Eliminate 2nd Driver w/ a Relay???

[ChiefRunningPhist]

So I have 2 areas, 12ft2 & 8ft2. I want to run em the cheapest way possible, while maintaining at least 160+ lm/w efficacy.

Thinking...
[8] Getian 2828 @37.5w ~170 lm/w
[11] Getian 2828 @41.0w ~160 lm/w

[1] 480h-36a

To the point of the thread, has anyone used a relay, timer, resistor circuit to eliminate the need for 2 drivers?

Can I buy a 480h-36a and wire up my 2 COB racks to a relay with a 8.63ohm resistor on the [8] COB rack (300w circuit) so that my lights automatically switch from either-or area when the relay is tripped?

The [11] COB rack is a 450w circuit and the [8] COB rack would be a 300w circuit . If I'm understanding correctly, using V = IR, I'd need an 8.63ohm current limiting resistor on my 300w circuit to reduce the current from 12.5a to 8.3a? 36v ÷ (12.5a - 8.3a) = 8.63ohm???

Is this right?

I'd run a lil cheap stupid Christmas light mechanical timer for my relay coil. Set it for 12 on 12 off so my area's were staggered?

Anyone try this???

 

[shimbob]

That little resistor is going to get HOT as it tries to dissipate 150watts.

 

[Hugo Phurst]

, has anyone used a relay, timer, resistor circuit to eliminate the need for 2 drivers?
Is this right?
Anyone try this???

Probably not
No
Probably not

Better & easier way would be to turn the driver off, switch area, turn the driver on.
And the chances of fire are less this way.

 

[Rocket Soul]

So I have 2 areas, 12ft2 & 8ft2. I want to run em the cheapest way possible, while maintaining at least 160+ lm/w efficacy.

Thinking...
[8] Getian 2828 @37.5w ~170 lm/w
[11] Getian 2828 @41.0w ~160 lm/w

[1] 480h-36a

To the point of the thread, has anyone used a relay, timer, resistor circuit to eliminate the need for 2 drivers?

Can I buy a 480h-36a and wire up my 2 COB racks to a relay with a 8.63ohm resistor on the [8] COB rack (300w circuit) so that my lights automatically switch from either-or area when the relay is tripped?

The [11] COB rack is a 450w circuit and the [8] COB rack would be a 300w circuit . If I'm understanding correctly, using V = IR, I'd need an 8.63ohm current limiting resistor on my 300w circuit to reduce the current from 12.5a to 8.3a? 36v ÷ (12.5a - 8.3a) = 8.63ohm???

Is this right?

I'd run a lil cheap stupid Christmas light mechanical timer for my relay coil. Set it for 12 on 12 off so my area's were staggered?

Anyone try this???

I think theres another way to go about this. As long as you can get a relay switching between the 2 circuits. But you would have to run all cobs at the same voltage/current, and use numbers to hit your desired wattage. Meanwell A-type drivers uses voltage adjustment. Which means you can sort of set a max wattage using the voltage control. How to go about this:
Lets say we start with your 11 cob setup. Dim using current adjustment to the desired level (or even a little bit above). Then dim down voltage little by little, very slowly until you see the first minimum drop in light levels. Make sure your happy with the levels. Lets say you get them to about 40w each. Youve now effectively setup the driver for 40w max wattage per cob by limiting its voltage: even if you tried only one cob on the driver it will still be 40w.
On the other side run 7 cobs, for about 280w.

Remember your relay needs to be rated for DC use, AC ratings arent ahe same.

 

[shimbob]

Alternatively, use a B driver and use a relay to connect/disconnect a pot to the dimmer wires.

 

[ChiefRunningPhist]

I think theres another way to go about this. As long as you can get a relay switching between the 2 circuits. But you would have to run all cobs at the same voltage/current, and use numbers to hit your desired wattage. Meanwell A-type drivers uses voltage adjustment. Which means you can sort of set a max wattage using the voltage control. How to go about this:
Lets say we start with your 11 cob setup. Dim using current adjustment to the desired level (or even a little bit above). Then dim down voltage little by little, very slowly until you see the first minimum drop in light levels. Make sure your happy with the levels. Lets say you get them to about 40w each. Youve now effectively setup the driver for 40w max wattage per cob by limiting its voltage: even if you tried only one cob on the driver it will still be 40w.
On the other side run 7 cobs, for about 280w.

Remember your relay needs to be rated for DC use, AC ratings arent ahe same.

I'm trying to set it up perfect for the 12ft2 or 11 COB rack, and then, put a current limiting device on the 8 COB rack so that all individual COBs no matter the 11COB rack or the 8COB rack are all about 40w.

When the driver that is set up and pushing 12.5a and 36v for the 11 COB rack is introduced to the 8 COB rack circuit, it undergoes some sort of current limiting while maintaining voltage.

I'm wiring the racks in parallel so I'd be manipulating current rather than voltage on the CV driver (480h-36 a), correct??

That little resistor is going to get HOT as it tries to dissipate 150watts.

A POT is just a variable resistor, no?

Alternatively, use a B driver and use a relay to connect/disconnect a pot to the dimmer wires.

Thinking about this. I'd need 2 relays then, 1 to switch the driver from rack to rack, and 1 to switch the POT circuit on the driver from close to open. Is this what you were talking about?

If done this way, the circuit without the POT would have the driver wide open because its not being limited, is this correct? B drivers run max unless a pot is added to limit? If this were the case, I'd need 2 POTs wired in. 1 to control the driver from being wide open, and 1 that was set to limit the driver when it switched circuits or lights.

Thoughts???

I might just try to wire it up and see.. But @Hugo Phurst I don't wanna burn the house down, not into hot boxing

If anyone has any experience with current limiting or resistors (sizes, heat issues, ect) blab it out! Lol

 

[ChiefRunningPhist]

A driver limits the wall electric from 110v & ~30a to whatever you need??

So to limit a DC current coming off your driver should require less tech or be relatively easy considering its stepping down a smaller amount than from your wall to your driver, and DC???

I could just hook up a 150w incandescent (or dome other power consumer) to the 300w circuit to equalize my per COB wattage between the circuits but if I limit the current couldn't I reduce my power consumption?

Isn't that what a current limiting resistor is for?

 

[CobKits]

just get a DPDT relay to throw power to the array of your choice

 

[CobKits]

A POT is just a variable resistor, no?

potentiometers arent used in the cob circuit, just as a control to send a resistance signal to a drivers dimming leads (which "A" drivers dont have)

 

[CobKits]

A driver limits the wall electric from 110v & ~30a to whatever you need??

not at all. it doesnt limit,it transforms AC into DC at a different voltage.

 

[ChiefRunningPhist]

not at all. it doesnt limit,it transforms AC into DC at a different voltage.

I'm just trying to get it figured out here, but why then are there different size drivers for the same voltage? I understand it's a rectifier, but don't they internally limit current, or draw specific current up to a certain point (ie limiting)? Transforms 110VAC into (...)VDC and then limits currents which is manipulated by built in POTs or wired in POTs? Or am I completely off?


After doing some more research, a 150w 9ohm current limiting resistor is going to get hot and need a heatsink which will cancel out the whole goal of trying to reduce costs.

Now I'm thinking 2 relays, 1 to switch between arrays, and 1 to switch between POTs (or resistors, 2 of them) wired in the B driver..

Thanks, good suggestion @shimbob

Anyone see any issues I may run into?

 

[shimbob]

Nothing wrong with running a B driver with the dimmer wires unconnected, it will run at full power just fine.