DIY LED driver for reef lighting

If the PWM signal is set at 100% and Vin is higher than Vout + .5v then, yes, you'll get the full nominal current specified by Rset.

You really shouldn't have to worry too much about complicated test rigs. Just pick the Rset value based on the formula in the datasheet, and adjust your DC supply's output to half a volt above the Vf of your LEDs.
 
I would add one further note to DWZM info above. You will have to adjust the potentiometer on the power supply if you decide to increase the amount of light you want the LEDs to output. THis is required becuase the CAT4101 won't be able to increase the voltage (to increase the current) without the changing the power supply.

Also if you decrease the light level you might want to consider turning down the voltage from the power supply. This is strictly for an efficiency sake. If you make a drastic change and the light start to blink on and off tis could mean the CAT4101 is going into thermal regualtion. In this case turning down the voltage may fix the problem.
 
Thanks for the refinement. I would refine your refinement (ha! :) ) by adding that the only changes that should trigger you to adjust the DC supply's voltage would be if you changed Rset or changed the LEDs in the string (to LEDs with higher or lower Vf at the target current.) If you "adjust the power" by way of changing the PWM signal, you should NOT adjust the power supply's voltage, otherwise you'll either be wasting power (if you turn it up) or clipping the IC's regulation function (if you turn it down). Again, it's important to remember that the PWM signal doesn't change the current the driver operates at, it only tells the driver to switch rapidly between zero and it's nominal current, giving the appearance that it's changing the current.
 
der_wille_zur_macht said:
If the PWM signal is set at 100% and Vin is higher than Vout + .5v then, yes, you'll get the full nominal current specified by Rset.

You really shouldn't have to worry too much about complicated test rigs. Just pick the Rset value based on the formula in the datasheet, and adjust your DC supply's output to half a volt above the Vf of your LEDs.
Click to expand...
So, for the XP-Gs I'm seeing a forward voltage of 3.2V @ 700mA. With 6 I'm looking at 19.2Vf + .5V is 19.7Vf?

When I set my Meanwell to 20V and the PWM to 100% I'm seeing 540mA.

Should I just ignore the mA reading I'm getting and assume I've got the V set right?
 
Where did you get the 3.2 V and 700 ma from? The chart in the data sheet? Those are typical numbers.

can you turn the meanwell down to 19.2 and measure the current in your string?
Can you measure the voltage across the string that the CAT4101 is providing?

My guess is either the CAT is not providing 19.2 or your string is not typical and it will measure 540 when run at 19.2 volts.
 
Few thoughts.

1) What value of Rset are you using? Make sure the value you're using is appropriate for your 700mA target current.

2) If you're getting less current than you think you should by way of mathematically calculating the Vf of the string, forget about the maths and just keep turning Vin up until you see your target current. It's pretty common that these LEDs have out-of-spec Vf at a given current.
 
If I knew exactly what an Rset was I might answer that question. :D

Makes sense that there could be some variance, though.

I'll have to test to see what voltage the CAT is giving.
 
Rset is the current sense resistor. It's the component that tells the CAT4101 IC what current you want to run at. Look at table 6 on page 7 of the datasheet - it tells you what resistor value corresponds to what current.

In my "triple" design for the CAT4101 the only three resistors on the board (R1, R2, R3) are the Rsets for the three CAT4101's respectively.
 
Then you should indeed be seeing a 700mA drive current. Put your multimeter inline with the LEDs, and turn up your DC supply until you get 700mA - or until it plateaus close to that value (the circuit is only accurate to within a few percent). At that point, pull the multimeter out of the circuit, and go!

It's worth nothing that you CAN run higher than that minimal point, but you'll just be turning the extra power into heat in the IC. Once you hit 3 or 4w dissipated (i.e. 4 or 5v more than it needs to be), the IC's internal thermal protection trigger will shut things down.
 
I was able to get it up to 700mA at one point. I didn't check the exact voltage at the time. And I'm not sure if I was driving blues or whites at the time. :D

Thanks. :cool:
 
IIRC the circuit has a 10% variation. Don't temember where I saw it, but that was why I set a 900 ma limit. So you maybe only able to get to 630 ma.
 
Was wondering if this PS would be ok for the 5v to vin on the cat4101?

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=260491050625&ssPageName=STRK:MEWNX:IT#ht_3678wt_905[/URL]
 
I think that would work. My only question is if it is regulated - I did not see that mentioned. Easy enough to check plug it in with nothing attached and measure the 5 volts. Since it is for computer I expect it is regulated.
 
That was my concern too, so I found one I had laying around and measured it, the voltage with no load is 5.35v. I connected four small computer fans to it (on the 5v side) and measured it again this time it was 5.28v. Would that be considered regulated ?? :rolleye1: If its good I would order some and be able to use it for the Cat4101 (5v) and also the 12v side for cooling fans and 12v for the Typhon controller. What da think? Thanks.
 
I think we need kcress :). I would say it is regulated, but since it is made to supply 2 amps when there is a really light load it does not regulate well. If that is the case then worst case you could get some power resistors (5 watt 10 ohm would use a 1/4 of the current) and use them as a load or find something else to power with 5 volts.
 
Doesn't sound very well regulated to me. It would be good to drop a larger load on it and see what happens. Find a load close to its rating and see what happens.

You can actually design a supply so that its own transformer provides some moderate regulation. I suspect that might be what's being employed here.
 
10.7 w ?? watts? ohms? other?

5 volts / 20 ohms = 0.25 amps times 2 for 2 resistors is 0.5 amps. That is only 1/4 of the rated current IIRC. Not sure it is enough of a load. Got some more? I think you should try and get over at least an amp.
 
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