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600mA 200V B+ (OTL) power supply questions

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I’ve been calculating power for a 540mA (total) 200V B+. So I have some newbie questions.

Using a LT3080, although currently modelling using 6000uF reserve, then a 140uF in a small low pass filter not to drop further. Input is 230V.

It’s good - however it’s pulling 5A from the Secondary. If my average calculation is right that is 550W average power which seems high for a 600mA 200V system.

I’m trying to gauge is this is normal or my LTSpice model - just seems out of kilter. Is that a normal ball park figure for 600mA power supply?
 
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Hi,
This is my model for the power, the LDO is using Tom Christians Maida implementation:
WjlXKAS.png


The resistor array for recharge I may switch to a current limiting circuit, a 900V 10A piece would be used capable of 156W max dissipation, in that position it was about 30-40W of heat during startup.
Later the small passive filter will changed to have one per channel to reduce cross talk over the B+ lines.

It should be noted that LTSpice has reordered the circuits after the graph following changes:
osKPwg2.png

The blue is the B+ line, the green is the Cap switch supply to the LDO, the red is the current that the secondary sees.

I've also attached the current LTSpice model file.
 

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Thanks Mona.

I was just a little concerned Vripple=Iload/2*50*C, targeting 1Vpp is the reason I have 6mF, I'll switch to 1mF and rerun.

The first 20 seconds recharge at 400mA limit, but let me have a look.

I may also move the filter too whilst I'm at it but it could destabilise the regulator.
 
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With 1mF of reserve, the ripple increases, the load is the issue. with 6A (not 5) that's an average of 4.2 A. I suspect that the majority of this is disappearing in heat..

qJXVBiC.png


Digging a little further:

WkPtd8c.png


The top shows the output of the switch S3, the blue shows the current just above D3 & D4.

Now this shows before the switch... and the pink shows the line above the capacitor C1:
jH7Y707.png


So .. ±6A is going in and out each side of the capacitor.
 
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Question

Looking over sch. The regulator passes (HT - Vdrop)*If

I'm not used to seeing this, maybe lack of familiarity, but wouldn't it be better to use the regulator for the "low" side, in this case? Floating the Earth to suit.
EDIT, nope that wouldn't help, thinking about it.

In order to minimise the Pd?

I.e. 30Vdrop*0.6A gives Pd of 18W in a resistor of 50R.

Far easier to handle.
 
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I think the problem is down to magnetisation - saturation. So I think the transformer model needs more magnetic flux.

The current for the transformer shows a spike with a flat between lobes. This causes a very high spike for current vs if the current was averaged over a wider sine wave.

Following up on my suspicion - Power Supply Simulations looking at figure 3 explains it..
 
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So a sine wave current is RMS and sqr(2) and have a crest factor of sqr(2) ie 1.414. This has 6A and 0.6A so has a crest factor of 10x if my mental maths at this lack of sleep is correct.

So the option here is - spreading the power spike. We only need 0.6A not 6A so as long as we get a minimum of 0.6A *1.414 it should maintain a continuous feed just like RMS.

So if we limit the current to just over 0.6A using something like this:
DumSxek.png


The current limit provides minimal resistance under 0.6A but high resistance above, so it's switching between a capacitive load it's happy to drive and a resistive load that restricts it's current flow. We see a closer to RMS (I'll remodel with 0.6A*1.414 in a bit):

CqkP6Qf.png


So this keeps our current spikes closer to RMS thus the power supply doesn't perhaps need to be specified to cope with the 6A crests but can instead be specced lower.. I'm just seeing if there's any issues with voltage drop (in theory there shouldn't be as we're covering the current and the ripple). I suspect in reality that the transformer would simply see maximum current spikes but short enough not to break the transformer windings, although probably not conducive to longer term lifespan. I could live with 160-200W but not 500-1KW+.

Result!

iMFmoRk.png


Red - current draw through the secondary
Blue - current draw through the current limiter output pin
Green - voltage at the load resistor (ie the amp)
 
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There's a couple of gotchas that need ironing out:
a) startup the mosfet needs power to work, so adding a 1uF to stabilise initial power to the mosfet quickly and a thermistor to help initial starting load may be useful.
b) The first few seconds see the mosfet dissipate 140W without inrush control. Once running it's dissipating 20W.
c) need to add diode protection for the mosfet.

I'll remodel for 600mA RMS, so this is 850mA peak current limit. Although the transformer secondaries are specified as RMS, I want to ensure we have close match..

RMS of 160W is 160*0.707 = 113W so this is worth continuing to investigate.
 
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You only need to worry about in-rush current, which you are doing with a soft start it seems. The peak current during steady state will depend on the transformer winding resistance which you may not be including in your simulation. The transformer does not have to be specced for the peak current, only the rms current plus a suitable margin. There are standard formulas for linear power supply design that allow for the spiky nature of the current waveform. For a capacitor input full wave bridge:

V dc = 1.4 x secondary Vac
I dc = 0.62 x secondary Vac (from the Hammond website)

So for 200V DC at 600 mA (= 120VA), you need a secondary rated at 143Vac at 0.97 amp ac (=139VA). You have to add regulator losses on top of this.
 
You only need to worry about in-rush current, which you are doing with a soft start it seems. The peak current during steady state will depend on the transformer winding resistance which you may not be including in your simulation. The transformer does not have to be specced for the peak current, only the rms current plus a suitable margin. There are standard formulas for linear power supply design that allow for the spiky nature of the current waveform. For a capacitor input full wave bridge:

V dc = 1.4 x secondary Vac
I dc = 0.62 x secondary Vac (from the Hammond website)

So for 200V DC at 600 mA (= 120VA), you need a secondary rated at 143Vac at 0.97 amp ac (=139VA). You have to add regulator losses on top of this.

Thank you. You're right that the secondary windings only have a small resistance added for stability. I'd calculated about 180VA including heaters in my initial calculations but the simulation results (and my rubbish data in) is probably compounding the issue needlessly.
 
220V-240V input regulator and I'd probably give it some current headroom. We have 230V power here ±10% and the Hammond transformers seem to be about 7-5.8% regulation.

Using the rough rule of thumb calculations, with what I've seen on the regulator model:

240Vdc/sqr(2) = 170V
660mA/0.62 = 1.065A
Or 181VA.

So with say 20% we're looking at around 220VA.

Hammond do 225VA 220V @ 1.02A but nicer still is the 300VA 240V @ 1.2A. Just would be throwing away a lot of heat.
 
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Seems todo hover around 242-246V at the socket it would be plugged into.

That's in the winter when the local supply has all the CH pumps going, cookers etc. That may change locally with holiday periods.

The CH heating, oven etc are all on their own consumer box lines, the fridge downstairs is the only big current switching going on the lower sockets. The sockets upstairs and downstairs are on separate rings back to the consumer box. So it appears the socket is well isolated from other spurs but has the two work-from-home offices on it so I'd expect some SMPS electrical noise on it over time but nothing that can't be filtered out.
 
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