440v to 12 v transformerless circuit

[D.A.(Tony)Stewart]

AG , sounds like an anachronism with external SMPS on Dial-up mode, but possible. Again, depends on all requirements , not stated.

 

[casemod]

Up to 500VDC I have successfully used a TNY254PN operating as a buck converter. This is good for up to 200-300mA output at 12V.

I'm guessing 440V is AC which would be a bit higher at 650VDC. You mention 1000VDC which I'm not sure where it came from (690VAC!?).

The only practical suggestion I have is to use a high voltage zenner diode (300-350V) in Series with a resistor, both rated for 10W. The voltage across the zenner will be constant with the excess showing across the series resistor. Wire a small 230VAC SMPS is parallel with the zenner and there you have it.

 

[Orson Cart]

a self driven buck ckt with 900v or 1kv Mosfet would seem to be the (simplest & lowest cost) way to go here, plenty of attention needed on the choke so that it doesn't arc over its windings with such high volts across it...!

 

[betwixt]

All points valid but the idea is is costs < 70 rupees (~US$1.50) and I would guess the MoSFET alone would cost more than that. If it's only an indication of the presence of the phase voltage difference, an opto-coupler running at say 1mA with a 1N4148 bridge and a pair of high value resistors would do the job. The LED could be in series with the optocoupler's interal LED.

Phase ---- 220K ---- bridge/opto/LED ---- 220K ---- Phase. The resistors would dissipate ~0.25W each.

Brian.

 

[D.A.(Tony)Stewart]

4W @$1.50us is a bit high, if you shop around in India transformer supplies are as low as $0.10/W in volume

keep in mind diode bridge to cap is a nonlinear load and power must be derated at least 30% on transformer.

 

[casemod]

a self driven buck ckt with 900v or 1kv Mosfet would seem to be the (simplest & lowest cost) way to go here, plenty of attention needed on the choke so that it doesn't arc over its windings with such high volts across it...!

Not so simple... Duty cycle would be 0.02, which makes regulation nearly impossible to achieve. A buck withp ulse skipping might do, but the inductor would have to be physically large. Easier to use a buck converter to step down Input voltage to 300VDC. Altough if the power was low a linear regulator could just suffice as well.

 

[Orson Cart]

0- 2% duty cycle is quite achieveable, even at 100kHz, 15V CMOS and a single high speed comp....hysteretic control would also work fine for a given ripple on the o/p...

 

[cts_casemod]

0- 2% duty cycle is quite achieveable, even at 100kHz, 15V CMOS and a single high speed comp....hysteretic control would also work fine for a given ripple on the o/p...

Defenitivelly achievable, I have previously designed such power supplies, but would not recommend using such a low duty cycle

The output will come in very narrow pulses with a very high current to compensate for the very low diode conduction time. For a 2% duty an output of 400mA and a theoretical efficiency of 100% thats 8Amp peaks for the freewheeling diode/capacitor/inductor, greatly dropping efficiency due to I2R losses. If the efficiency drops to 50% due to losses this shoots to 16Amps. Other issue is EMI.

 

[KlausST]

Hi,

8Amp peaks

Luckily it's not that bad.
The 2% is when the 650V is ON. If we calculate with 10W, then input average current is about 16mA. This divided by 2% gives about 800mA. Peak current depends on inductance, but i expect it to be less than 1.2A.

Klaus

- - - Updated - - -

Add:
It should be a low cost and also low quality application. So output ripple voltage is not that critical.
I tend to a burst mode regulation, then one could go to 10% min duty cycle.

Klaus

 

[FvM]

The initial specification was 150 or 200 mA output current. For buck converter toplogy with 0.02 duty cycle, this refers to 7.5 or 10 A input peak current (optimistically assuming low input ripple and near 100 % efficiency). Feasible but not really enjoyable.

 

[KlausST]

Hi,

how his can be?
in worst case if the coil is completely empty (whats not realistic for a buck) and the current linearly rises form 0A to 7.5A (thus the factor 0.5)
650V * 7.5A peak * 0.02 *0.5 = 50W.
but 12V * 200mA are 2.4W
I can´t folow the mathematics.

It seems you use the average output current to calculate the peak input current by dividing it by 0.02.
(200mA / 0.02 = 10A)
In my eyes you need the average input current to divide by 0.02 to get the peak input current.
(5mA / 0.02 = 250mA)
Or are you talking about a ns wide capacitive current peak? Or current peaks caused by saturated coils? Or a flyback?

Klaus

 

[Orson Cart]

Where do people get these high input current figures without specifying the buck inductor? for a buck the input current (peak if you like) = the output current, for a large enough inductor. So for 650Vin, 12V out, 300mA say, 100kHz, ton = 2% = 200nS, V/L = dI/dt, thus for +/- 100mA pk on the 300mA L = 650uH rated at >400mA pk.

The peak current input current is 400mA almost regardless of efficiency for 300mA out (the on time goes up for poorer efficiency, but this does not raise the current much)

The peak current in the buck diode is similarly 400mA.

For 50kHz, L can be 1.2mH (or larger) same currents ....

Only a tiny buck choke will give ridiculously high currents like 8A peak... this would be in-efficient...!

 

[FvM]

O.K., you are right, my error of reasoning. Sufficient large inductor means nearly constant current.

Sorry for causing confusion.