Convert AC LED Floodlight to Battery powered

[hydrogore]

Hi, I'm a complete newbie and hoping someone can tell me if what I'm doing is OK.

I'm trying to convert this LED floodlight to battery powered. I plugged in a KILL-A-WATT to measure the floodlight's voltage and amp consumption.  It shows that it's consuming 30V and 0.25A on it's maximum brightness. 

I connected a Black and Decker 40V 2.0AH lithium battery to the AC plug via some gator clips and metal tabs and it seems to power on fine and even the pot dimmer on the light works to dim. Although, I do plan on getting a proper battery holder with a negative and positive lead to solder directly onto the board inside the lamp. 

So, my question is there anything wrong with doing this?

Many thanks!

[mariush]

Inside the box will be a switching power supply which converts AC voltage to DC voltage and then there's a led driver circuit which controls and adjusts the current going to the actual LED.

To increase the efficiency you can try skipping over the AC to DC conversion by connecting the battery directly after that portion of the circuit.

It would be best to open up the thing and take some pictures of the circuit board inside.  Which may be potted or inside a plastic box.

[hydrogore]

Thanks for replying mariush.  I have taken a couple pics. Let me know if you require more in different angles.

[Zero999]

Hi, I'm a complete newbie and hoping someone can tell me if what I'm doing is OK.

I'm trying to convert this LED floodlight to battery powered. I plugged in a KILL-A-WATT to measure the floodlight's voltage and amp consumption.  It shows that it's consuming 30V and 0.25A on it's maximum brightness. 

That doesn't make any sense. A Kill-a-Watt measures the power consumption of mains voltage appliances, which will be around 120VAC, not 30V, assuming your country's flag is correct. Don't you mean 30W?

I connected a Black and Decker 40V 2.0AH lithium battery to the AC plug via some gator clips and metal tabs and it seems to power on fine and even the pot dimmer on the light works to dim. Although, I do plan on getting a proper battery holder with a negative and positive lead to solder directly onto the board inside the lamp. 

So, my question is there anything wrong with doing this?

Many thanks!

Good, the switched mode power supply works at a much lower DC voltage, than the mains, which isn't uncommon but isn't a certainty. No, there's nothing wrong with that set up. You could increase the efficiency slightly by applying power to the other side of the rectifier, but not by much. I'd recommend removing the mains plug and wiring it to the battery holder. Connecting a mains socket to a battery would create the risk of someone plugging a different appliance in, which will be damaged if it's only designed to run off AC.

[hydrogore]

Many thanks Zero999!

...Don't you mean 30W?

Yes, you're absolutely correct. My mistake it's 30W. 

You could increase the efficiency slightly by applying power to the other side of the rectifier, but not by much. I'd recommend removing the mains plug and wiring it to the battery holder. Connecting a mains socket to a battery would create the risk of someone plugging a different appliance in, which will be damaged if it's only designed to run off AC.

OK, if it's not going to make much efficiency gains then I'll take your advice and wire it directly to the holder. Yes, I did intend on removing the AC plug totally.  Another question, is there a risk of the battery getting damaged when it runs dead? I'm thinking the B&D battery pack has a battery management system that would shut off the circuit before over draining correct?

Thank you for your advice!

[Zero999]

Yes, the battery pack will be damaged if it's allowed to completely discharge. I think lithium cells shouldn't be allowed to drop below 3V per cell, which would be 30V, assuming your battery pack is 10 cells in series, giving just over 40V, when fully charged.

[hydrogore]

Ok I suppose I'll just have to keep a watch on it when the battery indicator on the pack shows 1 bar left to disconnect it before it drains out.  Thanks for your help!

[james_s]

The writing on the tape says 30VA, which is not 30V nor is it necessarily 30W. You'd need to know the power factor for that to be useful.

The power supply inside the flood is going to convert 120VAC to DC and then step that down to a constant current to drive the LEDs. I'm surprised it works at all with such a low battery voltage.

[Zero999]

The power supply inside the flood is going to convert 120VAC to DC and then step that down to a constant current to drive the LEDs. I'm surprised it works at all with such a low battery voltage.

It didn't surprise me. To minimise the size of the filtering capacitor, it's common for mains input switched mode power supplies to work off a much lower DC voltage, than the AC voltage they're designed for. My Rigol DS1054 will work down to 36VDC. I thought about making a battery pack for it, but lost interest in the project when I got the OWON SDS1102, with the built-in battery pack.
https://www.eevblog.com/forum/testgear/battery-pack-for-rigol-ds1054z/msg717702/#msg717702

[Siwastaja]

If it produces the same light output at just 30VDC, it has to take a lot more current at such low voltage, than it takes at 120V input, to produce the same output power. This means some parts will see higher current and heat up more.

Especially the bridge rectifier (assuming it has one) won't like it; when running on AC, of the four diodes, two are conducting 50% of the time, and another two the remaining 50%. On DC, it's always the same two diodes that are conducting, dissipating twice the average heat per diode. Add to that possibly higher than normal input current.

[Zero999]

If it produces the same light output at just 30VDC, it has to take a lot more current at such low voltage, than it takes at 120V input, to produce the same output power. This means some parts will see higher current and heat up more.

Especially the bridge rectifier (assuming it has one) won't like it; when running on AC, of the four diodes, two are conducting 50% of the time, and another two the remaining 50%. On DC, it's always the same two diodes that are conducting, dissipating twice the average heat per diode. Add to that possibly higher than normal input current.

On the other hand, steady DC is easier on the filter capacitor and the diodes won't be passing huge current spikes to charge it, twice per mains cycle. It's also highly likely the bridge rectifier is extremely de-rated as it is.

I did some tests on my Rigol DS1054 and found that working off lower voltage DC didn't cause any overheating of the power supply. It seemed to work more smoothly and with less noise on a battery, than the mains.

[Siwastaja]

Mains 100/120Hz ripple current rarely is problematic to the filter cap anyway; capacitors often die in switch mode supplies, but the large cap after the diode bridge seldom is the one dying. Current spikes from charging from a sinusoidal waveform also are far from "huge". Unless the capacitor is really badly undersized.

I don't see why a bridge rectifier would be "highly likely extremely de-rated". Extreme derating rarely takes place in cost optimized mass produced cheap electronics. I would say it's unlikely. BOM and purchasing is funny, though, sometimes a manufacturer can find a higher rated part for cheaper than a lower rated.

Many integrated diode bridges are not even rated as single diodes but list the maximum average output current assuming sinusoidal AC input. There may not be a DC rating available at all! So it's obvious for a designer to look at the average current number and choose a suitable part; if not generously derated, it's then not suitable for DC.

So the temperature of the diodes is definitely worth monitoring. For example, run for an hour at the hottest environment you'll ever use it at, use a thermocouple (or a thermal camera) to measure the diode temperature. Depending on package, the highest temperature may not be on the diode case but on one of the legs.

I know AC rated power supplies are widely used on DC and I do that myself, too, but it's good to know about the potential pitfalls.

The obvious, and very easy modification is to remove and bypass the bridge rectifier. After this, the device can't be connected to AC supply and correct DC polarity must be applied.

[Zero999]

The bridge rectifier typically used is rated to several hundred volts at 1A, often just four 1N4001s.

Look at the current waveform of a bridge rectifier and capacitor and you'll find there's a big inrush spike, then current is drawn in huge spikes, even though the current drawn from the DC side is a fraction of that.


Using a lower DC voltage results in less of a turn on surge and a steady current being drawn, rather than spikes, so it's easier on the rectifier, if anything.

The only potential with using a lower voltage is the driver transistor might have to work a little harder, but I've found it didn't cause a problem with my oscilloscope.

[Siwastaja]

You are simulating a circuit with extremely crappy power factor. If the device in question really works at 30VDC, that's a clear sign the diode conduction duty cycle is much longer, and the ratio of peak to average current more benign.

Besides, given that power loss in diode isn't I^2*R but more like I*Vf, diodes handle peak currents quite well; in other words, diodes are always rated for average current (look at any diode datasheet) because that defines the die heating, not RMS current like with MOSFETs or capacitor ESR. Average current for peaky or more steady current flow is the same, so is heat generated in the diode. So as long as the peak current is within specifications, it's OK.

But only using two diodes out of the diode bridge really doubles the average current, this is not insignificant.

You are right though that peaky current affects the capacitor power dissipation. Obviously the cap must be sized to handle it. Going for DC completely removes the capacitor ESR loss. But the capacitor ESR loss in such circuits often isn't that much to begin with.

Inrush current obviously is linearly related to the input voltage (AC or DC). Going lower lowers the inrush, but unless the OP plans to abuse the light by disco blinking it, it doesn't matter because it has to be designed for some quite high number of on/off cycles (often stated in the specs, even) at full rated voltage.

To recap:
* By going for smoother (DC) current instead of peaky AC, diode power dissipation lowers only slightly (due to the logarithmic nature of Vf ~ I), not by a lot as you seem to expect
* Diodes have to be rated for the full peak current during inrush at full line peak voltage, otherwise the devices would go up in smoke all the time
* Diode thermal rating is designed for the average current, not peak current
* Feeding DC to diode bridge doubles the average current two diodes see

But by all means plot the diode power and calculate the integral in your Spice simulation, and compare to the DC input. I expect you may see some 10% difference.

[Siwastaja]

See the attachment of simulating Zero999's circuit as is with diode power dissipation of 107mW, and feeding it DC voltage resulting in the same average output voltage, diode dissipating 164mW, increase of 53%.

[Zero999]

See the attachment of simulating Zero999's circuit as is with diode power dissipation of 107mW, and feeding it DC voltage resulting in the same average output voltage, diode dissipating 164mW, increase of 53%.

Fine I stand corrected on that. At lower voltages, the current will also be higher, so not good. I still think it'll be fine though, based on my experience. I'm wary of recommending that the original poster should modify it because it's working and given this is the beginners section, there's the risk of him making a mistake and ruining it.

[Siwastaja]

It often works OK because even if you increase the dissipation by 50-60% (although with a different power factor load, the difference can be up to +100%!), the neihgbor two diodes are dissipating nothing, so all depends on the thermal coupling and how the diodes share the cooling path; per diode dissipation is significantly increased, but the total of the four is slightly reduced.

Now everything really depends on how that SMPS deals with the constant lower voltage. If it acts like a PFC and pulls lower current because it sees lower voltage, that's fine, but then the obvious result is very dim light.

If the light is as bright as with the full voltage, it has to take significantly more average input current which really isn't good for possibly many other components than just the diodes.

I'd hazard a guess that even running an AC load with a DC voltage voids any warranty, and includes risk of ruining it anyway, so a simple modification like bypassing the diode bridge would be a good start for a beginner. Just remove the plug and mark the device properly so that no one accidentally powers it from mains supply ever again...