Converting 230V AC to 1500V DC

[newbie_hs]

I need to convert 230V AC to 1500V DC.
I know basic AC to DC conversion using rectifiers.
But how l convert to 1500 DC.
I need a compact solution(portable).
Can you please suggest a circuit idea for the same⁸l

 

[KlausST]

Hi,

1mA? 1A? 1kA?

What allowed voltage ripple? Number with unit.

Klaus

 

[danadakk]

Low current approaches :

https://www.voltagemultipliers.com/wp-content/uploads/2018/08/Multiplier-Design-Guideline.pdf

You need to answer Klaus's question s and I would add what accuracy/tolerance do you want ?
Does it need to be a regulated V ?


Regards, Dana.

 

[newbie_hs]

Current is 4000A .Ripple specification is not decided yet.

 

[FvM]

6 MW is unlikely to be supplied by 230 V, more likely 400 or 690 V three phase input.
Transformer with multi-phase rectifier (e.g. 6 or 12) would be the standard solution.

 

[KlausST]

Hi,

for 6MW I´ve only seen 10kV/20kV inpupt transformers.
6MW is nothing for a newbie. In my carrier I have experience with power regulation 2kV/3000A RMS but from a 10kV input / multi voltage output transformers.

Klaus

 

[BigBoss]

1500V DC, 4kA "Compact/Portable" solution ??
Is this a joke ??

 

[D.A.(Tony)Stewart]

I need to convert 230V AC to 1500V DC.
I know basic AC to DC conversion using rectifiers.
But how l convert to 1500 DC.
I need a compact solution(portable).
Can you please suggest a circuit idea for the same⁸l

Try again with more accurate details.

 

[KlausST]

Hi,

don´t use .PNG on photos.
Use .jpg and reduce file size to 100kBytes without losing thread related informations.

Klaus

 

[newbie_hs]

I need this power(6MW) only for 200uS.
May I know then portable solution will be possible or not

 

[dick_freebird]

Giving you a hint would only hasten your death. Pick another hobby.

 

[Easy peasy]

it can be easily done with a flyback ( or multiples ) having several outputs - say 3 of, stacked in series 500V each,

you can then charge the required caps on each output - to allow you to get the 4kA pulse for 400uS

there will be some droop of course - we have done 1.5kA this way no problem, you just need more 500VDC rated electro's

i/c = dv/dt, so for a droop of 50V over 200uS, at 4kA average, you need 16 milli - Farad total

or 48 milli-Farad on each 500V part of the series stack ( 48 mF = 48,000 uF )

To charge these caps is 10 sec say - total energy = 0.5 C. V^2 = 18 kJ ( / 10 sec = 1800 watts )

so 3 flybacks at 600W each, or choose a longer charge time, say 20 sec

this gives 3 sections at 300 watt each - much more do-able for a newbie doing a flyback.

For the caps - say we use 470uF, 550V rated, this is 103 caps per section ( 3 sections in series )

as these are about 35mm dia x 65mm high - this is starting to get a little on the biggish side already ( 25 litres just for the caps - say 20cm x 20cm x 80 cm to allow for the pcb and standoffs )

the good news is that a well laid out pcb for each section will easily handle the 4kA, ( 39A per cap - easily done repeatedly by good quality electro's - won't even get warm )

So there you go - 1 x psu for your rail gun prototype - just designed for you

- you 're welcome @newbie_hs


EP.

 

[D.A.(Tony)Stewart]

1500V/4000A = 0.375 ohms for source and load.

To achieve MPT dictates you need twice the power to generate Maximum Power Transfer at 50% efficiency and thus twice the voltage and loop resistance. So you need 12 MW stored but is that exponential 200 us or at least 1200 Joules transferred?


6MW only for 200uS or 12 MW stored
What is you load inductance and DCR? What should Q be to attempt critically damped?
Then what value of C is required to store twice this amount of energy to be delivered or 2.4 kJ stored at 3kV.
Now you can computed C min and ESR max.

But what value of L will have a rise time faster than 200 us critically damped with L, C & R since this determines the resonant frequency and decay rate? We must assume the transfer switch Rs is 0 for now.

So tell me what are your results?

 

[tepalia02]

@Easy peasy What else should be taken into consideration while following your suggested design? Filtering, heat dissipation etc.?

 

[FvM]

Load impedance hasn't been yet specified. Basic usage of suggested capacitor bank is ballistic discharge through SCR switch, self terminated at current zero crossing. Don't know if this fits the load requirements.

 

[newbie_hs]

The load inductance is 10uH

 

[D.A.(Tony)Stewart]

Let me try again. Let's model an impulse linear motor. Mass, length, friction, launch velocity are unknown.
Starting assumptions. 6 MW 200 us 1500Vdc L=10uH,
Many may be needed to accelerate an object with a=F/m over time t to reach launch velocity.

Start with a simple model.

1. Choose C with precharged energy to transfer 4kA @ 1500Vdc for at least 200 us
2. Magnetically couple linear motor to load mass to be simulated with Load R and back EMF C to match the source but achieve the requirements.
3. Since magnetic coupling length is limited during high acceleration in 200 us, determine how long L and the magnetic mass must be coupled. But for now assume it is long enough.
4. Storage C and Back EMF C are in parallel Zo=√(L/C), fo=1/{2π√(LC)} , τ=fo/2 for half cycle , ζs=R/Zo (?) for damping factor

Review Design and provide feedback

--- Updated Nov 2, 2023 ---

The load inductance is 10uH

Inductors are not real loads. They store and release energy. What is your real load?

 

[Easy peasy]

MPT does not apply when efficiently ( resonantly ) transferring energy from a source in a pulse or similar, i.e. C -> L,

or indeed C -> L -> C and so on,

it becomes the ratio of parasitic R to load R, such that very high efficiency of transfer can be achieved.

--- Updated Nov 3, 2023 ---

note a parallel L-C oscillating ckt does not suffer from MPT.

@d-a-tony-stewart

 

[D.A.(Tony)Stewart]

MPT does not apply when efficiently ( resonantly ) transferring energy from a source in a pulse or similar, i.e. C -> L,

or indeed C -> L -> C and so on,

it becomes the ratio of parasitic R to load R, such that very high efficiency of transfer can be achieved.

--- Updated Nov 3, 2023 ---

note a parallel L-C oscillating ckt does not suffer from MPT.

@d-a-tony-stewart

An impulse linear motor applies to this circuit where the mechanical acceleration like any motor applies a real R load across magnetically coupled inductance which dampens the resonance which otherwise presents a high Q resonance if a charged C is dumped across L. Then the fo natural resonance and Q of the linear motor is not the same as a lumped LC. High Q || circuits will NOT generate much current to an R load. Thus to maximize this surge current, the Zo impedance at resonance must be low enough to drive the coupled moving mass, R equivalent load.

MPT still applies to reactive loads where a conjugate impedance from source to load achieves MPT. Thus to perform this task , to achieve the pulse duration of 200 us and power transfer required for some estimated voltage, one may consider selection of the LC impedance and Q to absorb the energy in this duration into the moving magnetically coupled load R.

However as the object accelerates, it will induce a back EMF that reduces the voltage across the load R just as any motor current reduces as speed increases.

So I question if 1500 V, 4000A and 10 uH will actually transfer the estimated power to accelerate the distance assumed in this time. I suspect 10 uH is too small as the given data implies R= 1500V/4000A = 0.375 ohms and Tau=L/R = 26.7 us is not long enough time to transfer this power.

My attempt to make a linear motor simulation with above given does not work.



Generators supply negative power as in the case of the 390 uF cap precharged to Vi=1500V during a simulation reset.
I chose 375 mOhm to match the input power . Lower R, reduces to half-cycle time by damping which is too short already. Much bigger L is needed for L/R=tau value and a higher voltage. The back EMF was simulated by a series C in the effective load was just an estimate of some velocity induced back EMF with insufficient mass acceleration & velocity variables.

Feel free to improve the model

Note that 390 uF is set to an initial Vi= 1500V at t=0 and generates -7.316 MW power peak in a 78 us half-cycle.
The magnetically couple projectile effect load (TBD) must be less than 0.375 ohms to absorb 1500V/4000A but results in too short a time constant with 10uH inductance. making this much larger raises the reactive impedance Zo=sqrt(L/C) so the real values required to solve this linear motor problem needs better assumptions.

A critically damped impulse implies the desired Q to be applied ought to be the initial target for choosing values to generate 6 MW impulse power in 200 us. So the values of Vi, L, C, R can be computed assuming the assumptions are correct for an exponential decayed power.

One wonders how they computed the source power for the large hadron accelerator.

 

[Easy peasy]

This sentence appears to be cyclic / non-sequitur :

" MPT still applies to reactive loads where a conjugate impedance from source to load achieves MPT. "

if Tau = 26uS, this is less than 200uS / 5, so one can assume that nearly all the power will indeed be transferred - in your scenario.

In the example given we have 16mF and 10uH, which gives an unloaded Fo of 398Hz, an half cycle of 1.25mS, which is clearly too long for 200uS, however a loaded system with back emf will behave differently.

the Zo of the source here is SQRT (L/C ) = 25m-ohm, the peak current would be 60kA for a natural unloaded cycle.

if we linearly interpolate the rising current we reach 4kA after 41.67uS ( approx - actually sooner due to the shape of the sine curve ) again assuming unloaded with no back emf .

So - it would be interesting to observe the current shape, rise time and duration on a scaled down real world circuit/system, starting at lower voltages - to see what happens as things are scaled up.