170VDC to 185VDC boost circuit

Hi,

I simulated a boost circuit for 170VDC to 185VDC, but the efficiency is very low, it draws about 25watts when outputting about 9watts (50mA at

185V).

This is at a 200kHz PWM and 4% dutycycle (not sure I can drive the fet with the required 0.2uS pulse)

The dutycycle is so low because the boost voltage ratio is small I think. The inductor is running in discontinous mode.

I picked a high frequency to keep the inductor size small.

Any ideas on how to make this work better?

I also have a zener diode connected from the 185V output back to the

170V input to regulate the output. The controller will ideally keep the output voltage a bit under the zener voltage.

cheers, Jamie

Reply to
Jamie Morken
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On a sunny day (Wed, 29 Nov 2006 08:46:47 GMT) it happened Jamie Morken wrote in :

Are you making the 185 from scratch, or making 15V DC an putting it in series with the 170V? (I mean a separate 170 / 15 V switch mode in series). Saves copper / turns.

Reply to
Jan Panteltje

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.

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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.

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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.

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. 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

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helps us choose a small 500V FET like the VN0550, 99 cents at Mouser.
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--
 Thanks,
    - Win
Reply to
Winfield Hill

almost sounds like the induction is too low for the freq you're operating.

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Reply to
Jamie

Hi Win,

W>

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

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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.

I will use a CPLD or a microcontroller (lpc2148) to generate the PWM signal.

Is gate capacitance the important factor for both the turn on/turn off delay times as well as the rise and fall times? I guess an RF fet would work well for this application but is probably overkill?

I bet the layout wouldn't be this small with a transformer though! :)

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

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

cheers, Jamie

Reply to
Jamie Morken

Yes. Discontinuous mode isn't attractive at high frequencies.

--
 Thanks,
    - Win
Reply to
Winfield Hill

Jamie Morken a écrit :

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No. The extra power is dissipated in the zener diode.

BTW which schottky are you using that withstands your 200V or so output voltage? Schottkies are more low voltage diodes than anything else.

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Thanks,
Fred.
Reply to
Fred Bartoli

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?

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.

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.

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.

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
Reply to
Winfield Hill

Cool, how come the typcical example boost circuits don't do this as well?

I used the n-channel Si9420DY (Siliconix), 200Vds, 8.6nC gate charge,

1ohm Rds(on). I am using LTspice version 2.17a

I'd like to try building the circuit with transistors instead of a MOSFET driver IC. Any tips on this would be appreciated!

cheers, Jamie

Reply to
Jamie Morken

Hi,

I found some good mosfet drivers with fast turn-on/turn-off times (TPS28xx) they have like 19 transistors in them so I see why I couldn't get a comparable gate drive signal with a 3 fet (2 n-channel and 1 p-channel) circuit I made. :)

For a 500pF gate cap and a 10V gate drive voltage, that would be Es =

25nJ, and at 200kHz switching frequency, P = 5mW. The simulator tells me that there is about 13mW gate power used. I guess the gate driver itself will dissipate quite a bit more though.

cheers, Jamie

Reply to
Jamie Morken

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