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I have designed a digitally-adjustable 22W Flyback LED driver using FSL136MR which has a built-in 650V MOSFET. Here's the simplified schematic (Auxilliary supply regulators and some other components (e.g. bypass caps, PFC stage, etc.) are not shown):

schematic

simulate this circuit – Schematic created using CircuitLab

Vref1 and Vref2 come from DAC outputs of an MCU and they set the maximum output voltage and current, respectively. VDC = 390VDC coming from PFC stage.

As seen from the simplified schematic above, the circuit operates as a constant current or a constant voltage source depending on which control block wins.

Maximum allowed output current is 750mA (so the max. output voltage is set to around 30V by the MCU) and maximum allowed output voltage is 80VDC (so the max. output current is set to around 250mA by the MCU).

PROBLEM:

If I set the output current to 300mA and connect an LED strip (VLoad=75V, Pout=22.5W) the circuit works fine as a constant current source, chip and even the snubber network don't heat up much (around 75 degC which is way lower than the Tj-max=135 degC) and the total efficiency is greater than 85%, as expected.

However, if I set the output current to 600mA and connect a different LED strip (VLoad=36V, Pout=21.6W) the circuit works as a constant current source but the efficiency decreases to 72% and the chip heats up quickly (still the snubber network is cooler) thus the chip triggers its internal thermal protection.


I have checked the voltage across the built-in MOSFET's drain and ground. In the first situation (Iout=300mA, VLoad=75V), the drain voltage hits to maximum 510V (including spikes). In the second situiation(Iout=600mA, VLoad=36V), the drain voltage hits to maximum 460V (including spikes). This is expected because the drain voltage depends on the output voltage: The lower the output voltage, the lower the reflected voltage and thus the drain voltage. So in the second situation the MOSFET's stress should be lower, isn't it? But why does the chip heat up quickly even with lower output power?

Flyback Transformer Specs:

  • Core: PQ2620 (Ae = 120mm², Ve = 5.3cm³)
  • Primary Inductance: 1mH ± 10%
  • Primary Turns: 50
  • Secondary Turns: 43

I cannot find where the problem comes from.


EDIT

After reading comments I first scoped the primary current by scoping the voltage across a series 1R resistor. Even right before entering thermal protection the current had a well triangular shape.

I also started to think about the possibility of saturation. So I made another test with an 18V/25W LED load. Here's the results:

I started with Iout=300mA (VLoad=17.8V) and I gradually increased the output current by 20mA each step. Interestingly, the current could not go beyond 600mA even if I set it to 800mA! Is the core saturating? Can saturation be caused by load current?

Now the question is updated: What should I do?

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    \$\begingroup\$ What’s your current waveform on the primary? \$\endgroup\$ – winny Oct 1 '19 at 15:22
  • \$\begingroup\$ Is there any oscillation at the output? \$\endgroup\$ – rdtsc Oct 1 '19 at 15:36
  • \$\begingroup\$ Are you sure you have the feedback connected correctly. It seems to me that if you exceed a certain output current, the opto will turn off and the output will turn fully on. Maybe you have the comparators drawn incorrectly. \$\endgroup\$ – Andy aka Oct 1 '19 at 15:37
  • \$\begingroup\$ @winny The chip have no pin to measure it directly, I have to cut the primary track and solder a resistor. I'll do that ASAP. \$\endgroup\$ – Rohat Kılıç Oct 1 '19 at 15:54
  • \$\begingroup\$ @rdtsc No, the output current and the output voltage are almost pure DC. \$\endgroup\$ – Rohat Kılıç Oct 1 '19 at 15:55
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You are on the right track with your edit: you will have to decrease the number of turns on the secondary to run at higher currents, or get a core with larger cross section.

A flyback converter can be viewed as two inductors sharing the same core. You store energy in the inductor core by energizing the primary winding, then dump it into the load through the secondary winding.

If you are trying to get 600 mA out of the secondary, the average current in the secondary winding has to be 600 mA. If you are at, say, 50% duty cycle, the average current during the output pulse must be 1.2 amperes. And, since the output current pulse waveform decays via a ramp down to zero amperes during the output pulse, the peak current on the output must be around 2.4 amps. The energy that is stored in each pulse is proportional to the square of the peak current for an inductor.

Now see what happens if you start to increase the flux density in the transformer. The transformer becomes less efficient, and the output suffers. The flyback loop reacts by increasing the duty cycle of the FET "on" time, which reduces the the duty cycle of the "off" time, so the peak output current must increase further.

I would try to remove turns from your secondary winding until it just barely works with your highest-voltage load, then see how much current you can get out at lower voltages.

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