Range of ADC, can it be changed?

I am kinda new to handling ADC's, so bare with me all ye enlightened ones.

I need to digitise an input which varies from 0v to 1.5v using a ADC0804.
The range of this IC, i believe is 0-5v. Can we modify the range in which it works, say 0-1.5v so that i can have a better resolution.

Thanks.......

Is amplifying the 1.5v up to 5v not an option? That would certainly be easiest, you can attuenuate it again at the output if you need.

Actually i need as high a resolution as I can get by reducing the range and at the same time try to use a 8bit adc such as the adc0804.
My source gives an ouput from 0-1.5V

for example: if the range is 0-1.5V the resolution is going to be approx 6mV whereas as if the range is 0-5V then the resolution is 20mV.

So what i need to know is that if the Vref/2 can be set to .75V so that the range is only from 0 to 1.5V. I didnt not fully understand this from the datasheet.

If someone could help with this, would be grateful.

LijoeThomas said:

Actually i need as high a resolution as I can get by reducing the range and at the same time try to use a 8bit adc such as the adc0804.
My source gives an ouput from 0-1.5V

for example: if the range is 0-1.5V the resolution is going to be approx 6mV whereas as if the range is 0-5V then the resolution is 20mV.

So what i need to know is that if the Vref/2 can be set to .75V so that the range is only from 0 to 1.5V. I didnt not fully understand this from the datasheet.

If someone could help with this, would be grateful.

Click to expand...

An amplifier will work better. If you amplify a 0 to 1.5 volt signal by 3.33, you will now have a 0 to 5 volt signal. Each 6mV increment of your input signal is 20mV at the output of the amplifier.
The problem with reducing the A/D range is that it may have errors that don't scale. As an example, let's say it has a 6mV error at the point where the MSB changes. At 5V full scale, that's only a 1/3 LSB error. If you reduce the reference to 1.5V, the error is now a full LSB.
This type of error may or may not exist in your A/D, but in general, it is good practice to run the A/D at it's maximum full-scale range for best performance.

Hey Ron, Thanks for the idea. My input was initially 0-15V, which I had used an opamp to divide by 10, which gave me 0-1.5V.

I'm going to replace that with a divide by 3 [so the o/p is 0-5V] so that I still get the same resolution. Thanks for the solution.

I'm really grateful.....

LijoeThomas said:

Hey Ron, Thanks for the idea. My input was initially 0-15V, which I had used an opamp to divide by 10, which gave me 0-1.5V.

I'm going to replace that with a divide by 3 [so the o/p is 0-5V] so that I still get the same resolution. Thanks for the solution.

I'm really grateful.....

Click to expand...

If your A/D has high input impedance (many do), then you don't need an op amp. You can just use the resistive divider.

Can you draw me a sample resistive divider network? I'm not very sure if what I know is right.

ITs just two resistors wired in series, where the ends are connected to your 0 and 15V and the center between the two is the output...

Uo=Ui(R2/R1+R2)

For you R1 could be 2K and R2 1K so you get 1/3rd of the voltage on the output.

Heres the pic...

The ratio of the two resistors needs to be 2:1, as drrogla pointed out. The actual values depend on how much loading your A/D can tolerate, and the input resistance of your A/D. It may also depend on the conversion rate.
If you can tell us those three parameters, we can be more helpful.