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Help optimize 9V to 200V boost converter

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Artlav

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Greetings.

I'm designing a boost converter to power neon bulbs and such from a battery.
9VDC input, 180-220VDC output.
I'd like to know how much sense this design makes, and in which ways can it be optimised or improved.

Also, what kind of power output can i expect from this kind of a converter?
Mosfet gets hot at about 2-3W already.

boost_9_180.png


Timer runs at 50 kHz with 98% duty cycle, 2Ω resistor inserted to measure current, there is a 300µF capacitor at the input.

With a neon bulb:
Pin=0,81W
Pout=0,53W
Iout=2,45mA
Efficiency=65%

With 16.6k resistor:
Pin=2,43W
Pout=1,9W
Iout=10,6mA
Efficiency=78%

Issues:
-Gets hot over 2W output
-Want to improve overall efficiency and/or low efficiency with small loads

One thing i can not tell if it's bad or normal is the spikes in the current on the inductor and it's reversal, as measured across the 2Ω resistor between the mosfet and the ground.
With a neon bulb:
b_neon.jpg


With the resistors:
b_load.jpg


Then, the mosfet is warming up quickly at about 2W on output, but is still quite cold with 1.3W.

The gate is opening quick, thanks to the driver (200ns/square):
b_gate.jpg
 

Also, what kind of power output can i expect from this kind of a converter?
Mosfet gets hot at about 2-3W already.
For higher output power, you'll need an inductor with higher current rating. It clearly starts to saturate in the second waveform. The source resistor that hasn't a circuit function can be reduced.
 

As FvM states, the inductor is obviously saturating, which will lead to higher losses.

Getting good efficiency with extremely low or high duty cycles is generally difficult. For very large or small input/output voltage ratios, transformers with large turns ratios are used to avoid requiring these large duty cycles. Using a transformer with a 1:10 ratio would allow you to run closer to 50% duty cycle, which would help efficiency a lot. Also it reduces voltage stress on the FET by a factor of 10, allowing you to use a lower voltage FET with better switching speed.
 
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    FvM

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For higher output power, you'll need an inductor with higher current rating. It clearly starts to saturate in the second waveform.
Nice, that helped quite a bit.
Efficiency got to 75% at 0.5W, 80% at 2W, 85% at 4W, FET remains just warm.
And the current is now a clean, straight-lined sawtooth.

Getting good efficiency with extremely low or high duty cycles is generally difficult. For very large or small input/output voltage ratios, transformers with large turns ratios are used to avoid requiring these large duty cycles. Using a transformer with a 1:10 ratio would allow you to run closer to 50% duty cycle, which would help efficiency a lot. Also it reduces voltage stress on the FET by a factor of 10, allowing you to use a lower voltage FET with better switching speed.
Is there something inherently bad about high duty cycle?
As far as i understand it, with 50% it would just be idling half the time, and get less peak current in the inductor.
Switching losses are the same.
It should be roughly equivalent to twice the frequency at 98%?
 

Is there something inherently bad about high duty cycle?
The problem is that the output diode's conduction time (during which it delivers power to the load) is not much longer than its switching and recovery time (during which it delivers power to the FET in the form of dissipation). The high peak currents in the diode will cause very high reverse recovery losses, which will manifest as increased switching losses in the FET.
As far as i understand it, with 50% it would just be idling half the time,
Keep in mind that the FET needs to idle some time. That's when power is delivered to the output.
and get less peak current in the inductor.
This depends on the specific load, inductance, etc...
Switching losses are the same.
Not if your diode has significant reverse recovery charge (it probably does).
 

I noticed that you are using a fast recovery diode. 85% efficiency as reported now sounds fair for the large voltage ratio. I don't believe that you can improve it easily.
 

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