Buck driver for Luxeon with zetex chip?

I was thinking that maybe it is not such a bad idea to use a higher voltage, e.g. 9V and use a buck-converter for the Luxeon. The main advantage that I see is that the current draw should be considerably lower compared to a 2-cell boost configuration. By the time the 9V battery discharges to 4V or so I bet that it would be difficult to suck out much more power out of it, even with a buck-boost converter. Am I wrong?

I remember seeing some alkaline cell discharge charts that indicated that the cells deliver a greater total amount of energy if the current drain is kept low. Also, because the losses in the circuitry often are I square * R then keeping the current low could help make the conversion more efficient?

I know that the Zetez 300 chip can be used in Buck mode. Does anybody here have any experience with such circuit? What are the differences regarding component selection? Can the same fmmt617/618 transistors be used? Are there any known problems?

It uses an FET transistor - is there some special reason for this?
Also, I want to make a converter for a 1W LED instead of 3W, sorry for not making this clear in my first post.

I have same question in my thread: MOSFET or Transistor in Zetex circuit?

This is another circuit: 9V battery for mods?

Yes, but it seems that your question has not been answered either.

So once again let me ask: Why should an FET transistor be used instead of a bipolar transistor in a buck converter?

You save the energy that goes through the base of the bipolar transistor---as the data sheet states, you need 500mA through the base to drive it to 50mili-ohm.
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Specifications (portion):

FMMT617

RCE(SAT) 0.05 Ohm at 3A
IC CONT 3A
IPeak Pulse Current 12A
Base Current 500mA
Power Dissipation at Temp 25 C 625 mA

ZXMN2A01F

V(BR)DSS = 20V
RDS(ON) = 0.12 Ohm
Continuous drain current 2.2A
Idm Pulse drain current 8A
Power Dissipation at 25 C 625/806 mW


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So let's say using a FMMT617 to drive a 1W Luxeon, Vout 3.3V to 3.7V, Iout 350mW, what are those numbers mean? How much power (watt) will be consumed by the FMMT617?

And how about the MOSFET? I see that there also is a "RDS(ON) = 0.12 Ohm" resistance there. Does it mean that it is the working resistance (ON)? If yes, then would it consume some power while working?

Don't forget about the considerable losses of a schottky diode...

That is my another quest.

If we use the "maximum battery life solution", not only we can cut 2 components out of 6, but we can raise the efficiency about 6% to 7 %. Especially in a 1 AA boost converter, the average Vin is about 1.2V, at this point the efficiency is from about 82% raised to about 92%, about 13% difference at Vin dropped to about 1V (per charts provided by Zetex).

People say that the CAP(out) would reserve and release some energy during the inductor charging cycle, thus would increase the drive power to the LED and increase the LED brightness. Is there any energy being wasted and in what form without the CAP(out) connected?

What I am thinking is that what ever power reserve or release is all comes from the power source. That means to me the increase of drive power to light up the LED is just drawing more power from the battery to reserve to the CAP(out) and release later.

Did anybody do the measurement of Power(in) and Power(out) in both of the M.B. and M.E. circuits on same brightless level? Assuming we have 10% better eff. with the M.E., we can increase the current 10% higher to drive the LED, that makes the M.B. and M.E. are at the same power consuming level, what would be the brightness difference in these two circuits?

I read the specs at http://www.zetex.com/3.0/pdf/fmmt617.pdf and it does not say anything about 500mA base current as a condition for Rce = 5mR.

Yes, there is loss in the output cap, in the form of heat. The ESR is the bulk of the loss in all but ceramic, and then there is absorption losses to account for.

If you just omit the capacitor, the LED sees pulses, and LEDs get considerably less efficient when pulsed. I posted a thread devoted to this.

The datasheet does say this for saturation, 150mV Vce at 3A collector current requires 50mA base current. 0.150/3A= 0.05 ohms or 50 milli ohms It also turns out the base voltage is 0.9V during this time. 50mA * 0.9V = 4.5mW goes up as heat during the time it is on. For a 1W converter, that is alot, but if you are pulling 3A, you ain't doing a 1W converter... Take a look at the current gain on the part to determine how much base current is needed. There are some nice charts that come in handy.

Also consider the turn on and turn off transition time of the transistor or MOSFET, and how much power is consumed doing this. The longer the transistion time, the more power is burned up during that time the part is basically nothing more than a resistor. But using more power to drive it faster also burns up power in the driver.

Hi Newbie,

On the second chart of your Thread, I see that the average flux of the blue LED is about 75% at 50% duty-cycle. It seems to me that at 50% duty-cycle, the power consumption should only be about 50%(minus those loss in turn on/off of the switcher) at the full-cycle, but it yields 75% flux. So on contrary, it is (75-50)/50=50% more efficient to drive a LED with pulse current. And our human eyes will not sense any light pulse since the working frequency is so high.

Am I right?

Mike,

Wrong chart, thats from a LumiLEDs sheet.

It is definitely not more efficient whatsoever to drive LEDs with pulses.

Think about it. For an LED, when you turn the current down, the Lumens per Watt increases, and can easily increase 3x, 5x, or more. Where for a pulsed LED, you are hitting it with the full current, so the efficiency is nearly the same as running it at that current. Yes the slug temperature is lower, but the thermal time constant of the die is extremely short.

Sorry, you really are wrong, good try though.


Instead, here are actual measurements using equipment that is alot more expensive than the car I drive:

http://www.candlepowerforums.com/ubbthreads/showflat.php?Cat=&Board=UBB6&Number=821760

50mA is by all means different from 500mA, besides, the switching losses must be much greater in the boost version of the circuit where the peak current is probalby two to four times that of a buck circuit. I think that the peak current would only be 0.5A or so when driving a 1W LED from a 9V battery, so I still cannot see the point with the FET.

Newbie,

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Wrong chart, thats from a LumiLEDs sheet.

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Did you mean my understanding of the chart was wrong, pulled a wrong chart or the chart provided wrong info?

Newbie,

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The datasheet does say this for saturation, 150mV Vce at 3A collector current requires 50mA base current. 0.150/3A= 0.05 ohms or 50 milli ohms It also turns out the base voltage is 0.9V during this time. 50mA * 0.9V = 4.5mW goes up as heat during the time it is on.

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In your calculation, the 4.5mW should be 45mW, right?

On the FMMT617 spec page, I see this:

Vce(sat) min(mV)----max(mV)------Ic-------Ib
-----------8---------14---------0.1A------10mA
----------70--------100-----------1A------10mA
---------150--------200-----------3A------50mA

And from the chart Vce(sat) vs Ic, I estimated Vce(sat) should be around 30mV at 350mA(1W driver), so I got 0.086 ohm. On the same page the Base-Emitter Turn on Voltage = 0.84V(we only need to turn on the switch not to saturate while working?), so the power turned to heat would be 10mA x 0.84V = 8.4mW, which was much lower than the 45mW.

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mike101 said:
Newbie,

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The datasheet does say this for saturation, 150mV Vce at 3A collector current requires 50mA base current. 0.150/3A= 0.05 ohms or 50 milli ohms It also turns out the base voltage is 0.9V during this time. 50mA * 0.9V = 4.5mW goes up as heat during the time it is on.

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In your calculation, the 4.5mW should be 45mW, right?

On the FMMT617 spec page, I see this:

Vce(sat) min(mV)----max(mV)------Ic-------Ib
-----------8---------14---------0.1A------10mA
----------70--------100-----------1A------10mA
---------150--------200-----------3A------50mA

And from the chart Vce(sat) vs Ic, I estimated Vce(sat) should be around 30mV at 350mA(1W driver), so I got 0.086 ohm. On the same page the Base-Emitter Turn on Voltage = 0.84V(we only need to turn on the switch not to saturate while working?), so the power turned to heat would be 10mA x 0.84V = 8.4mW, which was much lower than the 45mW.

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Good chance the decimal got squirrely on me.

If you don't drive it into that saturation point, the resistance in the FMMT617 will be much higher, and the losses incurred the collector-emitter current much greater.
When you just turn on a transistor, it starts allowing current to flow from emitter to base, but the "resistance" is high. Increasing the current in the base-emitter junction to a point where you reach saturation levels puts the part into the saturation point, where you get that low resistance.

A transistor is not really a switch. X amount of current in the base-emitter junction produces X amount of current in the collector-emitter path.

The Hfe number will tell you the gain you get. The "magic" of the FMMT617/618/619 is its high Hfe at high current levels, as well as it's ability to still respond decent enough for a switching power supply. See the datasheet, Hfe vs. IC and take particular note of what happens to Hfe as it moves over temperature, as the die heats up due to current within the part. Keep in mind, this Hfe is with a Vce of 2V, not saturated....

So, in short, yes, you need to drive the transistor into saturation for low loss.

BTW, that wasn't Vce(sat) minimum, that was typical.

Now, since you chose a 100mV collector-emitter, when driving only 10mA into the base, add on the loss for the current flowing through the collector-emitter, P=E*I, so 100mV*1A= 0.1W dissipation here during the time it is on, on top of your 8.5mW, so 0.1085 Watts during the on time.

At this point, you may want to go look at the schottky diode forward voltage, and the amount of current you have flowing through it. As the Vin goes up, the schottky diode will be on for a longer portion.

Also think about this:
Say your input voltage was 2X your output voltage, so the duty cycle was half and half, or 50%. Lets say you were driving 1A to the emitter. Well, since the output diode is only on for 1/2 the time, to get 1A to flow in the Luxeon, you will need an average 2A to flow through the schottky diode for half the time...

Newbie, thank you for the detail explanation. I am still digging in your Constant Current vs. PWM dimming Revealed . Need time to digest.

Now back to the issue of FET or transistor. When comparing the MOSFET and bi-polar transistor(max brightness) verson circuits, I found that both circuits basically need the same amount of components, 1 IC, 1 FET/Transistor, 1 inductor, 1 resistor, 1 schotky diode and 1 capacitor. Looking at page 2 of this datasheet, looks like we can not get rid of the schottky loss, as well as the L, R, C losses.

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operator smooth said: so I still cannot see the point with the FET.

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Me too!

So, what we need to do is just compare the power consumptions of the MOSFET and transistor, both are driving the LED at the same power level.

Exactly.