Voltage dynamic range-Quantization

how does reducing voltage dynamic range in ADC makes quantization of the signal difficult?

Quantization is no less and no more difficult.

But when you reduce the signal range you must also consider if stray noise is less than the quantization level.

Certainly because if the voltage range is too low, it maximum value only reach the first ranges, not alls ?

For example, if you convert a 2.56 volts signal with a linear 8 bits digital representation that can handle a full range of 0 to 2.56V, you have 10 mV per step

If you apply to it a 0.256 volt maximum signal, the ADC conversion make a quantization that can only use only a little portion (say 25 or 26 steps) of the full 256 steps range that you have for 0 to 2.56 volts

=> the quantization precision is 10 times less precise with a 0.256 maximum volt signal than with a 2.56 maximums volts signal

And as say SunnySkyguy, more the amplitude of the signal is low, more the stray noise have a relatively more weight ...

thanks for the reply but I am totally new to this and am learning from scratch.. I know that quantization is rounding off and 8 bit resolution is 2^8=256steps.. but I couldn't quite understand what u meant by using only a little portion of steps of the full 256 steps..

If you use less than the full range for quantization then that's the same as using a converter with fewer bits of resolution. For example using only 1/2 the range of an 8-bit ADC would be the same as using the full-range of a 7-bit ADC. In effect, each bit would represent a larger percentage of the signal, giving poorer resolution.

now I get it thank u so much...

i was going through ADC resolution and came across the following post

"Resolution is that, the smallest change in input signal that you can detect.
If you use higher resolution, you can detect the signal to lowest quantity. For 10 bit ADC, and your reference say about 5V, You can detect the lowest possible change in signal is about = 5v/1024 => 0.0048828125V. ADC counts will increase on every 4.8mV. But if you use 16 Bit ADC, the lowest possible change in signal you can detect is about => 5V/65536 => 0.0000762939453125V. Here ADC counts will increase on every 76uV.
If you need detect very lowest input signal, go for higher bit ADC resolution. That's purely depends on your application"

It was told in 8-bit ADC for 2.56V the resolution will be higher compared to that of 0.256..... but 2.56V/256=0.01 and 0.256V/256=0.001..... it is told that resolution is when the ADC can detect the smallest change so for this case 0.256V will have a better resolution right... sorry if I am wrong but its quite confusing... please help

Both resolution and range are important. The maximum S/N ratio than can be measured is limited by this ratio. 10 bit is 1024 ~ 60dB range each bit is a binary factor of 2 or 6 dB per bit, thus 8 bit cannot measure any better than 48dB SNR and a noise free system is very difficult to achieve.

The reference voltage must be more accurate than the resolution unless self-calibration or sigma-delta conversion is used. Telephony ADC's use a logarithmic conversion method (A Law or u Law) to get more than 70 dB SNR with this method.

Minimizing Noise in any measurement your challenge and increasing signal strength and Vref is one way to increase this S/N ratio. Filtering is another way as well as shielding isolation etc. So consider that the number of bits is your best case noise free SNR and work out what you need for ADC resolution, and consider all the sources of noise and errors.