ADC result in Three Decimal point

micropad · Jan 27, 2013

Ian Rogers said:
The pic18 series have hardware 8x8 multiply.

This is the knowledge base question

In the hardware level multiplier, Can I have idea what happened inside

Please advice

Ian Rogers · Jan 28, 2013

Here is one kind of hardware multiplier

If there were 8x8 bits instead of 2x2... Just imagine how big it needs to be..

micropad · Jan 28, 2013

Ian Rogers said:
The other way is sample 5 times, then and add together... This gives a a reading 5115 ( Not quite 5 volts )...

Dear rogers,
Could you please guide me how did you get this 5115

Thanks in advance

dougy83 · Jan 28, 2013

micropad said:
Dear rogers,
Could you please guide me how did you get this 5115

Thanks in advance

The max ADC count (which corresponds to 5V if that is your reference) of a 10bit ADC is 1023. Multiply by 5 samples and you have 5 x 1023 = 5115 counts.

micropad · Jan 28, 2013

Dear dougy83,

5 x 1023 = 5115 is now understand

"Multiply by 5 samples'

Why it should say "samples "

Please advice

Ian Rogers · Jan 28, 2013

micropad said:
Dear dougy83,

5 x 1023 = 5115 is now understand

"Multiply by 5 samples'

Why it should say "samples "

Please advice

Because that is what it is known as You sample the voltage on the ADC module

Assume 2.5 volts

Sample 1 = 511
add sample 2 = 1023
add sample 3 = 1534
add sample 4 = 2045
last sample 5 = 2556

Insert the decimal place 2.556 ( a little high perhaps )... But a whole lot easier..

ericgibbs · Jan 28, 2013

hi micro,
A way I use is a simple resistive divider in the ADC input.

It drops the 5V down to 4.88V, which gives an ADC count of 1000 for 5V.

Then add 5 readings to give 5000 for 5Vinp.

E

Ian Rogers · Jan 28, 2013

ericgibbs said:
hi micro,
A way I use is a simple resistive divider in the ADC input.

It drops the 5V down to 4.88V, which gives an ADC count of 1000 for 5V.

Then add 5 readings to give 5000 for 5Vinp.

E

Yep!!! Good Idea..

micropad · Jan 28, 2013

ericgibbs said:
hi micro,
A way I use is a simple resistive divider in the ADC input.

It drops the 5V down to 4.88V, which gives an ADC count of 1000 for 5V.

Then add 5 readings to give 5000 for 5Vinp.

E

This is really interesting, Thank you so much

ericgibbs · Jan 28, 2013

micropad said:
This is really interesting, Thank you so much

hi,
Use 1% resistors or a 500R trim pot for the 240R

You have not said what the output impedance of your 5V test source is.???

micropad · Jan 28, 2013

ericgibbs said:
hi,
Use 1% resistors or a 500R trim pot for the 240R

You have not said what the output impedance of your 5V test source is.???

Dear ericgibbs,

Thanks for the reply

As per datasheet the impedance is The maximum recommended
impedance for analog sources is 2.5 kΩ.

I think in the picture it shows as "RS"

It would be much appreciated if you can explain more theory related to the picture

Thanks in advance

ericgibbs · Jan 29, 2013

It would be much appreciated if you can explain more theory related to the picture

Hi,
I did this LTSpice simulation some time ago in order to show the effect of RS [Source Impedance] and the sampling time, when the ADC input was a 0V to 5V step voltage.

Some PIC's are recommended for use at a 2k5 or a 10K maximum value of RS.

The attached sim plot shows a RS of 1R thru 12500R in 2500R increments.

The question I asked you was, what is your expected RS value.?

Basically what are you using the ADC for.??

E,

micropad · Jan 29, 2013

ericgibbs said:
Hi,

The question I asked you was, what is your expected RS value.?

E,

Dear ericgibbs thanks for the reply

I used 100k variable resistor for ADC, RIGHT pin for VCC and LEFT pin for GND and middle pin to RA2 of PIC ( Analog input)

Please advice

MikeMl · Jan 29, 2013

micropad said:
...I used 100k variable resistor for ADC, RIGHT pin for VCC and LEFT pin for GND and middle pin to RA2 of PIC ( Analog input)

Please advice

With the pot wiper at the middle, the Thevenin equivalent circuit is a 2.5V source in series with a 50K resistor (half the pot value). This clearly violates the maximum 10K source resistance spec for the PIC's A/D input pin. You should use a pot no higher than 20K. Note that the worst case spot as the wiper is moved occurs at the half resistance point.

ericgibbs · Jan 29, 2013

hi micropad,

What is the application for the ADC measurement.?

Examples: AC waveforms such Audio signals,,, DC levels such as batteries or power supplies.

Its important that we know what is the expected result from the PIC's ADC when used in the way you intend to measure which type of signal/s.

E.

micropad · Feb 9, 2013

MikeMl said:
With the pot wiper at the middle, the Thevenin equivalent circuit is a 2.5V source in series with a 50K resistor (half the pot value). This clearly violates the maximum 10K source resistance spec for the PIC's A/D input pin.

Dear MikeMl, Thanks for the reply

I drown a simple circuit which connected to RA PIN. Finally we have to find the Th-evening Equivalent cct.Values using Maths.

In my drowning three resisters represent single pot wiper

Can you please advice, I am in correct tract

Please advice
Thanks in advance

Nigel Goodwin · Feb 9, 2013

As long as you're only using one analogue input then there's no real concern about the source impedance - the problem isn't the input impedance of the pin, but the time it takes to discharge or charge the sample and hold capacitor via the source impedance. This only really causes problems if you're switching between different inputs, you have to allow substantial time delays for the charging/discharging to fully take place.

But as you're just using a pot, why not use a sensible value one anyway? (4.7K/5K would be good).

micropad · Feb 9, 2013

Nigel Goodwin said:
As long as you're only using one analogue input then there's no real concern about the source impedance - the problem isn't the input impedance of the pin, but the time it takes to discharge or charge the sample and hold capacitor via the source impedance. This only really causes problems if you're switching between different inputs, you have to allow substantial time delays for the charging/discharging to fully take place.

But as you're just using a pot, why not use a sensible value one anyway? (4.7K/5K would be good).

Dear Nigel Thanks for the reply and advice.
Actually I try to implement and understand the The Thevenin theorem using real practical situation which is we learnt in the class room as a theory. In the above my post 56, I tried to drown a equivalent circuit related to preset. What do you think
Am I in the correct drowning??

Please advice
Thanks in advance

Nigel Goodwin · Feb 9, 2013

As you're no idea what the input impedance of the ADC pin is I don't see what you hope to achieve?.

ADC result in Three Decimal point

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