explain efficient please.....

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Robocop

Robocop

Mammoth Killer
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I often hear that one circuit is more efficient than another in regards to lights. Can someone explain what this means in simple terms?
Is it correct to say that a light that gets 1 amp at the tailcap and actually provides 800 mA to the led is more efficient than a light that gets 1 amp at the cap but only 500 mA to the led?(with both circuits using the same power supply)
Does a more efficient circuit simply push more power using the same voltage? Does a more efficient circuit simply mean that it will suck a battery down more than another?
Just curious as I hear this term often yet do not know exactly what it means as far as boost/buck circuits go.
Thanks...
 
Power P is the product of voltage U and current I: P=U*I. Measured therefore in VoltAmperes, but more commonly called Watts.

Efficiency is simply the ratio of Pout and Pin: Pout/Pin.

Your comparison of currents is therefore only valid if the voltages are the same -- it works pretty well for LEDs, though, since LEDs need their forward voltage (or maybe a little bit more) to even light up.
 
To add on to what elgarak said your typical boost circuits use extra current to produce high enough voltages in the range of the Vf of the LED etc, which buck circuits try to use extra voltage to produce more current at the lower Vf level. Some lights use resistors or voltage regulators instead of a buck circuit which at certain levels of voltage above Vf tend to be less efficient than a buck circuit is.
 
Since energy can neither be created or destroyed, efficency is a measure of how much of the energy is actually converted into the desired form. In a light, this would ultimately be the measure of light produced from the electricity used. However in many of these lights and especially the LED lights, the electricity itself is controled and regulated prior to being provided to the LED. In this control and regulation, some of the energy is lost as heat. A 90% efficient converter that is managing 10 watts of power will be sending 9 watts of electricity to the LED and emitting 1 watt of heat.
 
Volt amperes is only power for DC, and for AC where V and I are precisely in phase. In the real world of AC, that is almost never the case. Transformers and electric switch gear is rated in VA, but there is no assurance at all that it is power.

Consider this for a moment. I feed AC to a super conducting inductor. V will be sin(theta), I will be cos(theta). The two are at quardature. The integral over time is zero. NO ENERGY TRANSFERRED! Average Power is in fact ZERO although volt amps could be very high. This is why the Electric company is not fond of reactive loads. They get I^2 x R losses in the transmission system, but because you consume no energy, the watt hour meter doesn't budge.

Both AM stereo and Independent Side Band tranmission are dependent on the fact that the average power for two signals modulated at quadrature is zero. The Telephone company used to make very extensive use of that fact, as did Milgo 4800 baud modems in the 1970's...
 
Hi Robocop,

In simple terms, your example is correct. Assume you have two circuits and test them both with the same power source and the same LED with a stable Vf. If both circuits draw 1 amp from the power source, but Circuit-A delivers 800mA to the LED while Circuit-B delivers 500mA to the LED, then Circuit-A is more efficient.

In the real world, however, you can almost never test two different lights with the same LED. Also, it is highly unlikely that two different circuits will have the same current draw. Also, the Vf of most LED's will go up as the current rises.

Elgarak gave the formula for measuring efficiency above, but I'd like to expand on that with an example.

P(ower) = V(olts) * I(current)
Efficiency = P(out) / P(in)

So in a light, you must first measure the voltage of your power supply and the current being drawn by the circuit. For this example, we'll say that our power supply is a AA battery.

The battery, when the circuit is on, has a voltage of 1.3V and is delivering 1.2Amps of current. So, it is delivering (V*I) watts of power, or:
1.3V * 1.2A = 1.56 watts

Now you measure the voltage across the LED and the current being delivered to the LED. Assume that we're measuring a Lux-1 and we measure a Vf of 3.5V and it is being driven at 350mA.
3.5V * 0.35A = 1.225 watts

Now, our P(in) is 1.56W and our P(out) is 1.225W, so we divide P(out) by P(in) to get the efficiency:
1.225W / 1.56W = 0.785 = 78.5% efficient.

You will usually see these numbers represented like this:
Vin: 1.3V
Iin: 1.2A
Vout: 3.5V
Iout: 0.35A
Eff: 78.5%

The additional 0.215 watts of power in this example are being converted into heat in the circuit.

pb
 
Ok these big words tend to confuse me at times however thanks for the explanations. I am still a little shaky as to the meaning of what flashaholics consider to be a super efficient circuit.
Lets say that we have 2 circuits both using a 1.5 volt power supply and identical bin luxeons. Lets say both circuits show 1 amp at the tail cap. One of these circuits supplies 800 mA to the luxeon while the other only supplies 500. Would this be the meaning of a more efficient circuit for the one putting more mA to the luxeon ?
Using the same example above the mA to the luxeon on the weaker circuit is exactly half of what it is pulling at the tail cap. Would this be a 50% efficient circuit ?
I keep thinking of efficient as less restriction allowing more of the total mA to actually reach the luxeon...is this correct thinking?
This is almost reminding me of a high power motor in a vehicle. You can have a 700 HP motor however if only 400 HP is reaching the tires then your motor is not very efficient...well there are many variables in that example but you get my point.
What parts on a circuit make it more efficient than another? I am also wondering this...I may have a circuit that can make 1 hour of very bright light from a single 123 cell. After this hour I get 2 more hours of very low level light until the circuit shuts off at a low voltage. I then may have another circuit that makes 8 hours of much dimmer light but for much longer on a 123 cell. Even though the dimmer light is less bright I am getting light for longer all the way down to the end of the battery voltage. Would this dimmer circuit not be more efficient as it is using the total amount of energy in the cell?
I began thinking about this after I was trying to decide between the JIL DD light vs the JIL with the boost circuit. In terms of actual efficient lights is the DD JIL light not a much more efficient option? Is it better to say that all DD lights are more efficient?

Sorry for all the questions but this is how I have learned so much here....just ask the pros.
 
Robo,
If in your example, you have two circuits that are both drawing 1 amp of current from the same type battery and one of these circuits then provides 800 mA to a LED while the other would only provide 500 mA to the same LED then the one providing 800 mA is more efficient. However realistically, it ismore likely that you would want to compare two circuits that both are set to provide the same amount of current to the LED and if you were to measure the current they draw from the battery, the one drawing the lower amount of current would be the one that is more efficient. It accomplishes the same work as the other but requiring less energy to do so.

As others have stated, power is the volts X amps. A converter's efficiency is a function of its conversion of input power to output power. In your example of 1 amp drawn from the battery and 500 mA going to the LED, you can't get a feel for the efficiency unless you know what the input voltage as well as output voltage is.

If your battery is a CR123 and you are pulling 1 amp out of it, its voltage is probably around 2.5 volts under load. So you have 1ampX 2.5 volts of 2.5 watts of power being consumed by the converter. If the converter is delivering 500 mA to the LED and the LED is a Luxeon with a Vf of 3.5 volts at 500 mA then the power to the LED is .5 amps X 3.5 volts or 1.75 watts. 2.5 watts in and 1.75watts out is not very efficient! You are loosing .75 watts of power in the converter. 1.75watts/2.5 watts= .70 or 70% efficient.

A DD system will always be more efficient than a converter because there is no power loss in the conversion (no converter).However, in DD, you have no control over the current that is being provided to the LED and it wil depend on the battery's voltage and the Vf of the LED. You may have a case of the drive level being higher than you would like or possibly lower. As the battery expends its energy and its voltage drops, so too will the current to the LED. If you can get a good match of LED Vf to battery, you will have a very efficient system. Unfortunately, it has been discussed and verified that LED's Vf can change over time, usually dropping, so a DD system will likely end up over time with the LED driven harder than initially was the case. A low value resistor can be used to put some control on the current to the LED and in a well matched system of LED to battery, the use of the resistor may be more efficient than a converter and stil give you some control over the LED.
 
Hi Robocop,

Efficiency in a circuit has nothing to do with how far down it drains the battery.

All of the circuits we use to drive LEDs in flashlights must take a component of the battery power and convert it to another component to drive the LED. A boost circuit, like my example above, must take extra current from the battery and convert it into a higher voltage at a lower current.

The efficiency is a measurement of how much power was lost in the conversion.

In your example above, you have two circuits. In order to measure efficiency, you must first measure all of the components. I'm going to use 3V instead of the 1.5V in your example.

Circuit-A:
Vin: 3v
Iin: 1A
Vout: 3.5v
Iout: 0.800A

Circuit-B:
Vin: 3v
Iin: 1A
Vout: 3.5v
Iout: 0.500A

So the efficiencies are:
Circuit-A: (3.5*0.8)/(3*1) = 93.3%
Circuit-B: (3.5*0.5)/(3*1) = 58.3%

Now.. This is a bad example because identically binned luxeons will not have the same Vf at different drive levels. If you run a Lux-III at 500mA and it has a Vf of 3.5v then it is likely to have a Vf near 3.7v when being run at 800mA.

A direct drive light is always the most efficient because 100% of the battery power is being delivered to the LED. Discounting any losses from the resistance of the wires connecting the battery to the Lux, a DD light can be considered 100% efficient.

A DD light isn't always the best choice when it comes to driving LEDs, but that's a different discussion.
 
Hi there Robo,

I'd like to add to the other posts too...

Sometimes numbers like eff=80%, eff=86%, etc. dont
seem to have too much meaning when they stand alone like
that so here's another way of looking at it...

If you have a circuit that is 80 percent efficient that
means if you drive it with one cell and you go out and
purchase 10 cells to use with it, by the time you drain
down all 10 cells running that circuit you will have
completely wasted 2 out of those 10 cells because the
circuit can only use 80 percent of the energy it receives
for providing an output...the remaining 20 percent is
completely wasted as heat.

If the circuit was only 70 percent efficient, then 3 out
of 10 cells are completely wasted.

If the circuit was 90 percent efficient, then 1 out
of 10 cells are completely wasted.

Since these cells cost money, it's obvious the lower
the efficiency the more cells are wasted and so more
money is wasted.

From this you can quickly see that the higher the efficiency
the less waste energy (and thus wasted dollars) there is.

Using the efficiency spec for circuits we can compare
circuits to see which one is highest and thus wastes the
least energy.


Take care,
Al
 
Another thing to remember is that for some circuits the efficiency shifts a lot depending on Vf, Vin and drive current. It can go from 95% to 45% caused by only a small change of voltage. Just look at Sipex SP6685 (look at bottom of page 4).

Sigbjoern
 
Thanks for the detailed responses and I think I may have it figured out a little better now. I believe it is way too complicated to grasp the details however I do get that some circuits convert less power to make more light than others.

I believe it is the actual term "efficient" that had me confused. I was thinking more along the lines of less restricted such as a more free flowing circuit that simply allowed more power to reach the emitter.
Lets try this....If I have two circuits and 2 identical amounts of power going to each one. Both circuits are a boost circuit so they must somehow boost the amount in and increase it before it leaves the circuit. Now the path this power takes before it leaves is what determines the efficient term? If one circuit somehow can do more light with the equal amount of power it is more efficient....I think I said that correct.

I remember reading a thread some time back that was talking about a "magic" circuit or something along those lines. As far as losing some power to heat goes...well I am not even going to try that one as I assumed heat was simply something that was going to happen anyway.

Regardless I am really starting to enjoy the more technical side of this hobby. At first it was simply trying to have the brightest light in the smallest package. I am now learning more about run times and the efficiency of certain applications. I just read a huge post on how a boost circuit works in general and it was very interesting. The whole thing about a field collapsing and then re-generating hundreds of times in a second...man that is confusing yet amazing to read about.
Thanks for the assist on this and once again MrAl has somehow explained something to me in terms I can grasp.
Thanks again...
 
I think a simple explanation is efficiency is related to operating range of the circuit, voltage source range, and the light device (bulb, led, etc.) all combined. If you use a circuit that cannot take complete advantage of a battery then it may only use perhaps half or so of the batteries capacity before it shuts down because the voltage input is too low or the range the circuit is most efficient in is outside the average range the battery can provide.
A light source (bulb, led) that doesn't properly match the output perameters of the boost circuit (vf etc) may not be able to perform as efficiently as one that does because the bulb itself is more/less efficient at certain ranges.

An example would be a 3.6v Vf LED driven by a boost circuit that operates from 1.5 to 3.2v input running off a single cell. It may run fine till the battery drops to perhaps 1.2v or so then... quit leaving the battery not completely used up. If you used a lithium ion in the same design it would be outside the range of 3.2 at start and may try to overboost then drop within the boost range and work then fall ofo (if protected around possibly 2v or so.
 
Robocop, basically, all circuits that regulate output in some way are inefficient to some degree, as in they take some juice to function.
Some are a lot more than others but in all cases, it means some juice from your batteries won't make it to the led.
If a circuit is very efficient, most of it will make it; if it's not efficient, a lot of juice will end up wasted in one way or another (mostly in heat). There is always a trade-off in level of regulation and efficiency.
Of course most of us think basic regulation is well worth it if it prevents the light to fall into a level of brightness that's useless for a task for most of the runtime as it is the case with direct driven incandescents wasting the juice in another way.
Of course, the circuit is not the only thing that's inefficient; the led for instance will also get more efficient in the next generation.
The experts explained it much better than I can.
 
Hi Lynx,

Unfortunately, no.

A boost or buck circuit's efficiency is simply the measure of the amount of power which is NOT lost as heat in the circuit while being converted to the required output power.


---------------[BEGIN THEORY]------------------
Assume that you want 350mA to an LED which has a Vf of 3.5v, and you are using a power source which can provide a steady 3 volts. A boost circuit which can only operate between 2.9v and 3.1v and draws only 408.3mA would be a 100% efficient circuit.

Why? Becuase:
P(out)/P(in) = 1 or:
(3.5v * 0.35A) / (3v * 0.4083A) = 1

It matters not that the circuit stops operating below 2.9v or above 3.1v. It does mean that any such circuit would not be the best candidate for a flashlight, but the circuit is still 100% efficient at converting 3v into 3.5v at 350mA.

The crux of efficiency is that all that matters is how much power is retained when the conversion is done. To determine this, you must measure only four things while the circuit is operating:
1. Input voltage
2. Input current
3. Output voltage
4. Output current

The deal is, that it is impossible for a boost or buck circuit to be 100% efficient. There will always be some losses since the components require electriciy to operate. This operational overhead is converted directly into heat by the components. So efficiency of a circuit is a measure of how much of the source power is NOT converted into heat by the circuit.

Again, let me emphasize that a circuit's ability to fully deplete a battery has absolutely nothing to do with the efficiency of the circuit.

------------[END THEORY]---------------

------------[BEGIN PRACTICE]------------

In practice, circuit efficiency is an elusive beast. The measurements must be taken and they must be taken correctly. One cannot assume that since your circuit is designed to output 350mA and the Lux is binned with a Vf of 3.5v that the output current is 350mA or that the output voltage is 3.5v.

The reality is that the circuit may not be very well regulated, and may be putting out more or less than 350mA. Furthermore, the Luxeon datasheets specify a range of Vf for a given bin at a given current. Each individual LED will have a slightly different Vf at a given current and a single LED will have a different Vf at different currents.

So a circuit which is designed to output 350mA through a Luxeon with a Vf bin in the 3.5v range may well be outputting 355mA at 3.4v through that LED.

Furthermore, you cannot assume that a 3v battery, like a CR123A, or a 1.5v battery, like a AA is actually delivering 3v or 1.5v while the circuit is powered. Batteries have a tendancy towards a drop in voltage while under load. So an alkaline AA delivering 10mA of current may well deliver 1.5v, but that same battery delivering 1A of current is likely to have a voltage much lower. Possibly as low as 1 - 1.2v instead of the assumed 1.5v.

Furthermore, a circuit which is 90% efficient when powered by 3v may only be 70% efficient when powered by only 1.5v. In these cases, to measure true efficiency over the life of the battery, you must take multiple measurements and take an average of the efficiencies.

To complicate matters further, measuring current in a pulsed current application can be problematic. Most all boost and buck converters draw current from a battery in pulses and deliver it to the output in pulses as well. These pulses are generally very fast. This confuses most multimeters which are unable to properly read and average those pulses. To correct this, it is customary to insert a very low value resistor in series with the input and output, then measure the voltage across that resistor. You then use ohms law (V=IR, or I=V/R) to determine the actual current.

------------[END PRACTICE]---------------

-----------[BEGIN "WHAT DOES IT ALL MEAN"]------------

So what does all this mean?

Basically, as MrAl pointed out, circuit efficiency is a tool that we can use to determine how much energy is being wasted as heat before it ever reaches the LED.

There are factors other than circuit efficency which we must use to determine whether a given light is good or not. The operating range of the circuit is one other factor, as is the LED used, heatsinking of the LED, the optics used to collimate the beam, and the list goes on.

However, circuit efficiency does not measure these things. It does not measure how much light, or how much more light a given circuit will provide. It simply measures how much battery power is wasted before it reaches the LED.

----------[END RANT]-----------

pb
 
Well this has been an interesting read for me and I know now that there are many factors to consider in an efficient circuit.
I see many circuits that will work well until the voltage drops below a set level. Is it correct to say a voltage regulated circuit will increase output as the input voltage drops to compensate and maintain a set voltage out?
Are some circuits smart circuits that will somehow know when voltage drops and maintain an efficient output?
I assume it is wishful thinking to say there is a circuit that can maintain a set level of output through the entire life of a battery. When I say entire life I am saying from a 123 cell that starts at 3 volts and this magic circuit pushing out 3.7 volts even when the 123 cell is only giving say about 1 volt.

I was looking at it all wrong in that I assumed a circuits ability to bump current and drain a battery dry determined the efficient part of the circuit. I always thought my Inova X5 to be the most efficient light I own as it will make light from cells that were long too weak to power other lights. I assumed that by getting x amount of use from a specific amount of energy was a good way to determine efficiency.

I am at work now so I will have to sit down with this tonight and read over the details of this thread. I enjoy learning new things and appreciate the time.
 
I have been reading stuff like this for months and there is a lot of variables. The majority of inexpensive lights use either a boost or a buck or direct/resistored drive circuit.
A few use a combination of boost or buck and regulation. Some are voltage regulated and some current regulated. Essentially you end up with boost circuits below Vf, buck well above Vf direct drive very near Vf and resistored slightly above Vf.
The cheapest way is to use more cells to go above Vf and use resistor(s) to drop it to Vf or slightly above for a little overdrive to keep brightness up.
 
There are a number of modern boost or buck chips that will achieve 96-98% efficiency, which I have verified in the lab myself. The hard part is handling higher power in a minimal space, which means smaller inductors/capacitors which will lower the efficiencies.

Besides just looking at the converter, it is always important to look at the whole system. Efficiency of the power comming out of the battery, the converter, the efficiency of the LED which is affected by heat, and even the quality of the path that carries the electricity- springs, contacts, wires, flashlight body, and the connections in between these parts. Tag on the efficiency of the reflector and the bezel lens transmission, and you have a good start.

There are a number of examples where the circuit drops the output to keep efficiency up. The ARC4 and EDC are examples as such. I made one of shiftd also.
 
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