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

 

[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.

 

[BlackHelicopter]

You convert the AC to a higher voltage using a step up transformer (i.e. more windings on the secondary) and then to DC with a full wave rectifier.

 

[newbie_hs]

You convert the AC to a higher voltage using a step up transformer (i.e. more windings on the secondary) and then to DC with a full wave rectifier.

Can you please suggest any transformer less solution

--- Updated Nov 9, 2023 ---

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.

May I know any transformerless solution for the same problem

 

[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

Why you need to convert this is more important than your question. This is because no answer may satisfy your curiosity without us knowing the power, energy and safety requirements as well as the cost and availability. It is also useful to say your age and skill level for such high voltage can be unsafe.

 

[KlausST]

Hi,

Can you please suggest any transformer less solution

it would help a lot to state what´s the idea behind "transformerless".

Like: EasyPeasy´s solution is transformerless.
Still it uses at least two inductors. The difference is: in a transformer the two(or more) inductors/coils are combined.
Thus the transformerless solution is not smaller nor cheaper nor less complicated ...

So I wonder: Why transformerless?

Klaus

 

[Calm down]

If the current is not high, double voltage rectification can be used