# 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.

NOTE: I hve the basic electronic skills and the system needs to be relativly small.

active| newest | oldesthttps://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.

https://www.instructables.com/files/deriv/FCF/0G7G/G4PBXIVS/FCF0G7GG4PBXIVS.LARGE.jpg

From left to right, the components in that picture are:

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 V

_{RMS}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.

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

^{2}/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.Its just like a normal, low voltage supply, but this one can blow you across the room.

Steve