3 phase High voltage ac to low power 5V dc out

Thread Starter

Vindhyachal Takniki

Joined Nov 3, 2014
560
1. Have 3 phase ac input with neutral, where each phase voltage goes from, 85V-600Vacrms
2. Need output at 5V/500mA

3. I was checking it: https://www.st.com/resource/en/application_note/cd00074286-threephase-smps-for-low-power-applications-with-viper12a-stmicroelectronics.pdf 
its page 8 has this, that is to make dc from 3 phase
1600958804264.png

I checked ST/s VIPER267K, which has BVDSS of 1050Vdc .
But at full 600V each phase, the output dc after diode, can go upto 1100V (simulated on multisim)

4. https://training.ti.com/sites/default/files/docs/Very_high_Voltage_bias_BK_13-Sept_complete.pdf
page 22 shows external fet operation, but again limited to 1200V, very close, so cant use

5. Anyone have experience in this low power application where size/size constraints? Any reference design?
 

Tonyr1084

Joined Sep 24, 2015
5,091
If you want 5V @ 500mA you don't need to use all three phases. You can pick any single phase and use that to power a 240V in / 5V 500mA out wall wart. Use any one of the three phases in and the neutral.
 

MaxHeadRoom

Joined Jul 18, 2013
21,451
If the AC is varying where "each phase voltage goes from, 85V-600Vacrms"
Then you are going to need some form of extreme regulation.
Max.
 

Thread Starter

Vindhyachal Takniki

Joined Nov 3, 2014
560
1. Yes if its single phase 85V-600Vac, I can use VIPER267K from ST (1050V BVDSS).

2. If I use like this to make dc from all three phase with neutral what would be peak dc at out? Multisim simulation shows would go to 1100Vdc when each phase at max 600Vac?
I think it would be less, no peak dc should remain no?


1601012571076.png
 

Thread Starter

Vindhyachal Takniki

Joined Nov 3, 2014
560
1. Yes easy to use by single phase and neutral.

2. But if I use like this? The output dc will go to max 850Vdc?
Advantage is here, if one phase get missing/broken, system keeps on running.


1601045888654.png


3. After that dc out, can i use VIPER26K, used ST-edesign suite, and put these inputs:
a) Vin: 85V to 600Vac
b) Vout: 5Vdc/0.47Amps

1601045911305.png
 

Tonyr1084

Joined Sep 24, 2015
5,091
If you're going for 5VDC then why the heck do you want to start with 850VAC? It's like building a bridge for cars, trucks and trains and all you're going to do is send a mouse over it. It's WAY overkill.

Since you're set on doing it your way I will step aside and allow you to proceed without having to deal with anything further I may have to say. Other than when was the last time you had a phase fail? AND IF you had a phase fail, you have a 1 in 3 chance of being on the wrong phase. Nevertheless, you're the project engineer on this one.

Have a good day.
 

MisterBill2

Joined Jan 23, 2018
7,053
I just came across this thread. Five volts at 500mA is only 2.5 watts, which is not a lot of power. The power source is described as 3-phase plus neutral, varying from 600 volts down to 85 volts RMS volts, not peak volts.
No mention of frequency is given, but I am guessing that it is an engine driven alternator that has a wide speed range, and thus a wide frequency range.
We have no hint as to the current available or of any efficiency requirement. I do see a concern that one or more phases will fail in a random manner, thus the appeal of taking power from all three phases.
But with those voltages safety does become a concern, and so the first step will be adequate isolation
If the frequency were fairly stable then a capacitive voltage drop arrangement would be more efficient, but if the frequency rises with the voltage then resistance is the better choice.
So the circuit will start out with three resistors, one connected to each phase, and each resistor feeding the primary of a small transformer for each phase, with the other end of the primary connected to the neutral.The transformers will be fairly high impedance and have a fifteen to one step-down ratio, so that the secondary winding output PEAK voltage will not be lower than 8 volts. Each secondary will feed a half wave diode rectifier tied to a single common filter capacitor. THis capacitor will be shunted by a 12 volt Zener diode with an adequate power rating to hold the voltage at a maximum of 12 volts. Then am LM309K voltage regulator will provide the stable 5 volts DC.
No high voltage rectifiers, no high voltage DC capacitors, and much less probability of a really nasty shock.
 

Thread Starter

Vindhyachal Takniki

Joined Nov 3, 2014
560
1. I have attached schematic and BOM.
2. Each phase have range of 85V-600Vacrms , Neutral is present. 600Vacrms is max considering all the worst case scenarios
3. Freq: 50/60Hz , both
4. Output required: 5V/500mA
5. Then a isolated dc-dc converter is conencted, to make 3.3V & proper isolation.



1. Selection of MOV is 750V ax max. What should be ideal max voltage should be selected? On some places I read 10-25% higher than maximum operating is ok. So 660V or higher should or work.
But all I saw that MOV degrade over time, so better to have atleast 20% higher?

2. I opened some another brand supply for this. They also placed a combination of C1& C3(two 1uF in series), across U1 also? Is it necessary?

3. I want to select two series across U1 : https://www.digikey.com/en/products/detail/vishay-semiconductor-diodes-division/P6KE350A-E3-54/2145264
Unidirectional will select. U1 has minimum breakdown voltage of 1050V, so have to protect it in any case.
Issue is it this TVS combination have breakdown start from 666V (two in series). Clamping will be 964V.

The maximum dc voltage when 600Vacrms is reached, will be 850Vdc across U1. But TVS will start conducting at 666V.
But if higher conducting voltage TVS is selected then its clamping is very high, which deafeats the purpose to protect the 1050V switcher.

4. Any other protection which can be added so that switcher canbe protected?

5. Any other suggestions?
 

Attachments

MisterBill2

Joined Jan 23, 2018
7,053
The circuit shown now looks quite reasonable, although I do have one more concern, which is about noise on the input AC power. Harmonics will tend to charge the input capacitors to higher voltages, and that is something that I had not considered adequately. Of course I have no idea as to the purity of the input power, but it is something that should be considered. Series choke coils in the input circuit could reduce such a problem if it exists.
For the second part of the system, the 5 volt supply, the physical arrangement of the PC board matters and so I suggest considering the layout recommended by the IC manufacturer in the application notes.
 

MisterBill2

Joined Jan 23, 2018
7,053
For the circuit in post #7, those protective varistors should be after the current limiting resistors. Other wise they can not do much protecting, because they will burn and fail when they start to conduct. And, given the very small amount of power the load requires, I suggest a higher value of resistance as well.
Protection against spikes can be assisted by adding an inductance in series between the junction of the diode strings and the input to the DC filter. But the insulation of that inductor must be adequate for the high voltage.
 

Thread Starter

Vindhyachal Takniki

Joined Nov 3, 2014
560
As in ckt in #10,

1. The output required is 5V/500mA i.e 2.5Watt

2. Consider switch losses, buck convert efficiency, what should be the value of R1,R4,R7,R15 in terms of valueand wattage, I should select to be on safe side?

Currently I have selected 33ohms, 4Watt, 5%


1603285552539.png
 

MisterBill2

Joined Jan 23, 2018
7,053
As in ckt in #10,

1. The output required is 5V/500mA i.e 2.5Watt

2. Consider switch losses, buck convert efficiency, what should be the value of R1,R4,R7,R15 in terms of valueand wattage, I should select to be on safe side?

Currently I have selected 33ohms, 4Watt, 5%


View attachment 220189
With this connection arrangement the varistors are protected from a massive burnout if a spike switches them into conduction. A much better and safer arrangement.
And a second suggestion for additional protection would be a fuse in the 15 volt connection to the 5 volt switching circuit, followed by a zener diode or varistor across the input to that switcher, with the current ratings being proper for the switcher supply and the voltage rating of the zener diode or varistor being a bit below the max input rating of the switcher supply. This would offer some more protection in the event of a massive spike such as lightning, or accidental grounding of one phase leg somehow. Basically cheap insurance against an unfoeseen disaster.
 
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