dx 7880 regulator efficiency

lolzertank

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I'm planning to get a Romisen RC-G2 and replace the driver and led with a cree xr-e q5 and a this 20 mode driver(sku 7880). Does anyone know the efficiency for this driver? Also, could you post the current going to the led? I know the specs say 800ma, but I want to hear some real measurements since I'm not so sure i trust dealextreme's.
 

VegasF6

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Hello, and welcome to CPF.

There was a similar question to this posted just a few days ago in fact. I would suggest you do a search at the top of the board here for 7880. Do a little reading and come back with any specific questions.
 

lolzertank

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Do you mean the first post on this thread? https://www.candlepowerforums.com/threads/188227
"3W/5W" 1x AA. $3.96. Boost board; output voltage and current not constant, approximately 0.8-1.5 V in. 1.0 V in gives 0.3A out, 1.4 V in gives 0.7 A out. Run on two fresh NiMH it gets very hot - not advised. 20 modes in three groups.

Combining what it says with other posts saying it draws about 2-3 A from the battery, would this mean this board has about a 70% efficiency? Hope my math is correct...:thinking:
 

VegasF6

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It seems people have had different experiences with that board. I have seen reports of outputs at 600 mA as well. And I now see on the driver thread there are actually 2 different links to it, 1 mentioning 1xAA and one says .8-7V in. I wonder if that one is a typo?

There are probably different ways to figure efficiency but what I do is simply figure watts going into the circuit and watts going out, and then a percentage of such.

So, if it takes for instance.

1.2Vx3A=3.6W in.
3.4Vx.8=2,72W out.
2.72/3.6=75.55

Rizki_p who is also a member here reported "Am i the only one who get 0.3A output and 0.8A draws from the battery, mine doesnt draws 3A like most people indicated. " on the DX comments there.

Efficiency will be effected by resistance in the circuit as well including how you measure the current. Others have mentioned a better way to figure the current at the led by measuring the voltage and somehow figuring the drop using ohm's law, I don't recall how. I think you had to measure it against a known value resistor. I have been measuring it through a shunt meter which does introduce resistance and skew your #'s a bit.

This link here shows 2 different equations for efficiency:
http://www.mobilehandsetdesignline.com/howto/207000658

The #'s I gave above are pretty rough figures. There are other members much more versed in electronics that could explain better.
 

lolzertank

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Thanks everyone for your help. I've ordered it, but its backordered. I'll probably take my own measurements when I get it. Hopefully it won't output 0.3 A like Rizki_p's. 75% efficiency doesn't sound TOO bad for a 1xAA boost circuit.
 
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