# questn is abt the resolution concept of ADC..

Discussion in "General help Guidance and Discussion" started by ammi_paranjpe Dec 19, 2009.

Sat Dec 19 2009, 07:56 AM

what does it mean actually?like 8 bit resolution,10bit resolution............and what is step size?if analog vtg is 5v then then hw to mesure step size?...................plz help...i m really nt getin this....

Sat Dec 19 2009, 02:28 PM

If you have an A/D converter reading an analog voltage, with a reference voltage of 5 volts, your reading will be "referenced" to this 5 volt value.

For instance, if you input voltage is 4 volts, then you are reading 80% of your reference voltage, right? (4/5 is 80%)

The resolution, or "how well you can resolve" this reading, is described in bits. An 8-bit A/D divides up the reference voltage into 8-bits, or 256 equally spaced voltages, from 0 to 5 volts. This corresponds to about 19.5mV per bit.

So how accurately can you measure the 4 volt signal? Well, 80% of 256 is 204.8, but you don't get any fractional bits when you read, so you will read 205. And what is the value of 205? 205/256 = 80.078125% of 5V, or 4.00390625V. Not too bad, right? Less than 0.1% error...

But what if your input voltage was 4.01V? 4.01/5 = 80.2%, which works out to 205.3, but remember? No fractional bits, so you would read 205 - But that's the same as before, right? So you would again read 4.00390625V even when the input was 4.01.

Here is where the step size comes into play - the size of the voltage step between bits is the step size.

With a 5V reference, the step size is 19.5mV, because 5V/256bits = 19.5mV

What about a 10 bit A/D? 10 bits is 1024 steps, so...

with a 5V reference, the step size is 4.9mV, because 5V/1024 = 4.9mV

What about a 12 bit A/D? 12 bits is 4096 steps, so...

with a 5V reference, the step size is 1.2mV, because 5V/1024 = 1.2mV

... and so on, and so on...

Clearly, these values are targets, and things like noise and voltage stability will play an important factor in how accurately you can read a signal, but I hope this clears up the concept...

For instance, if you input voltage is 4 volts, then you are reading 80% of your reference voltage, right? (4/5 is 80%)

The resolution, or "how well you can resolve" this reading, is described in bits. An 8-bit A/D divides up the reference voltage into 8-bits, or 256 equally spaced voltages, from 0 to 5 volts. This corresponds to about 19.5mV per bit.

So how accurately can you measure the 4 volt signal? Well, 80% of 256 is 204.8, but you don't get any fractional bits when you read, so you will read 205. And what is the value of 205? 205/256 = 80.078125% of 5V, or 4.00390625V. Not too bad, right? Less than 0.1% error...

But what if your input voltage was 4.01V? 4.01/5 = 80.2%, which works out to 205.3, but remember? No fractional bits, so you would read 205 - But that's the same as before, right? So you would again read 4.00390625V even when the input was 4.01.

Here is where the step size comes into play - the size of the voltage step between bits is the step size.

With a 5V reference, the step size is 19.5mV, because 5V/256bits = 19.5mV

What about a 10 bit A/D? 10 bits is 1024 steps, so...

with a 5V reference, the step size is 4.9mV, because 5V/1024 = 4.9mV

What about a 12 bit A/D? 12 bits is 4096 steps, so...

with a 5V reference, the step size is 1.2mV, because 5V/1024 = 1.2mV

... and so on, and so on...

Clearly, these values are targets, and things like noise and voltage stability will play an important factor in how accurately you can read a signal, but I hope this clears up the concept...

ammi_paranjpe like this.

Tags ADC workingHow ADC worksAnalog to digital converter
Sun Dec 20 2009, 01:25 AM

thank u so much..........nw i gt it.......u even cleared ma doubt regardin referance vtg also.....

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