Ignore

[Doug S]

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Doug S said:

Arithmetic: 0.3W/2.54W does not equal 0.031


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Very true.

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Doug S said:
Conceptual: This is not the correct formula since the schottky is not conducting 100% of the time. To avoid all of the complicated duty cycle versions of the correct formula, an easy approximation is to take the loss fraction to be (Vfschottky)/(Vfschottky + Vfled) We understand, of course, that this *slightly* overstates the loss since it is not acting on the full input power but instead the input power reduced by the losses up to that point.


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Your loss equation for the schottky diode at first glance to be inadequate, and appears to be quite wrong looking at it further and quite significantly underestimating the loss in the schottky diode???

I'll have to differ on a point though. The average current comming out of the output side is 1A. Thus the current in the diode could easily be 2A, depending on the input voltage which affects the duty cycle, at which point the diode drops more voltage across it. Remember, to get 1A output, since the current flowing in the schottky diode is only on for a portion of the duty cycle, the current will increase significantly above 1A- directly in porportion to the duty cycle.

http://rocky.digikey.com/WebLib/Diod...a/1N5819HW.pdf

The new fancy dancy Zetex schottky is in the same boat:
http://www.zetex.com/3.0/pdf/ZHCS1000.pdf


At 2 Amps the Vf jumps to 0.5V. So the losses are P= V * I
2 Amps * 0.5 Volts = 1W of loss. If the duty cycle was 50%, then we'd have an average power loss of 0.5W. Remember the Vf of the LED was 3.9 and the current 1 Amp = 3.9W output.
So, 0.5W/3.9W= 0.128 or 12.8% loss from the schottky diode alone.

Sounds alot worse.


It gets complicated yes, to get more specific, but here is a common approx. found in many switcher datasheets, and works out to pretty darn close to reality, even though the math isn't specific:
http://www.linear.com./pc/downloadDo...15,P1597,D1509

Page 14, starts right at the top.

"and the average power dissipation (PD) in the diode is:
PD = (IOUT)(Vf)
where Vf is the forward voltage of the diode at ***peak***
current. "

So, if you want to make things a tad more complicated, but still keeping things simple, figure out the Vf drop at peak current, and use the average current out.

So, we'd have 2A peak with a Vf drop of 0.55, and an average current out of 1A. So 0.55 * 1A = 0.55 Watts lost in the schottky alone, or 0.55W/3.9W = 0.14 or 14% loss due to the schottky, so we start at 86% efficiency before we even account for MOSFET or Inductor losses. Tack in the MOSFET loss, and add in the Inductor loss, and you are in the 70's % efficiency range quite quickly.

BTW, for those curious, the duty cycle is 1 – (Vbatt/Vout)
In our case, 2.2V off the NiMH, and we need to make 3.9V.
so 1 - (2.2V/3.9V) = 44% So the MOSFET is on for 44% of the time and the schottky diode is on 56% of the time. So the diode current would be about 1.8 Amps. So the schottky diode drop would be about 0.5V.

Of course, that is if I am still thinking clearly at 1:42 am in the morning.






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greenLED said:
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dat2zip said:
In fact I have a derivative of the Nexgen that will do more than 1A from a single 123 cell and it is fully regulated constant current design.


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/highjack alert!/
Also, am I understanding correctly that driving a circuit at 3v with 1x123 is "better" (whatever that means electronically) than using a 2xAA config? Can someone please explain why this is so, or point to a source I could check? I'm trying to educat myself on the electronix world.
Thanks!
/Highjack off/

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greenLED

Lets take an imaginary converter.

Consider the sag of the NiMH under high loads and also the 123 cell. Think about the internal resistance of the battery. Then think about how the cell voltage sags over time under load. Consider the starting voltage. Remember power in is voltage * current = power. You want power.

But remember there is resistance in the MOSFET/Transistor to ground. Higher voltage means you need less current for the same power. The majority of loss in the MOSFET is due to Power = Current^2 * resistance. So, a small decrease in input voltage means you need more current, but the losses in the mosfet are due to Current being squared, so it takes it's toll quickly.

The same type of thing happens as the battery sags.

In every battery, you will find some internal resistance. This causes the cell voltage to sag such as when you put a 1 Amp load on it. A NiMH that I have here charged two days ago. When it isn't under load, the voltage across it reads 1.287V. When I put a 1 Amp load on it, it's output voltage drops to 1.190V. If I was utilizing two of these, the voltage would be 2.574 Volts, and under a 1 Amp load, the voltage would drop to 2.380 Volts, and they would sag from this over time as they were drained.

But, if I set everything up, such that I needed 2.574V on the input at 1A for 2.54W going in, and then the voltage dropped to 2.38V, then I'd need to draw 1.07A to maintain the same power going in. As the cell is used, the battery voltage continues to drop. Lets look at the point As the cells drain down to their 1.8V ( 0.9V each) cut-off point, I'd need to draw 1.41 Amps to maintain a flat light output level down to the battery cut-off point.

Now, let say your input MOSFET had 0.1 ohms. With 2.54V off the battery and 1A flowing into the input, there would be Power = Current^2 * resistance loss, or 0.1 Watt of loss. Keep in mind we are not including the output MOSFET/schotty diode loss or the power needed to drive the MOSFET gate, nor the power needed to make the control circuits function. So then we take the power lost and divide it by our input power. 0.1W/2.54W= 0.04 or 4% loss due to the resistance just in the input MOSFET.

Now, when we drop on down to 1.8V at 1.41A, things change.
Power = Current^2 * resistance. The MOSFET had 0.1 ohms of resistance. So 1.41^2 * .1 = 0.20 Watts of loss. We have 2.54W input, so the power loss in the MOSFET due resistance/input power, or 0.20W/2.54W = 0.0787 or 7.87 % loss.

Then we go on to the schottky diode, such as used in the BB and these Zetex converters, on the output side of things. There are some super low loss schottky diodes that will have a Vf of only 0.3V when 1A flows through them. Well, power also equals Power = Voltage * Current. Since we are always putting out 1A in a constant currrent converter, this loss is easy to figure. 0.3V * 1A = 0.3W So, we had 2.54W, so we take this 0.3W/2.54W = 0.031 or 3.1% loss.

So with a new battery, only accounting for just two of the loss factors (there are many more), we'd have 7.1% losses with a fresh cell, and 10.97% losses near the cut-out.

You still need to add in losses due to the inductor DC resistance, eddy current losses in the inductor, B-H losses in the inductor, power lost driving the gate charge in the MOSFET, power lost in the circuit that drives the gate, losses due to charging and discharging the MOSFET body diode (internal to the MOSFET), losses from driving the capacitance in the schottky diode, losses from operating the control circuitry in the chip, losses due to the ripple current in the capacitors, and several other minor loss factors. But for know, we will keep it basic and simple.

Remember how we wanted 2.54W on the input? Well if I start off with a 123 cell that has 3.2V, remembering that Power = Voltage * Current, swapping things around Current = Power/Voltage, so the input current needed is 2.54W/3.2V= 0.79 Amps. Guess what, you have less amps flowing in the input MOSFET, and the losses here drop down to only 2.5%. If the batteries had the same internal resistance, this lower input current necessary, would also mean you'd have less input voltage drop due pulling less current out of the battery. Then the 123 Lithium battery typically has a rather flat discharge curve, so the current drawn towards the end wouldn't go as high, both from the less internal loss and from the higher voltage at the end of battery life.

It's late, and I hope you were able to follow things through all of this.

Now, lets look at things further. Most Luxeon LEDs are around 3.9V when you have 1 Amp flowing in them. P=I*V (I is Current), so in reality, we need 3.9 Watts.

For a moment, imagine you had an imaginary unrealistic absolutely losses converter.

Under load, my two NiMH cell voltages added up to 2.380 Volts. Okay, so how much input current do we need to produce 3.9W? Power/voltage= current 3.9W / 2.38V = 1.64 Amps. Humm, I go back and measure my NiMH voltage at 1.64 Amps and find that the Output voltage of the two NiMH cells drops to 2.28 Volts. Ouch. So I go back and take 3.9W/2.28V= 1.7A I'll need to pull from the batteries. So I again go back and measure the cells, this time at 1.7A, and I find the output voltage of the cells drops to 2.268 Volts. Going back and doing back and repeating again, 3.9W/2.268 Volts = 1.72A, and we get to 2.2648V. Okay, lets stop here.

The next major loss in the circuit is most likely the inductor, but we won't go into things that deep right now.

Hopefully you get the basic idea...when you start with a higher voltage, you need less current on the input, and the losses in the converter (for a boost) will be less. Basically a win-win type thing.