Efficiency of 1000W flyback

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Hello,

Using flyback topology for that power level is a bad idea.
We don't use flyback over about 200W because it's not magnetic efficient and cause high stress on primary switch.
Prefer to use push-pull, more easy to design, smaller and much more efficient.
With a good design,you could achieve easily over more than 90% efficency.

Frex
 
Hi,

Although forward converters can be used for more output power than flyback,
1kW in this mode is again a bad idea.
You will have also hight stress in primary switch and the magnetic core will be big
with high losses.
It's nevertheless possible...

You should really look for using push-pull converter type.
You will have probably much more chance to get a functional smps...

Regards.
 
Use good ol' half bridge, or two transistor forward. Flux walking is not an issue there.

Though, there is a 7kW (at least that's what the service manual says) flyback in one of the big Lab Gruppen amps... can't recall off the top of my head which one it was tho. So it is possible... it's just not very efficient, unless you use big and expensive parts. On the other hand halfbridge will get you to 1kW for cheap.

I'm applying the finishing touches on my adjustable split supply which goes from +/-20 to +/-60v, and in testing i have achieved 650W continuous output. I ran it for >1 hour no problem, cooling was a standard 80mm fan thermally controlled and it didn't even reach the max speed. Primary heatsink temperature was under 75C, it's the transformer and inductor that didn't let me continue further, i designed those for 600W max and they started getting hot. Efficiency was 85%, with the primary switches being bog standard bipolar 13009 devices, in TO-247. More specifically the Fairchild E13009L. Switching frequency 60kHz. I remember Eva pulled 4kW from a 13009 full bridge... so 1kW could be doable in halfbridge if you cool them well.

Still, i would use MOSFETs or IGBTs instead.
 
ok

but i prefer full bridge to half bridge...cheaper as less cost for film capacitors....also far less fet curretn...and over four times less dissipation in a fet compared to halfbridge.

Half bridge difficult to layout nice cuzz of huge film caps in primary side
 
but i prefer full bridge to half bridge...cheaper as less cost for film capacitors....also far less fet curretn...and over four times less dissipation in a fet compared to halfbridge.

So why don't you use full bridge then? I agree, i was just too lazy to design something 100% from scratch - i actually used an old ATX power supply as my base, and built upon that.
 
thankyou, though i hear of flux-walking problems with push-pull smps?

There are current-mode converter IC's that solve this problem by current limiting on each half-cycle. If you get closer to the limit on one side, it shuts off earlier and moves the flux closer to balanced.

For old-style push-pull inverters feeding a choke after the rectifier diodes, there is sometimes not enough volt-seconds available to reset the core. If you replace the inductor with a tapped inductor and feed the tap from the diodes and put a reverse-biased diode on the "new" end (the other end goes to the filter cap as before), the added diode gets slammed down a diode drop below ground but the tap remains at a high enough voltage to reset the core. I did this on a converter mounted on the Space Shuttle and used for the Hughes GEOS series of satellites in about 1976 and it eliminated a saturation problem immediately (before the era of current-mode converters).
 
My schematic drawing software is a colossal pain to use, but I can explain it in a simple fashion using nodes, just like some of the original circuit analysis programs (and I used ECAP, ICAP and SYSCAP before SPICE became available, so circuit analysis is not a foreign idea to me). At the output of the push-pull transformer, there are rectifier diodes that connect ordinarily an inductor that goes to a filter capacitor.

Node 1 = D1 cath, D2 cath, L1
Node 2 = L1, C1
Node 3 = C1, GND

This is the normal connection. The revised connection replaces L1 with T1, an autotransformer with three leads and adds diode D3:

Node 1 = D1 cath, D2 cath, T1 tap
Node 2 = T1 end, C1
Node 3 = C1, GND
Node 4 = T1 other end, D3 cath
Node 5 = D3 anode, GND

The tap on T1 does not have to be any specific ratio. In the Hughes case, it was a 28 volt output with the tap held 10 volts above ground, including the effect of the D3 diode drop. This was adequate to correct for the imbalance but not high enough to get a significant drop in current during the "diodes off" state. The current tends to follow a constant NI, that is, number of turns times current tends to remain constant, so when the rectifier diodes D1 or D2 stop conducting, the current through T1 drops in proportion to where the tap is. During this time, the T1 other end (Node 4) is slammed down to one diode drop below ground and the tap is held at a higher voltage just by autotransformer action. Moving the tap to a higher voltage would have enabled more imbalance to be corrected, but the sudden change in current in T1 may have caused an unacceptable amount of noise, so we left it at 10 volts.
 
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