This is a subject I touched on in an earlier video where I tried to run about 175 volts worth of LED (5 COBs) off of a driver only rated for 143 volts max. The result in that test was that the driver maintained its full voltage (~143v), but put out next to nothing for current. What I set out to test this time was what would happen if you were trying to pull just a handful of volts more than the driver was rated for, as opposed to trying to pull 30+ more than rated. By adding resistors in series with my 4 COB lights, I was able to slowly increase the voltage drop of my circuit to meet and exceed the max rating of my driver and see how it reacted.
For this test, I used the following gear:
- 1x HLG-185H-C1400B (same driver as last time)
- 4x Cree CXB3590 COBs wired in series
- Assorted 10 watt resistors used to increase the load on the driver
I used a screw terminal strip for all my wiring to make it easier to move things around and test voltage and current. The white wires on the bottom of the terminal strip are the positive and negative leads coming directly off of my driver. The black cable you see coming in from the left of each picture is the feed out to my 4x Cree COBs in my grow tent, and I used a number of jumpers and alligator clips to rig the circuit up to allow me to test voltage/current and add resistors.
First, I tested voltage and current for the circuit with only the 4x CXB3590s connected (they are actually what provided my lovely lighting for all these pictures). I determined voltage was 140.9V and current was 1,391mA. The data sheet for this driver states that the constant current voltage range is 71V to 143V. With the 4 COBs on their own we have 2.1V of head room before we hit the driver max.
I ordered a handful of 10 watt resistors for this test. Their impedance values were 1 ohm, 2 ohms, and 5 ohms. When measured, they read fairly close to what they were rated for:
Expected Voltage Drops by Resistor
Since I determined the current flowing through the 4 COBs in their normal configuration was about 1,390mA, we can calculate how much voltage drop each of the resistors should (in theory) add to the circuit.
Using Ohm’s law, we know that Voltage is equal to Current times Resistance (V=IR). We have our current and resistance values, so we can plug them in to find out how much voltage drop will be added to the circuit by each of the resistors. Note that these numbers won’t be totally accurate since the resistors are not exactly 1 ohm or 2 ohm or 5 ohm, but they’re close.
- 1 Ohm Resistor: V=1.39*1
- 2 Ohm Resistor: V=1.39*2
- V= 2.78
- 5 Ohm Resistor: V= 1.39*5
We can also check the total power, in watts, dissipated by each resistor. Power is equal to Current times Voltage (P=IV), so:
- 1 Ohm Resistor: P=1.39*1.39
- 2 Ohm Resistor: V=1.39*2.78
- 5 Ohm Resistor: V= 1.39*6.95
So, since the 4 COBs were producing a voltage drop of 140.9V by themselves, if I were to add a 1 ohm resistor, this should increase to 142.3V. If I add a 2 ohm resistor, voltage drop should increase to 143.7V. If I add a 5 ohm resistor, voltage drop should increase to 147.9V. Again, these are theoretical figures as well, as the voltage of the COBs will slowly decrease, the longer they’re on and the hotter they get.
Adding a 1 Ohm Resistor
To make sure my math was good, I first tested by adding a 1 ohm resistor in series with the 4 COBs. I measured 142.6V across the circuit, which is an increase of 1.7V over my initial reading of 140.9V with COBs only. This is close to the calculation above, which predicted an increase of ~1.4V. This would indicate that the resistor’s true impedance is about 1.2 ohms, and not 1 ohm exactly. Nevertheless, the system is responding as predicted. Current was again measured at about 1,390mA. With a 1 ohm resistor, we are still under the driver’s max voltage rating of 143V.
Adding a 2 Ohm Resistor
Adding a 2 ohm resistor, by our initial calculations, should push the voltage drop of the circuit beyond the rating of the driver. We expect a 2 ohm resistor to drop 2.78V across itself, and when I hooked it up, I was getting a total circuit voltage of 143.1V which is .6V less than expected. I measured the impedance of the resistor to make sure it wasn’t lower than 2 ohms, which it was not – it read 2.2. This must mean that the driver is holding its max voltage steady and not allowing it to increase.
The next step was to check the current. Something has definitely changed, as the current is now 80 milliamps lower than it was for the 4x COBs alone and the 4x COBs + 1 ohm resistor circuits.
Okay, interesting. Now let’s add a 5 ohm resistor to see if this effect is increased.
Adding a 5 Ohm Resistor
A 5 ohm resistor should bring the circuit voltage drop up to 147.9V. Similar to the 2 ohm test, the driver held voltage steady at its max – it hovered around 143.1-143.2V.
Current output with the 5 ohm resistor added dropped drastically, from 1,390mA all the way down to 1,050mA.
Well, let’s pile on a bunch of resistors in series and see what happens.
7 Ohms of Resistance (9.7V Added)
10 Ohms of Resistance (13.9V Added)
12 Ohms of Resistance (16.7V Added)
So, what I’ve learned from this is that when you exceed the upper limit of a Mean Well HLG’s constant current voltage range, the driver refuses to increase its output voltage. It holds it at maximum, and begins to lower its current output – the more voltage drop/resistance you add to the circuit, the lower the output current becomes. Also, as the current lowers, the voltage drop across each resistor decreases as a result of Ohm’s Law. The driver behaving in this manner is like the reverse of Mean Well’s CV+CC drivers, which operate in constant voltage mode until they’ve reached max current draw, at which point they hold the current steady and start to vary the voltage.
If your COBs are right on the cusp of exceeding the max voltage for your HLG constant current driver, you should be fine. If you’re a volt or 2 over, you are no longer getting constant current functionality and may end up losing 100-200mA of output current, but mine didn’t seem too upset that I was trying to push it. However, by presenting a load that should have drawn a mere 5 volts more than max, I lost 25% of my output current, which is a big hit.