50 watt led driver using lm3447

Hello eng riad,

I would like to get more information.  From this email I'm guessing that your transformer ration from primary to secondary is 7 turns to 16 turns?  Can you verify this?  At 250 VAC the peak input voltage is 250V * 1.414 = 353 volts.  At 7 to 16 that would equal -807 volts (this is why I ask).  The secondary side rectifier would have to be rated to 807 volts + the output voltage + any ring voltage which I would guess would need rating over 1000 volts.

I would guess the TVS diode you are talking about is the primary side clamp for the leakage inductance?  The power rating of this depends on the output power but more important the transformer design.  If the coupling of the primary to secondary is poor the TVS clamp will have to dissipate a lot of power.  If the coupling is good the clamp may still need to dissipate up to five watts depending on operating conditions.  There are also other ways to clamp the MOSFET drain during MOSFET turn-off.

It would be good to get a lot more information about your design.

im sorry for late

first we ask  about how to measure leakage inductance of transformer which is contain of secondary coil and auxiliary coil and primary coil

second we ask also about TVS diode we change tvs 170 voltage with TVS SMCJ200A  clamping voltage is 324v

and remain the same result TVS heated and burned  

so we ask about are any other parameter unlike spike of voltage will affected on TVS DIODE      

third : we try to measure the inductance of transformer we found that :

1- primary coil inductance L(1,2): 0.256mh

2- auxiliary coil inductance L(4,5): 0.00405mh

3- secondary coil inductance L(10,6): 0.018mh

all measure with open circuit

4- when short circuit between (4,5)pin and (10,6) and measure leakage inductance

we found that L(leakage inductance)   (1,2):0.033mh

and when all pin (4,5,6,10) and measure leakage inductance L(1,2): 0.0708mh

this is parameter that your request before

please advice me 

thank you  

sorry for 

im sorry for late

first we ask  about how to measure leakage inductance of transformer which is contain of secondary coil and auxiliary coil and primary coil

second we ask also about TVS diode we change tvs 170 voltage with TVS SMCJ200A  clamping voltage is 324v

and remain the same result TVS heated and burned  

so we ask about are any other parameter unlike spike of voltage will affected on TVS DIODE      

third : we try to measure the inductance of transformer we found that :

1- primary coil inductance L(1,2): 0.256mh

2- auxiliary coil inductance L(4,5): 0.00405mh

3- secondary coil inductance L(10,6): 0.018mh

all measure with open circuit

4- when short circuit between (4,5)pin and (10,6) and measure leakage inductance

we found that L(leakage inductance)   (1,2):0.033mh

and when all pin (4,5,6,10) and measure leakage inductance L(1,2): 0.0708mh

this is parameter that your request before

please advice me 

thank you  

sorry for 

Hello again,

I did the math to calculate the power dissipation for the clamp.

If the leakage inductance is 0.033 uH and the primary inductance is 0.256 uH the power dissipation for the clamp would be around 7 watts.

If the leakage inductance is 0.0708 uH and the primary inductance is 0.256 uH the power dissipation for the clamp would be around 15 watts.

The SMA package can only dissipate approximately one watt depending on how well it is thermally connected to the board.

Leakage inductance that high is bad for a flyback converter.  Looking at the transformer drawing, if it is built that way, I don't think the leakage would be that high but I have seen it too high because it doesn't match the drawing.  The secondary between the two primary windings is good.  They also need to lay over each other.  If they didn't evenly split the winding to have good coupling (how the secondary lays over the primary on the bobbin, the coupling gets worse, this increases the leakage inductance.

Regardless, at 50 watts the clamp needs to dissipate more power than what you have for a clamp.  Moving the clamp voltage up will only slightly affect the energy because it's trying to clamp the leakage energy of the transformer which is from an inductor.  I would try, like I said before, placing five 33 volt TVS diodes in series for the clamp to see if that works, that allows up to five watts of dissipation (again depends on heatsinking the parts).  If it is still too hot I would investigate the transformer manufacturing since the leakage must be too high.  Leakage loss will directly affect the efficiency of the converter.

Hello,

Units are correct, mH is millihenry, uH would be microhenry per how I use the units.  I also calculated the operation of the transformer separate from the design tool and the peak current is very similar so I know the units are correct.

Hello,

The current rating for a TVS is for a surge event.  The TVS in your circuit is being used in a different manner than original intent.  A TVS is a sort of Zener diode that can handle large pulses of current/energy.  It is used to clamp voltages of circuits during a surge event such as a lightning strike.  This prevents overvoltage of components in a circuit. 

In your application you are using it as a zener diode than can handle large pulses of current.  At 200 volts and 3.5 amps of current that is 700 watts.  This is for a very short time so the average power is much much lower.  The current rating does not matter in this application as long as it is above what the leakage current spike is, 3.5 amps.  The problem you are running into is the average power.  The SMC package is capable of one to two watts if mounted on an FR4 board.  Your leakage energy average is more than this and why the part is running too hot.  Trying five SMCJ33 parts allows five times the power dissipation.  It may still not be enough in your case especially if the transformer is not optimized to reduce leakage inductance.

Hello,

I don't think your leakage measurements are correct but if they are you will have to dissipate a lot of power in the TVS clamp circuit.  This is something you have to work with the transformer manufacturer.  I would take one apart to see how the winding fit next to each other but in you case I would just try the five SMCJ33s in series to see if it still runs too hot or not.

The leakage inductance does not need to be 1% or less, that would be ideal.  The higher it is the more power needs to be dissipated in the clamp.  It will also affect the efficiency.

Hello,

Increasing from five 33 volt TVS to eight 33 volt TVS diodes not only adds three more parts for power dissipation but there will also be less dissipation in the clamp due to higher voltage, 5 X 33 versus 8 X 33.  There are other methods to clamp and burn the leakage energy such as a diode, capacitor and resistor.  More important though is how much you have to dissipate in this clamp.  A discontinuous flyback requires the transformer be wound to keep the leakage inductance low.  I would look at the transformer design to see if you can lower it significantly.  I don't know how much power you are burning in the 5 X 33 volt TVS diodes but if it's a watt each the leakage is going to be responsible for about 10% of your efficiency loss.  It would be best to try and reduce it more if possible.  What I do is take one of the transformers and take it apart to see how it is manufactured.  If you know what you are looking for it becomes apparent if the windings are coupled well or could be improved.

Rsnub.  I'm assuming this is the damper for the triac dimmer, R17 and R2?  With a 1 uF damper capacitor at 265 VAC input (high line) and 60 Hz (worst than 50 Hz) the resistor will need to dissipate 1/2*C*V^2*(60Hz*2) which is over eight watts during triac dimming.  It will also need to dissipate around four watts continuous from the AC input waveform.  The resistor, if 390 ohms, needs to be able to handle pulse power rating of 375^2/390 = 360 watts (V^2/R).  This is the minimum pulse rating of the resistor, I would much higher to prevent damage from surge events.

Power rating for resistor needs to be over 12 watts continuous, 360 watts peak rating if using 390 ohms.  In the application it will be lower but an equation really doesn't help with this since it's related to any triac dimmer you connect to the input.  This, unfortunately, requires testing with many triac dimmers to see what the input current to the power supply looks like when triac dimming using all the available triac dimmers you want to be compatible with.  The resistor value will be much lower than 390 ohms.  That value is more optimum for a much lower capacitor value such as 0.1 uF to 0.33 uF.  The pole frequency will be lower if the capacitor is 1 uF and the idea is to try and damp the input current waveform from the resonance the triac rising edge causes with the LED drive EMI filter (mostly the X-capacitor).  I would revisit the EMI filter to possibly add another LC set so the values can be reduced.

Hello,

73% is low.  Normally a lot of this would be known during the design cycle when calculating for all the component values and the transformer.  There are many areas that can dissipate additional power.  The easiest method to figure it out now would be a thermal camera to see what is dissipating power.  Some of the areas I would check are the MOSFET, the main output rectifier, the transformer, the leakage clamp circuit and for incorrect operation.  It can be a combination of things causing low efficiency.

Thanks,