DC to DC Converter

Voltage regulator chips to do the job are widely available, and very cheap too.

Expect some power loss in the chip as no energy conversion is ever 100.000000% efficient.

GA yourself, nice links too.
Del

getting .23A out of a .1A ckt seems to be the same as the discussion of putting 10 windmills in a tube and getting 10 times the power generated.

Even if .1A X 11V =.11VA, power wanted is .23A X 5V = 1.15 VA.

Power cannot be created, one is short .05 VA and conversions are, at best 85 or so %.

So where does the power come from?

Karl Mavis

You are quite right, but I believe you have just uncovered an error in the arithmetic (which I hadn't bothered checking).
If you just assume we are just talking about approx 1watt that's near enough.
You seem to have generated your own arithmetical error too.
Even if .1A X 11V =.11VA

Errrr .1 x 11 = 1.1 even in cat land where maths is generally ignored

And 1.1 is close enough to 1.15, but if this really offends you, he could probably get away with say a 4.5v supply at .22A which is less than a watt.
Del

A nearly valid excessively accurate assessment, Karl. You are quite correct that power cannot magically appear out of nowhere, and that the losses from the switching supply itself were not taken into account. But the apparent additional 50 milliwatts is actually a quantization error that happens when one forgets the importance of significant figures and precision. 50 milliwatts is less than the uncertainty of 73 milliwatts if one assumes that the single digit 5V is actually the two digit 5.0V. So the equations should be

(0.10±0.01)A*(11±1)V=(1.1±.21) W and

(0.23±0.01)A*(5.0±0.1)V=(1.15±0.07) W

Once you include the measurement precision to the numbers, there is no mystery power. You will notice though that the input power value is less precisely known than the output power. This happens because 0.10±0.01A and 11±1V are each less precisely known than 0.23±0.1A and 5.0±0.1V.

Now as far as losses from the switching supply, you are correct the input power must always be greater than the output power. But I doubt the OP has a supply that cannot provide more degraded power than the 1.1W of non-degraded power. The application note I cited does explain how to utilize degraded power.

can it be done using zener diode as voltage regulator and the out put is 5V 0.23A?

No...see #2.

It would need to draw 0.23A from the 11v rail which can only supply 100mA

I watched Ad video before talking about DC to DC issue. which posted by a IC company who have already manufactured this kind integrated big chips (actually it's size is small). its efficiency is high close to 95%. You can google them out.

A purpose-made dc-dc converter such as a Recom RS-1205S or a Traco TMR1211 will happily convert your supply. There will be some loss in power output as PWS says (the price for conversion you might say).