Max power through a hipFlex- Mtbr.com

# Thread: Max power through a hipFlex

1. ## Max power through a hipFlex

Hello

I cant seem to get onto the Taskled website and need a little info. Basically, can I draw 56W through the hipFlex?? ie 2.8A and 19.8v output? I realise this will need uber heatsinking! Will this be reliable?
If not, can you recommend another driver to use?!

I want to make a one off silly bright light to satisfy my craving then I will calm down!!

Thanks for any help
Andrew

2. Andrew
Drop me an email and I will send you the hipflex manual if you want it.

3. TaskLED.com is backup for me and I can get the manual here:

4. Thanks Chris and Chris!

I have the hipFlex.pdf and it says...
1 to 5 P7 (in series) 24 V
1 to 5 MC-E (4P) 24 V
1 to 3 MC-E (2P2S) 24 V

However, can I run 6 MC-E in 4P6S?? 4 Strings of 6 dies? If the Vf of the die is ~3.3v my total Vf will be ~19.8V. At 2.8A so each die sees 700mA. I'll be using a 22.2v Li-ion to power it.
Therefore my total power is ~55.8W.

Is this feasible or is this simply too much for the hipFlex? Or would 2 drivers be a better option? If so, any ideas?!

Thanks again

5. However, can I run 6 MC-E in 4P6S?? 4 Strings of 6 dies? If the Vf of the die is ~3.3v my total Vf will be ~19.8V. At 2.8A so each die sees 700mA
each string will see 700ma ......... not each die

6. Hendo - I get 700mA per die given an ideal current source... can you please explain your logic? The breakout is symmetric and parallel, so how do you find each die in a string does not carry the string current?

7. 4 Strings of 6 dies?
Is this feasible or is this simply too much for the hipFlex?
hipFlex max current = 2.8A / 4 strings (of 6 dies) =700ma per string (of 6 dies)

8. Originally Posted by HEY HEY ITS HENDO
hipFlex max current = 2.8A / 4 strings (of 6 dies) =700ma per string (of 6 dies)
Cool. So how much current do you calculate per die if not 700ma? I thought current would be conserved along a string of dies when wired in series?

Thanks,
-b

9. Originally Posted by ca_rider
Cool. So how much current do you calculate per die if not 700ma? I thought current would be conserved along a string of dies when wired in series?

Thanks,
-b
Yeah, I worked it out at 700mA per die too.

So do you reckon I hipFlex can handle 56W?? I may have to wait for George to get back!

10. I may have to wait for George to get back!
lol, you don`t believe me eh!
we can work this out easily
hipFlex spec ........ max volt in =24v, max current out = 2.8A
if you want to drive 6 x MC-E then ahhaah ....
......4p6s THIS IS NOT 4 STRINGS (this statement threw me)
each MC-E 4 die parallel------ 6 MC-E
=one string of 6 leds, does this make sense?

nightmare2wire.jpg

Each die (and string) would indeed see 700ma

Rather you than me! Very fiddly to solder all that up......GREEDY BOY

12. the (4) MC-E dies will be joined in parallel
the 6 will be joined in series

22.2v Li-ion to power it.
....... fully charged voltage is?? .........

13. 6 series li-ion should be 25,2V off the charger.

Originally Posted by HEY HEY ITS HENDO
the (4) MC-E dies will be joined in parallel
the 6 will be joined in series

....... fully charged voltage is?? .........

14. yep, thats what i was thinking too brum ...
The IN+ hole (bottom right corner) and IN- (middle right) are the input power connections to hipFlex. 24V MAXIMUM input voltage.
piesoup .....looks like your gonna need 2 drivers and a 5s2p battery ( 5cell x 2)

15. Hendo, I still don't understand your response to my question. Can you please elaborate?

Given the proposed hardware, I would arrange it as 6p4s. This gives 2.8A/6_leds=467mA each. Referencing the MC-E pdf, this gives a 3.35V drop across the leds. With 4 sets of 6 leds in parallel, this yields a total voltage drop across the array of 4*3.35V=13.4V. The total electrical power drawn is 0.467A*6_leds*4_sets*3.35V_per_set=37.5W. Note the power and voltage requirements are satisfied by a single hipflex and the specified battery pack. Referencing the hipflex pdf, 467mA yields 125% luminous flux which is 370lm for group K using all 4 elements. So in this case total luminous flux would be 125%*6_MC-Es*370lm_per_MC-E=2775lm.

Cheers,
-b

16. Originally Posted by bumphumper
nightmare2wire.jpg

Each die (and string) would indeed see 700ma

Rather you than me! Very fiddly to solder all that up......GREEDY BOY

Spot on, thats how I drew it too. It will be some fiddly soldering but I think I can manage that! With individual die access I will be able to achieve that circuit.
So, I would still have a P out of ~56W and 4410lm. 78.75lm per Watt.

However, I have got M bin which will be 125% * 6 * 420lm = 3150lm. 84lm per Watt. I think I may be satisfied with that!! Thanks ca_rider, I may go with your proposed set up!

Yes a fully charged battery pack will be 25.2v so I may use a 5 cell li-ion, with a max of 21v. I'll most probably end up using the light at a lower current most of the time so the driver wont be working flat out all the time.

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