Introducing CobKits.com - specializing in DIY and Citizen COBs

[CobKits]

So the 600-36b would only provide 1800 ma even if only 1 single 1212 was hooked up to it? It wouldn’t blow or burn out the cob from pushing to much current through it?

yes you would be here on the driver output curve

 

[GrowLightResearch]

When I decided between 4 and 3 strips for my lights I worked out efficiency at full and veg power, cost of the unit and how much money it would cost to run for a year.

You should also look at the economics of adding a strip and decreasing the current. The decreased current may allow you to add an additional strip increasing the total irradiance and decreasing the power consumption. The benefit being improved uniformity and possible reduction the distance between the fixture and canopy. The power savings from decreasing the distance and improved efficiency (lower current) can more than pay for the additional strip.

 

[GrowLightResearch]

again at 36V the cob can only operate at one current. it cant possibly take more current at that forward voltage

What?? Really?

So if I hook up 1 Citizen 1212 to a MW 600H-36B it will only draw 1800 ma?

So just so we’re clear there is no way to over power (current/amps/watts) an led/cob with the driver we are talking about?

It will likely burn instantly if the blue and white dimmer wires are left open. You would need to put a 10KΩ 5% resistor across the blue and white dim wires to limit the 16.7 Amp max current to 1670mA.

To set the current to 1800mA use a 10.7KΩ 1% (1.8A÷16.7A x 100,000Ω)

Resistor value = desired current in amps ÷ 16.7 x 100,000.
or
Resistor value = desired current in mA ÷ 16700 x 100,000.

 

[pop22]

sounds like you need to spend more time reading, there is a lot of information on RIU explaining this.

What?? Really?

GLR, it makes more sense to tell him to match the driver to the cob(s). And your wrong on the dimming unless your putting the resistors inline with the cobs. Meanwell drivers dimming is PWM and the use of the potentiometer regulates this circuitry, it does not directly influency the current output. A 10k resistor will dim the driver 10%. This is why you can't dim these drivers to 100%. Might pay for you to actually read a Meanwell spec sheet.




It will likely burn instantly if the blue and white dimmer wires are left open. You would need to put a 10KΩ 5% resistor across the blue and white dim wires to limit the 16.7 Amp max current to 1670mA.

To set the current to 1800mA use a 10.7KΩ 1% (1.8A÷16.7A x 100,000Ω)

Resistor value = desired current in amps ÷ 16.7 x 100,000.
or
Resistor value = desired current in mA ÷ 16700 x 100,000.

That is what I had assumed, I don't know how a driver can know the watt/amperage limitations of any given led..

 

[pop22]

This is how it works: Notice the PWM and PFC curcuitry, it DOES DETECT cob voltage/amperage requirements, it just takes some understanding of electronics., that includes you GLR, for all your math, cobby has already proved you wrong on more than one occassion.....

This is why Meanwell provides data sheets so that you can make an informed decision as to what driver you need. If your building your own light, you need to educate yourself, your not building with legos......

 

[CobKits]

So just so we’re clear there is no way to over power (current/amps/watts) an led/cob with the driver we are talking about?

sure, if you had a cob where 36.0V is out of its safe operating range. but not with the cob/driver combo you listed

 

[GrowLightResearch]

Notice the PWM and PFC curcuitry

circuitry*

your not building with legos

you're*

Notice the PWM and PFC curcuitry, it DOES DETECT cob voltage/amperage requirements,

Where is it detecting amperage? How does it know not to provide more current than the CoB can handle? Why does it have a Dimming current limit circuit (e.g. voltage or resistance on white and blue wires)?

PFC is power factor correction which is IV phase correction for the input voltage.
PWM is the current control circuity which needs an input from the dimming wires and if left open or if the resistance is equal or greater than 100KΩ the HLG will supply full load, max current.

where 36.0V is out of its safe operating range

The "safe" operating zone would be where the load has a capacity greater than the HLG's max wattage. In the case of this CoB and HLG 600H the single1212 CoB would need to be rated at greater than 16.7 amps or 600 watts. Even at that the question was asking if the HLG would exceed 1800mA. The answer is yes it will.

cobby has already proved you wrong on more than one occassion.....

occasion*
No he has not. Show me. Never happened. Just because you think something is true does not mean it is true. Example you were wrong here.

 

[pop22]

thanks for the grammar and spelling lesson. You want an explanation of how the Meanwell works, write to the company.and run wide open only adds maybe 5% more current. I have 2 Meanwell drivers run without resistance in the dimming circuit and they work just fine, so I do not see what your point is with that comment, the driver is still NOT going to dump all available current into the cob, only what the CV/CC circuitry will allow.

circuitry*

you're*



Where is it detecting amperage? How does it know not to provide more current than the CoB can handle? Why does it have a Dimming current limit circuit (e.g. voltage or resistance on white and blue wires)?

PFC is power factor correction which is IV phase correction for the input voltage.
PWM is the current control circuity which needs an input from the dimming wires and if left open or if the resistance is equal or greater than 100KΩ the HLG will supply full load, max current.

And you need to take your own advice about what you think is true.


The "safe" operating zone would be where the load has a capacity greater than the HLG's max wattage. In the case of this CoB and HLG 600H the single1212 CoB would need to be rated at greater than 16.7 amps or 600 watts. Even at that the question was asking if the HLG would exceed 1800mA. The answer is yes it will.


occasion*
No he has not. Show me. Never happened. Just because you think something is true does not mean it is true. Example you were wrong here.

 

[ChaosHunter]

@CobKits Will you be offering any kind of framing options ?

 

[CobKits]

yes but maybe another month or two

 

[GrowLightResearch]

thanks for the grammar and spelling lesson.

You are welcome, it is difficult to take someone serious when they cannot spell or know to use a spell checker.

You want an explanation of how the Meanwell works, write to the company.and run wide open only adds maybe 5% more current

I KNOW how a CC drive works. I was asking questions in an attempt to get you to think it through. I am an electrical engineer that has been designing power switching circuits since 1979.

The problem with what you say about the 5% is there is no way the driver can know what the current limit of the load is. By leaving the dim wires open you are telling the driver to supply 16.7 Amps.

I have 2 Meanwell drivers run without resistance in the dimming circuit and they work just fine

Yes it is possible for the driver to work fine without resistance on the dimmer wires. That is, as previously explained, only when the load is capable of handling the max output of the driver. The case here is a 600 watt driver with a max output of 16.7 Amps and a single CoB. I assume the CoB is rated at less than 600 watts. Even if it could handle 600 watts, as previously stated, the question was about limiting the driver to 1800mA. And in the case of the CoB being rated over 600 watts I cover my ass by saying it would "likely" burn. Then I went on to explain how to limit the current to 1800mA.

only what the CV/CC circuitry will allow.

The driver is capable of providing 16.7 amps, correct?
If you configure the driver to supply 16.7 amp it will deliver 16.7 amps correct?
By having a resistance of greater than 100,000KΩ (in this case infinity), the driver is configured to supply 16.7 amps.

 

[nfhiggs]

I KNOW how a CC drive works.

It's not a CC driver they are talking about. The HLG-320-36 is a CV 36 volt driver. Its output voltage does not change until load current increases into the CV region. Under all current conditions its output will not exceed 36 volts

 

[GrowLightResearch]

Under all current conditions its output will not exceed 36 volts

We are not talking about exceeding the voltage. We are talking about exceeding 1800mA with a 600 watt HLG driver.

 

[GrowLightResearch]

The HLG-320-36 is a CV 36 volt driver.

The HLG is definitely a CC driver. The HEP is a CV power supply.
All CC drivers have a voltage range where the CC does not work.

The HLG-36 can provide up to 37.8V. Between 36 and 37.8V the HGL will output max current if not limited with a resistive load or by the dim control.

 

[nfhiggs]

We are not talking about exceeding the voltage. We are talking about exceeding 1800mA with a 600 watt HLG driver.

And to exceed that current you have apply more voltage or allow it to overheat. Just having the current capacity does not mean it will be used.

 

[1212ham]

What?? Really?

It will likely burn instantly if the blue and white dimmer wires are left open. You would need to put a 10KΩ 5% resistor across the blue and white dim wires to limit the 16.7 Amp max current to 1670mA.

The HLG-36 can provide up to 37.8V. Between 36 and 37.8V the HGL will output max current if not limited with a resistive load or by the dim control.

Read posts #3329 & [URL='https://www.rollitup.org/t/introducing-cobkits-com-specializing-in-diy-and-citizen-cobs.916763/page-167#post-14005013']#3335. Coby is not unfamiliar with citi 1212's and Meanwell drivers! [/URL]
[URL='https://www.rollitup.org/t/introducing-cobkits-com-specializing-in-diy-and-citizen-cobs.916763/page-167#post-14005013']The 1212 datasheet voltage/current curve shows about 1800mA at 37.8v.[/URL]
How would 1800mA instantly burn a cob rated at 2760mA?

 

[GrowLightResearch]

How would 1800mA instantly burn a cob rated at 2760mA?

Without the exact part number I am at a disadvantage. As far as I know citi 1212 CoBs can have Vf of 9.2, 34.6, 35.7, or 52v.

Let's assume it is a 35.7v.
At 1080 mA and 25°C Vf ranges from 32.8v-38.5v.
At 1080 mA @ 25°C the Vf will typically be about 36.5v.

An increase of case temperate can drop the Vf another 2.5v.

So the Vf could be as low as 30.3v. Furthermore Citizen does not guarantee the min and max values meaning Vf could go beyond these values.

Additionally Vf is a function of forward current. Forward current is not a function of Vf. You cannot control current with Vf.
It would be ill advised (insane) to use Vf to regulate current.
The proper way is to put a 10K resistor on the dim wires.
Otherwise the CoB is likely to to burn out.

And to exceed that current you have apply more voltage or allow it to overheat. Just having the current capacity does not mean it will be used.

It is a current source. You do not regulate current with voltage.

 

[nfhiggs]

It is a current source. You do not regulate current with voltage.

LOL... whatever. There is no current flow without voltage. Current flow in a load results from the voltage potential not the other way around. You most certainly DO control current (through the load)with voltage. Where did you go to school?

 

[GrowLightResearch]

you most certainly DO control current (through the load)with voltage

Not with a CC driver! PLEASE just stop! You obviously do not have the background to argue this.

Why is it if the CC current were to be a function of the load voltage, if you change only the load from an LED with a Vf of 24v to an LED with a Vf of 34v the current remains the same? But yet the voltage changes. Why is that?
E = I x R
Voltage (E) is a function of the LED's dynamic resistance (R) based on the LED's IV characteristics and the constant current flow (I).

It is because the forward voltage is a function of constant current (CC) flowing through the LED's dynamic resistance!

 

[nfhiggs]

Not with a CC driver! PLEASE just stop! You obviously do not have the background to argue this.

Why is it if the CC current were to be a function of the load voltage, if you change only the load from an LED with a Vf of 24v to an LED with a Vf of 34v the current remains the same? But yet the voltage changes. Why is that?
E = I x R
Voltage (E) is a function of the LED's dynamic resistance (R) based on the LED's IV characteristics and the constant current flow (I).

It is because the forward voltage is a function of constant current (CC) flowing through the LED's dynamic resistance!

You are simply ridiculous in your inability to grasp this very simple and BASIC concept. The CC driver controls the load's current by varying it's voltage output as needed. If the loads temp increases and resistance drops, the driver adjusts the VOLTAGE downward to maintain the current at the same level. Do you not understand the fact that current in a load is the direct result of applied voltage? This is really VERY basic stuff.