LED bulbs don't like MSW inverters?

[IanB]

I obtained a little Ryobi 150 W inverter with the idea of using it to run a table lamp or two in the event of a power outage. I thought that if an LED bulb uses 7 to 9 W it could run for quite a long time on a 4 Ah battery.

However, I've noticed that while these bulbs do indeed draw the rated power from the mains according to my power measuring device (a Kill A Watt clone), they draw twice as much at the input of the inverter, giving an overall efficiency of 50% or less. Since my measuring device doesn't work properly on the output of the inverter, I cannot tell where all the extra power is going. Is it wasted in the bulb, or is it wasted in the inverter?

(I am measuring the efficiency by running the inverter off a DC power supply and measuring the input current at a fixed 20 V.)

Some LED bulbs don't work at all on the inverter, and I have heard stories of other LED bulbs failing prematurely.

For comparison, incandescent bulbs of a similar 6 - 25 W power rating run the inverter at a reasonable efficiency approaching 90%, but this is not helpful when the total power draw is so much higher.

I have one 14 W CFL that draws 17 W at the inverter input, and although that's a reasonable efficiency it is still 17 W.

Questions:

• Does anyone know what's the deal with modified sine wave inverters and LED bulbs, and why they don't work efficiently?
• Does anyone have experience of Kill A Watt devices on MSW outputs, and knows of one that works correctly and gives proper readings?

[Siwastaja]

How did you actually measure? Did you just measure Pout/Pin and are surprised that the efficiency of the inverter is lower at lower load? That is to be expected because it has some minimum power draw to operate at all, hence efficiency must approach 0% when load power approaches zero. Or did you measure efficiency of incremental power, (Pout2-Pout1)/(Pin2-Pin1), where Pout2>Pout1, or Pout1=0? Even then, the problem might be that the inverter enters some power saving mode at zero output load.

[IanB]

I attach two graphs. One shows the measured performance with resistive loads (incandescent bulbs), and one with non-resistive loads (various LED and CFL bulbs).

The resistive loads show a consistent pattern with an ultimate inverter efficiency of about 92% and a static power consumption of about 2 W.

The LED and CFL bulbs show a very inconsistent pattern with an operating efficiency often less than 50%. There is quite a contrast between the two load types.

[Edit: I have added the no load point to the first graph]

[IanB]

How did you actually measure? 

See my post above. I measured Pout/Pin for various loads, both resistive and otherwise, and looked at the trends. For similar sized resistive and non-resistive loads the efficiencies are markedly different (~80% vs ~50%).

[Geoff-AU]

The bulbs that draw more power than expected probably use a capacitive dropper technique.  Your inverter is a modified sine wave (yuck) which may be pushing more power through the dropper (and also likely why your AC power meter doesn't like being on the output of the inverter).

[IanB]

The bulbs that draw more power than expected probably use a capacitive dropper technique.  Your inverter is a modified sine wave (yuck) which may be pushing more power through the dropper (and also likely why your AC power meter doesn't like being on the output of the inverter).

It's doubtful they use a capacitive dropper (60 W equivalent bulbs use more sophisticated circuits), but it is true that they are somehow being upset by the modified sine wave. Simple non-dimmable LEDs generally rectify the mains to DC on a reservoir capacitor and then put a current regulator in series with a long string of LEDs so that the regulator only has a small voltage drop across it. For a true sine wave the rectified voltage (V peak) would be about 170 V, but for a modified sine wave it would be lower. The lower peak voltage probably explains why at least one LED bulb fails to light up.

For the other bulbs, if they are drawing more current due to the ugly wave form then they are probably running hotter and stressing their internal components, which is not good.

I think nearly all consumer level inverters use a modified sine wave. If you browse available products it is nearly impossible to find a pure sine wave output in this space. But this limits the possible loads to purely resistive loads (barely relevant), or to laptop/phone power supplies. Apparently, switched mode power supplies handle a modified sine wave OK.

[NiHaoMike]

Have you considered just using DC LED lamps?

[tszaboo]

How did you actually measure?

See my post above. I measured Pout/Pin for various loads, both resistive and otherwise, and looked at the trends. For similar sized resistive and non-resistive loads the efficiencies are markedly different (~80% vs ~50%).

So what was the power factor for the LEDs?
Or better question, if you have a capacitor on your AC to reduce the voltage (and some linear driver from that) then what happens if your input is a sine wave with a lot of harmonics?

[IanB]

Have you considered just using DC LED lamps?

Yes, but USB lamps do not come in the same form factor as mains lamps, so they are not entirely equivalent. Nevertheless, I am exploring this option.

[IanB]

So what was the power factor for the LEDs?
Or better question, if you have a capacitor on your AC to reduce the voltage (and some linear driver from that) then what happens if your input is a sine wave with a lot of harmonics?

These are commercial, off-the-shelf LED bulbs, either from Philips or from Feit Electric.

The power factor for such bulbs is typically around 0.8.

I have no way to test for arbitrary mains waveforms.

[NiHaoMike]

Yes, but USB lamps do not come in the same form factor as mains lamps, so they are not entirely equivalent. Nevertheless, I am exploring this option.

Take a look at 12V LED strips.

[IanB]

Yes, but USB lamps do not come in the same form factor as mains lamps, so they are not entirely equivalent. Nevertheless, I am exploring this option.

Take a look at 12V LED strips.

LED strips are not lamps you can simply place on a table and light up a room.

But it does prompt the thought that USB power supplies can negotiate higher voltages than 5 V these days, like 12 V and above, so there should be space for USB powered devices like lamps that can take advantage of this.

[james_s]

It's the dimmable ones that don't like these, the weird waveform confuses the heck out of the dimming circuit. Best case they run dim, worst case they flicker and flash. Smart LED bulbs seem to work fine though, I have a couple of lamps with Hue bulbs that are plugged into modified sine UPS's and they don't even glitch when the UPS kicks in. Many non-dimmable cheap LED bulbs will work fine too.

[IanB]

I'm sorry my answer doesn't fit exactly to your question but it is generally bad practice to be using inverters for such low current DC devices in the first place and trying to fix the problem could be a matter of a simple capacitor needing placing across the inverters outputs to clean up the modified square wave or it could need completely disassembling the LED lamps.

Either choice isn't very good because you might not be able to get the LED device back together and a capacitor across the inverter might damage something.

Funnily enough this vendor lock in is why I gave up on Ryobi stuff that and because they are so expensive yet corded tools are so cheap. Can't use the batteries for anything except for what Ryobi sells for them.

Which leads me to my final point, a dedicated lithium power station with an inverter or 12v output would make you better off.

I'm fine with the idea that there are better answers out there, and I agree that it is surely better to keep the 20 V DC from the lithium battery and use it more directly, rather than inverting it up to 120 V AC, only to down-convert it again for an LED light bulb.

My question is more of curiosity that the inverter is very inefficient when driving such loads compared to resistive loads. This is not something I've heard anyone talk about, and so I wondered if anyone with an electronics background might have an idea of exactly why this is the case?

As far as low voltage solutions are concerned, I will be doing some experiments with USB power and seeing how that works out. I will try to find out how efficient the USB 20 V DC to 5 V DC conversion is compared to the mains inverter. But that will have to wait until I have some USB loads to measure with.

I have no problem with Ryobi in general. They make some quite decent tools, and the prices are not so bad if you look out for the special deals and bargains that crop up from time to time. I really like the plug-and-play idea of big lithium batteries and multiple uses that you get from cordless power tools like these.

[IanB]

It's the dimmable ones that don't like these, the weird waveform confuses the heck out of the dimming circuit. Best case they run dim, worst case they flicker and flash. Smart LED bulbs seem to work fine though, I have a couple of lamps with Hue bulbs that are plugged into modified sine UPS's and they don't even glitch when the UPS kicks in. Many non-dimmable cheap LED bulbs will work fine too.

I'm trying both dimmable and non-dimmable samples. I just picked up a 9 W (60 W) LED bulb from Dollar Tree for $1.25 (non-dimmable). It measures 7.8 W on the mains, and 20 V, 0.62 A to the inverter, for an effective efficiency of 63%. That puts it top of the efficiency table for all the LED bulbs I have looked at. Next best is the Feit Electric non-dimmable at 55%. The worst is a Philips dimmable bulb at 42%.

[IanB]

It occurs to me that maybe I should include the power factor and apparent power in my table and see how that relates to the effective efficiency.

[IanB]

Its designed for specific use cases, in this case powering something higher power than a low power LED light.

Actually, no, I don't think it is. The inverter has a maximum power output of 150 W, and in any case it couldn't do that for very long on a 75 Wh battery.

The advertised use cases for the inverter are things like laptops (and maybe only laptops). There's no point plugging in phone chargers when there are USB charging ports already on the inverter, and I struggle to think of any other low power mains devices you might want to power apart from lights.

[IanB]

Its designed for specific use cases, in this case giving something higher power than a low power LED light better efficiency than a low power device either through budget constraints and not needing a heatsink on the mosfets or because the customers want the best lifetime out of the inverter when powering moderate to high power consuming devices.

I'll mention again, it is not about the low power, it is about the type of load. The inverter is more efficient driving lower power incandescent bulbs. A 6 W incandescent bulb worked out at 74%, while a 6 W LED bulb managed only 46%.

[IanB]

Sounds like you need to borrow a thermal imaging camera to see where that additional heat is going.

Or a cheaper way using a thermocouple device.

Yes, it would be interesting to know whether the additional heat is in the inverter or in the bulb. I don't have an easy way to figure that out (no thermal imaging camera), but I might see if an IR thermometer can help at some point.

[tszaboo]

My question is more of curiosity that the inverter is very inefficient when driving such loads compared to resistive loads. This is not something I've heard anyone talk about, and so I wondered if anyone with an electronics background might have an idea of exactly why this is the case?

I don't think it's the inverter which is inefficient. Let's make some assumptions. The LED bulb will have a capacitor in series with the phase,  a rectifier bridge (probably a smoothing cap) and a constant current driver, and all the LED in series. So in your capacitor, which is sized for 50Hz with zero harmonics will have a certain voltage drop. Let's say 50V. And the linear driver will have a few volt's  (3 for the sake of this calculation) of drop. The drop on the capacitor is not a loss, because it's not a "real" current (that's why you have power factor) If you feed it with modified sine wave, with a lot of harmonics, the capacitors voltage drop will be lower, because reactance is less at higher frequency. So the linear driver's voltage drop will increase. Let's say the capacitor will only drop 30V, then suddenly the linear driver will have to dissipate 23V, and your efficiency went down by 50% or so (on 110V mains). Here is a driver like that. https://www.onsemi.com/download/data-sheet/pdf/nsi45015w-d.pdf Figure 14.

[IanB]

That makes sense. So it's likely the bulb will be running hot inside and may have its life shortened. I might be able to check that by looking at the temperature of the bulb base where the electronics are.

[james_s]

I don't think it's the inverter which is inefficient. Let's make some assumptions. The LED bulb will have a capacitor in series with the phase,  a rectifier bridge (probably a smoothing cap) and a constant current driver, and all the LED in series. So in your capacitor, which is sized for 50Hz with zero harmonics will have a certain voltage drop. Let's say 50V. And the linear driver will have a few volt's  (3 for the sake of this calculation) of drop. The drop on the capacitor is not a loss, because it's not a "real" current (that's why you have power factor) If you feed it with modified sine wave, with a lot of harmonics, the capacitors voltage drop will be lower, because reactance is less at higher frequency. So the linear driver's voltage drop will increase. Let's say the capacitor will only drop 30V, then suddenly the linear driver will have to dissipate 23V, and your efficiency went down by 50% or so (on 110V mains). Here is a driver like that. https://www.onsemi.com/download/data-sheet/pdf/nsi45015w-d.pdf Figure 14.

Is anybody still making LED bulbs that have a capacitor in series? Every one that I've taken apart has either a small SMPS of some sort or a linear LED driver IC with enough LEDs in series that it doesn't have to drop much.

[2N2222A]

Do you have a way of verifying that the amount of light being produced is not greater when used on the square wave inverter power?

Do your LED lights have potted control circuits that prevent you from investigating it?

Regular 120VAC peaks at 171V, but those inverters peak at 130V to 150V depending on input Voltage and load. Very few have internal voltage regulation. Most adjust the duty cycle of the AC output directly to maintain close to 120VAC RMS.

[james_s]

Do you have a way of verifying that the amount of light being produced is not greater when used on the square wave inverter power?

It's trivial to take a relative measurement, a photodiode or small photovoltaic cell and burden resistor connected to a multimeter will make an effective light meter. 

[tszaboo]

I don't think it's the inverter which is inefficient. Let's make some assumptions. The LED bulb will have a capacitor in series with the phase,  a rectifier bridge (probably a smoothing cap) and a constant current driver, and all the LED in series. So in your capacitor, which is sized for 50Hz with zero harmonics will have a certain voltage drop. Let's say 50V. And the linear driver will have a few volt's  (3 for the sake of this calculation) of drop. The drop on the capacitor is not a loss, because it's not a "real" current (that's why you have power factor) If you feed it with modified sine wave, with a lot of harmonics, the capacitors voltage drop will be lower, because reactance is less at higher frequency. So the linear driver's voltage drop will increase. Let's say the capacitor will only drop 30V, then suddenly the linear driver will have to dissipate 23V, and your efficiency went down by 50% or so (on 110V mains). Here is a driver like that. https://www.onsemi.com/download/data-sheet/pdf/nsi45015w-d.pdf Figure 14.

Is anybody still making LED bulbs that have a capacitor in series? Every one that I've taken apart has either a small SMPS of some sort or a linear LED driver IC with enough LEDs in series that it doesn't have to drop much.

I've seen E14 socketed bulbs that had dropper in them, some 5 years ago. The socket is very small to place any magnetics in them. Of course these come in different shapes and sizes. These were these sort of lamps like below, and they are quite bad to be honest (no smoothing cap in mine, 100Hz blinking and failed bulbs). According to Big Clive, these filaments have 70V forward voltage or so, so the two in series is only 140V. Size dropper cap and linear driver accordingly I guess.