blue shark first thread I think that links to a parallel configuration.

Wayne