How do I make a 160 DC volt power supply?

Alright, for a science project I need a 160 DC volt power supply.  How do I make one?
NOTE:  I hve the basic electronic skills and the system needs to be relativly small.

Picture of How do I make a 160 DC volt power supply?
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The  discussion for this question seems similar to this one:

https://www.instructables.com/answers/Ok-I-need-a-diode-bridge-to-change-110-ac-to-110d/


That's where the attached picture came from.

Erm... a  question no one has asked yet is how much current, or equivalently how much power,  must this thing supply?  I mean if it's a small amount of power you might be able to run it off batteries.  For example the circuit for a camera flash produces DC voltages, roughly in the range 0 - 300 V, but with just a few watts of power.

Another question might be how tight does the regulation have to be? What magnitude ripple voltage is tolerable?

Of course I don't want to bore y'all with too many details though.

And I definitely don't want to know what its for.
full_bridge-ac_to_dc.jpg
Here's a link to a bigger version of that picture:
https://www.instructables.com/files/deriv/FCF/0G7G/G4PBXIVS/FCF0G7GG4PBXIVS.LARGE.jpg

From left to right, the components in that picture are:
  • an AC outlet and power cord
  • screw terminals and voltmeter set to scale 200V AC
  • bridge rectifier
  • electrolyic capacitor, 460 uF @ 350 V
  • screw terminals and voltmeter set to scale 200V DC
  • a neon bulb in series with a 22K(red-red-orange) resistor
  • a big 10K (brown-black-orange) resistor


Originally this picture was posted to this question:
https://www.instructables.com/answers/Ok-I-need-a-diode-bridge-to-change-110-ac-to-110d/

And basically this was just to settle a little mathematical argument between me and this guy Jim, about the DC output of this circuit being the square-root-of-2 (1.4142...) times the RMS AC voltage.  He thought the result was VRMS divided by square-root-of-2, instead of multiplied by it.

The 10K resistor, plus neon bulb in series with 22K, these are just there as a small load.  They are there to demonstrate this circuit is capable of supplying a small amount of current.  How small?  Well, 160 V divided by 10 K is 16 mA. That's the current throught the 10K resistor.  The current through the neon bulb is probably about 2 mA.  The 10 resistor is the "big load", and the neon bulb is just there as an indicator to show the thing is turned on.

Another reason I wanted the load current to be small, is so that the voltage ripple would be small.  How small?  The formula I'm using to calculate ripple is this one:

ΔV = (I*T)/(2*C)

Substituting I= 20 mA, T=(1/60)s, and C= 460 uF, gives:
(20e-3)*(1/60)/(2*460e-6)
= 0.36 V

The reason I mention voltage ripple is because you might be wondering how big of a filter capacitor to use for your application.  The answer depends on how much DC current your load is drawing, and how big of a voltage ripple you're willing to tolerate.  Put those two numbers into the formula above, and you can figure out how big of a capacitor, C,  is needed.
Here's the same thing, but as a circuit diagram. The first diagram was drawn by hand, by me.  The others were found at the locations linked below.

http://sub.allaboutcircuits.com/images/05188.png
from this article
http://www.allaboutcircuits.com/vol_6/chpt_5/6.html

http://en.wikipedia.org/wiki/File:RC_Filter.png
from this article:
http://en.wikipedia.org/wiki/Rectifier
rectifier-filter-diagram.jpgallaboutcircuits-rectifier_and_filter.pngwikipedia-rectifier_and_filter.png.png
Something else worth mentioning about this picture.  The 10K "big load" resistor really is kinda "big" in terms of the power it is dissipating.  Since P=V2/R, this is 160*160/10000 = 2.56 W.  A typical 10K resistor is rated at 0.25 or 0.5 W, and you'd see smoke if you tried this with such a (too small) power rating.
Kalrag (author) 7 years ago
Good I wont tell you what it is ill probably get some negative posts. And all I know is that I need 160v.
A cold fusion set up I guess.
Re-design7 years ago
You can make an Isolation transformer by taking two 12 volt transformers and conecting the 12 volt secondary lines to each other, 120 to one of the 120 primaries and you will have slightly less than 120 on the opposite primary. Do not connect any of them to ground.
framistan7 years ago
When 120 volts is rectified and a filter capacitor added.... then you will have 169 or so volts. This is because the 120 volts actually goes as high as 169 volts PEAK... and down to minus 169 volts peak. The capacitor keeps the voltage at whatever the peak voltage is. Use an isolating transformer as steveastrouk suggests for better safety. The thing can be built small if you can find a small isolation transformer.
Kalrag (author) 7 years ago
Let me put it this way a step by step instruction that will not be a full project on its own.
If you are in the USA, take a 110 V ISOLATING transformer - that gives 110V on the other side, isolated from the mains, add a bridge rectifier, and smooth the output with a 250V rated electrolytic capacitor.

Its just like a normal, low voltage supply, but this one can blow you across the room.

Steve