Re: 170VDC to 185VDC boost circuit

Jamie Morken wrote...

Winfield Hill wrote:

Hmm, I wonder if you're properly driving the MOSFET? Or did you
select too large a FET with too much capacitance? You'll want to
use small, low-capacitance MOSFETs like those offered by Supertex,
e.g., the 250V 7-ohm TN5325, in stock at Mouser for 40 cents each.
http://www.supertex.com/pdf/datasheets/TN5325.pdf

For even lower capacitance you could use the TN2130, with Coss less
than 15pF (compare to 60pF) at 25V. It's a 44-cent sot-23 surface-
mount part. http://www.supertex.com/pdf/datasheets/TN2130.pdf

Thanks! That was the problem! The efficiency went way up. Here
is the current schematic:

"http://www.rocketresearch.org/new/170V-185V%20step-up%20boost%20converter/step-up%20boost%20schematic.jpg";

You could use a lower-voltage C1 cap and return it to the 170V input.

What MOSFET are you using in your spice circuit. What spice?

What would be a good way to drive the fet gate? From the simulation,
it requires a ~1Amp, 10Volt 0.25uS square-wave pulse at 200kHz PWM.

Yes, you need a MOSFET driver IC to allow a high gate current, for fast
switching, to keep the MOSFET dissipation low.

You should a consider using a slower switching frequency. You lose
energy Es = 0.5 C V^2 on each cycle, and you spend power P = f Es,
so in addition to reducing C (the MOSFET's Coss) as you've done,
you should also reduce f.

But this type of application, which is akin to 185V in and 15V out,
cries out for using a transformer rather than an inductor. That is
to say, consider the inductor having a secondary winding: Instead of
operating at 50mA it can operate at 50mA * 15/170 = 4.4mA, isn't that
attractive? The FET will switch at much lower currents. Furthermore,
you can likely lower the switching frequency, and hence the switching
losses, and still keep the transformer small.

I bet the layout wouldn't be this small with a transformer though! :)
"http://www.rocketresearch.org/new/170V-185V%20step-up%20boost%20converter/step-up%20boost%20layout.jpg";

In principle a transformer need not be much larger than an inductor,
it's just another winding on the same core. In my suggestion, the
transformer will handle much less current than your inductor had to
handle, so that in an optimized design it could be much smaller than
your inductor.

Another thought, if you can use a fixed 15V step-up, you can use
a fixed duty cycle on the MOSFET switch and avoid a dc-dc regulator.

Isn't the zener diode on the output a good regulator, since it feeds
any extra power back into the 170VDC supply?

In your circuit it sets a maximum output, but it doesn't send the
current back into the source (thereby saving it), rather the zener
simply turns the excess energy stored in the inductor into heat.

For example, if the transformer winding is 15/170, or say 34:3 turns
(with some turns-ratio adjustment for losses, etc), you can simply
run it at fixed a 50% duty cycle in a voltage-delivery mode.

BTW, using a transformer instead of an inductor means the FET could
see 340V, so http://www.supertex.com/products/selector_guides/101/100
helps us choose a small 500V FET like the VN0550, 99 cents at Mouser.
http://www.supertex.com/pdf/datasheets/VN0550.pdf

340V! You mean for working with 240VAC? :)

No, but as an primary inductor storing energy when the FET is on
and delivering it to the load when the FET is off, it'll swing to
twice the supply in the second half of the cycle. 2*170 = 340.
The MOSFET is off, and it has to handle the transformer's swing.
But that's really just a piece of cake. It's as easy as pie. :-)


--
Thanks,
- Win
.

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