Transformer -- rated power

Al Gore's house.

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You can't have a sense of humor, if you have no sense.

If that 450 W load is with incandescent lamps with slightly above 10 lm/W efficiency, the total light output would be about 5000 lm.

To produce that amount of light, the most economical solution would be to use normal fluorescent lamps. The standard series have CRI index above 80, while colour corrected versions (with slightly lower efficiency) in the /9xx series have CRIs above 90 and are available in colour temperatures at least in the 2700-6500 K range. The system efficiency is about 40-50 lm/W, thus you should be able to maintain the current illumination level at about 100 W, thus saving about 350 W.

Since you apparently live in a hot climate needing air conditioning for several months a year, the reduced heat load inside the house will also save in the air conditioning bill, perhaps with a similar amount as the actual savings in the lamp consumption.

Despite various claims by various LED manufacturers of over 100 lm/W efficiencies, these are laboratory values for the bare LED without electronics and at 25 C junction temperature. :-).

The system efficiency for any realistic LED setups is at best in the same range as fluorescent lights.

White LEDs are usually made with a narrow band blue emitter and some phosphors converting some of the light to a broadband yellowish peak, but still letting some of the original blue through. The colour temperature is controlled by altering the ratio between the narrow blue peak and the broad yellow peak.

The highest efficiencies are claimed for those cool white (over 5000 K) with a strong blue peak, but unfortunately the CRI is very low (60 something) and thus not really suitable for general illumination.

Those warm white (2500-3000 K) LEDs have a weak blue peak and a strong yellowish peak, with lower efficiency, but with a CRI around 80 i.e. in the same order as office fluorescent tubes, which are considered barely acceptable for domestic use.

I do not understand, why anyone would consider LEDs for general lighting today, but of cause, LEDs are usable for spot lights etc.

John --

But I measure 330ma with my VOM when I put it in series with the DC24+, into the

6 series LED'S.

Because my meter (an old analog simpson 260) will only show the peak?) If I wanted to measure accurately what would I use? Some sort of fast sampling meter that would show the real "average" current? (A recommended meter I'd appreciate -- would like to buy one -- cheapy at MCM or something?)

So if it's 60ma, I can actually (and safely) use a lower rated transformer like

100ma -- very cool.

If I had the right VOM I probably wouldn't have even asked the question but I could never figure this one out -- thanks so much.

the

meter that would show the real "average" current? (A recommended meter I'd appreciate -- would like to buy one -- cheapy at MCM or something?)

100ma -- very cool.

If the switching regulator is smart, and it probably is, and if it's efficient, ditto, power out (to the LEDs) is about equal to power in (from the 24v supply into the switcher.)

Power out = total LED voltage times LED current

So if your six LEDs need (I'm guessing) 15 volts at 330 mA, they need about 5 watts of power. So the power supply has to furnish 5 watts, because there's no free energy in the world.

So the power supply has to furnish 5 watts / 24 volts = 208 mA. Allow for 250 maybe, to make up for inefficiencies. The "gear ratio" is

24/15. If your LEDs drop more voltage, the current ratio get bigger, approaching 1.

If you only drove one LED, you'd be delivering maybe 2.5 volts or so at 330 mA, which is only around 0.8 watts. So the 24 volt supply has to furnish 35 mA or so. Big gear ratio.

could never figure this one out -- thanks so much.

Your VOM is probably pretty accurate on indicating the actual power supply current.

John

circuit, using my VOM.

help. the transformer question in particular.

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OK, here's your original post:


I've been playing around with LED drivers and need to understand
something better.

Ok, you have a 1 watt led rated at 330 ma and with the driver my 500
ma 24v xformer runs very cool especially compared to NOT using the
driver and making a direct DC connection to the led, which makes the
xformer nice and warm.

Thanks so much to you guys for the tutorial. I'm printing this out for future reference.

I have seen some very small transformers in some of these drivers that have 120v input, in fact I also bought another part off ebay that's a 1w driver, good for up to 6 leds and the transformer is about a half inch square and doesn't even get warm.

For the heck of it, I replaced my 500ma transformer with one rated for 180ma and bench testing it -- 24 hours so far and warm but no problems.

You are reading 330mA through the LED string -- yes? If so, then as the voltage is stepped down from 24V on the input side to whatever is used on the output side, the current is stepped _up_ from whatever is used on the input side to the 330mA on the output side. If the converter is reasonably efficient, then the current consumption _from the transformer_ should be fairly low.

Have you measured the supply-side current? Doing so should clear up much confusion.

Less than 330mA is required, if it's a switching LED driver.

Of a sort -- you need to learn more about electronics.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

If it is a switching driver, then you want to start by looking at power equations:

1 Watt to the LED, divided by the efficiency of the driver, gives the input power. Assuming that it's 80%, that's 1.25W at 24V. That's a bit over 50mA, which is significantly less than 330mA.

If you are slapping that transformer directly across the LED, then you are doing _bad things_. First, 24V at 500mA is what the transformer is _rated_ at, not what it necessarily _does_ at any given moment. It's like the spedometer on your car -- just because it goes up to 100 miles per hour neither means that your car _will_ go 100 miles per hour, nor does it mean that your car can't go _more_ than 100 miles per hour.

So when you put that LED straight on the transformer leads, the transformer is doing the current limiting, and is probably delivering more than 500mA to the LED. Lots-o-current = lots-o-heat.

330mA is 330mA, but 330mA _output_ does not necessarily mean 330mA _input_.

'tisn't. With a switching supply the average input current does not, in general, equal the average output current. And the input and output currents should be somewhat smoothed, and therefor close to average.

Voltage step _up_ from 24V would require more current demand. Where is your basic physics? Without any energy storage, power in = power out. That means that whatever power goes into your driver, it comes out in the form of electrical power to the LED and heat. If you're burning up 1W in the LED and 1/4 watt in the driver, then you can't be sucking up more than 1.25W from the transformer. If you are, then you should call the Physics Cops.

Yes and no. Yes it is based on dissipated heat, but also on how much heat is generated for a given current. So no, it is not "not really current".

Not necessarily. But if you're really only using 50mA on the input, then yes.

The transformer isn't rated for any one power supply in particular. But the folks that rate it do hope that the user understands the rating.

I'm sitting on my hands on this one.

Learn some really really basic physics, then apply them.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

What all of it comes down to is the fact that you CAN use a lower rated transformer and that's what inspired my question in the first place.

You CAN do it (although maybe you shouldn't) because it's being done with all of these LED drivers that I'm seeing. The transformers in fact are pitiful.

I knew that a switching supply could reduce heat but I didn't realize that it could reduce it so much -- it's kind of confusing because although it isn't steady state, it's still ON and OFTEN (depending on the duty cycle).

By the way Tim, no need to be rude or condescending -- take John Larkin for example -- the guy is always polite.

I'm mis-age and have loved electronics all of my life but sometimes I can be a real D- student -- it just depends on what I'm trying to digest.

Bottom line, I know now why I can use a CHEAPER transformer.

something better.

24v xformer runs very cool especially compared to NOT using the driver = and making a direct DC connection to the led, which makes the xformer = nice and warm.

as well for full brightness.

(because it's on and off based on the duty cycle) but again -- the = current is still there.=20

"step up" going on creating less current demand.

dissipate? (and not really current?)

Pretty much. There is a little bit more that i am aware of, such at the thermal limitations of the insulation. Maximum rise in the transformer temperature (only partly due to insulation issues, many bobbins have much lower temperature limits) and other transformer construction factors dominates. The magnetics can only do so many VA at a given frequency and similar limitations in the conductors as well as the insulation are roughly balanced in modern transformer design.

something smaller say 250ma or so? BECAUSE --

(continuous=20

Not really, just ignorant. A much more tractable situation.

?-)

is

True, but i would rather use a reasonable zener with a husky bipolar and = a heatsink, maybe the same as the LED.

so WHY does my 500ma transformer feel so cool with the switching driver?

(always on) power supply?

less than the 330ma required?

This idea might help, milliwatts is milliwatts. Measure the transformer current with the controller and without. Should be over 5:1, maybe over

10:1. Thet difference translate directly to copper losses in the transformer, and perhaps slightly more/less core losses.

?-)

transformer and that's what inspired my question in the first place.

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What do you mean by "transformer"?