calculation help for digital to analog converter

[maxketcham]

50 mv = 0.05v
15.000 v
15/.05= 300
binary of 300 is the value i posted above
count the numbet of bits used = 9

total accuracey is 2^9 bits
or
111111111
or 511
or 0~511
(512 bits)

It is not 512 bits as you stated. it is 9 bits for the example you just stated which would be 512 individual steps in the analog to digital output. the actual output would then be 15/512 or .29 volts per step if 15 volts wer the voltage maximum and O is the minimum range

 

[iant]

if you want to nitpick fair enough

:doh:Let me try that again
now divide 15 by 511 and your accuracey shows
0.02935 - this is the value of each bit in Volts
well within the requested 50mv

OR using 512
it is 0.02929
or 29.29mV instead of 29.35mV as I mistakenly stated
I acknowledge in certain circumstances it is important
29.29 x 512 = 14.996
29.35 x 512 = 15.07
But set the topvalue in a PLC and it is 511 not 512 that you need to set
29.29 x 511 = 14.967 (9 bit)
29.35 x 511 = 19.997 (9 bit)
to input dec 512 in a plc is a 10 bit code not 9
sorry

 

[Peter Nachtwey]

maxketcham, your answer is correct IF you think the question is a valid one. The instructor may say you are right, I would say you are both wrong in a subtle sort of way. I ask you and the instructor how you will design a DtoA converter that goes from -5 to +10 volts.

The practical answer is that one designs the DtoA converted to have an output range of -10 to +10 volts. Now do the math.

 

[iant]

Of course you are right Peter - I did know that
Just was not going to go that far
I, as usual, assumed it was a study question
- you know the type - that answer is something close to something.
as this was a student's question I chose not answer as per a field device.

 

[maxketcham]

maxketcham, your answer is correct IF you think the question is a valid one. The instructor may say you are right, I would say you are both wrong in a subtle sort of way. I ask you and the instructor how you will design a DtoA converter that goes from -5 to +10 volts.

The practical answer is that one designs the DtoA converted to have an output range of -10 to +10 volts. Now do the math.

I wondered when anyone was going to bring that point up, @ 20V (+/-10) It works out to @39 mV per step so by luck the right answer was found. Though the student still doesn't understand why. That is why I sent the link. it explains how and why they work. If he can read, and think, he'll figure it out himself.
I only used your simile to keep from confusing him further

 

[iant]

it could be +/- 5 volt and they add a battery later

 

[iant]

20/0.05 = 400
20/512 = 0.039
or
20/511 = 0.0391

or if you want to be able to set 0.05 mV
using practical devices - Cheap (12 Bit) analogue (not AB)
20/4096 = 0.00488 v
or
20/4095 = 0.00488 v

 

[garoken]

wow really thanks you guys for all this amazing answers it really help me alot as i am really struggling at this questions because i am pretty weak at this plc subject

 

[iant]

garoken, glad we can help
I hope you excuse our occasional stirring though.
we are being friendly

 

[garoken]

Hi sorry another question for a range 4to20ma is the range 20 ma or 16ma because my friend told me it was 20 while I calculated 16 so I am not sure which is the correct please help....

 

[maxketcham]

the range is 16 ma. the range would be maximum value - minimum value

 

[iant]

Hi sorry another question for a range 4to20ma is the range 20 ma or 16ma because my friend told me it was 20 while I calculated 16 so I am not sure which is the correct please help....

tell your mate to prove it to you.
and lots of luck to your mate:beer:

 

[garoken]

so for the question can you guys please help me correct my mistakes because i am not sure if i am correct anot and tommorow is my exam let say i am given a speed sensor with the reading of 0 to 1500rpm m it has a analog output of 4-20ma and a adda resoulution of 12bit and a plc reading of 3500 i am ask to calulate the speed and ma so for my answer
to calculate speed
first 12 bit = 4096
so i use 1500 divided by 4096
which equals to 0.366211 rpm
then use 0.366211 x 3500rpm
which equals to 1281.74rpm
so the other question if the speedometer show the reading of 200rpm, whats value is the plc showing and what is the voltage at the input
for the value the plc is showing
= 200 divided by 0.366211 rpm = 546
for the voltage at the input
output range =4-20ma = 16ma
so 16 divided by 4096 =0.003906
then for the ma 3500x0.003906=13.67ma
then back to the voltage at the input
=546bit x0.003906
=2.13ma
then use 2.13ma plus 4 because 4 is a positive value which is equals to 6.13.
so is my answer correct or is there any corrections needed??

 

[iant]

your calculated value
0.366211 is (RPM/BIT)
lets check that first
0.366211 * 4096 = 1500 All good
so lets work out the mA for the RPM
16/1500 = 0.01066667 mA per RPM
so x 200 = 2.133 mA above zero
so add the ofset of 4mA
6.133mA
same values different approach

 

[garoken]

but there is one thing that i do not get that is the first question to tell me to calculate the speed and ma i know the speed but i don't get how to calculate the ma