7-segment decimal point management

10 bit = unsigned 3 digit
11 bit = signed 3 digit
12 bit = signed 3½ digit and finally
signed 3½ digit + 3 D.P.= 12 bit + 3 decade gain option

doraemon said:

Hello!



I agree, but if you do this, you will never have a 5V reading even if you measure the reference.
With 10 bits, I agree that 1024 corresponds to 5V, but the problem is that with 10 bit, there
is no 1024. So dividing by 1023 instead of 1024 has its advantages.

Dora.

Click to expand...

Hi dora,

i can see your point. But isn´t the "0" reading and the "1023" reading somehow meaningless?

With "0" you know the input value of an ideal ADC (for a real ADC it is even worse) is below 1/2 lsb. With a 5V Ref and 10 bit ADC it shows you the input is below 2.5mV. But is it true 0.0V or is it -10mV or is it -100mV? you can´t say.
The same is with 1023.
If you really want to measure 0V and 5V you need some headroom on both sides.


With average or RMS it is even worse, because aou don´t see that the ADC is on its limits.
In my company we decided to say all values below a certain level ( maybe 1% FS) as underflow and all values above a certain level (maybe 99% FS) as overflow.
The thresholds depend on the external circuit and the ADC.
If a single value of ADC readings (with average and RMS) is out of range we decide the RMS value as "unreliable".

For a hobby project such considerations may be overkill, but there are areas (airbag control, power plants, medical measurement devices .....) our company has to proof that the output value is reliable.

But you are right. By working with 5VRef*ADCvalue/1024 one will never see 5V.

Klaus

Hello!

The "0" meaning of 1023? Sorry, I don't get it. Do you mean the 5.0 reading?

Anyway, I know it's not very clean, but at least if you plug a plain potentiometer on GND, VCC and ADC,
and if it displays something between 0 and 5V, then it will display 0 on 0 and 5 on 1023.

But I'm not sure it's word in terms of quantisation noise (never calculated, just thinking):
- A plain ADC does a truncation if you consider that 1024 (that does not exist in 10 bits) is 5V (or
more generally the reference). Therefore, if you want to calculate for any analog value, you will
find the average error being 0.5 LSB.
- If you map 0->0 and ref -> 1023, then in the lower range, you will truncate, and in the higher range
you will ceil (??) the value. This means that except for the first interval which corresponds to a truncation
and the last one which corresponds to a ceiling (truncation +1), all the intermediate values will have
an average error which is less than 0.5, and which will be 0.25 in the middle.

Well, that's pure speculation and again I didn't calculate it, but this somehow proves that the average
error is smaller.

Dora.

Hi Dear KlausST,

I hope you are fine.
Thank you for your co-operation.
The ammeter I built is successfully running and I am enjoying my project.

Hi,

Thanks.
Nice to hear that it's running fine.

Klaus