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I have one voltage which vary logarithmical.
How change the code below to show voltage on LCD in logarithm.(not linearity)
Code:loop:
ADCIN 0, adval ' Read channel 0 to adval (0-1023)
adval = (adval */ 500)>>2 ' equates to: (adval * 500)/1024
LCDOut $fe,1,"DC Volts= "
LCDOut DEC adval/100,".", DEC2 adval ' Display the decimal value
Pause 500 ' Wait .5 second
GoTo loop ' Do it forever
I searched the form for logarithm and found this,
http://www.picbasic.co.uk/forum/show...ight=Logarithm
anyone to know to help me
Build a table containing log values derived from a HLL running on a PC. Embed this table into your program. You could use maybe 10, maybe 50 points in the table and interpolate between them to get a log reading. This method is not precise but workable since I assume you only need to visualise the log reading on a display. For precise log values, you have to do it the hard way.
Jerson
Hi, Savnik
Jerson is right ... the simplest Pic way is to build a retrieving table.
IN PbP that's called LOOKDOWN2 ... ( catches the closest values, sup and inf ... just think to use a clever index !!! )
Then just have to add a linear approx. between table values ...
Buuuuuut, My old brain tells me it is very easy to do it with 3or4 couples of linearising resistors and diodes ...
or to use a logarithmic AOP !!!
Last and simplest solution: The App Notes of ICL 7106 or 7136 show a logarithmic scale design ...
Which is closest to your requirements ??? ...
Alain
I have make a table.Quote:
Originally Posted by Acetronics
I want to measure the volts at the input of pic , and show on LCD the power
Code:POWER VOLT POWER VOLT
1 7,1 26 36,1
2 10,0 27 36,7
3 12,2 28 37,4
4 14,1 29 38,1
5 15,8 30 38,7
6 17,3 31 39,4
7 18,7 32 40,0
8 20,0 33 40,6
9 21,2 34 41,6
10 22,4 35 41,8
11 23,5 36 42,4
12 24,5 37 43,0
13 25,5 38 43,6
14 26,5 39 44,2
15 27,4 40 44,7
16 28,3 41 45,3
17 29,2 42 45,8
18 30,0 43 46,4
19 30,8 44 46,9
20 31,6 45 47,4
21 32,4 46 48,0
22 33,2 47 48,5
23 33,9 48 49,0
24 34,6 49 49,5
25 35,4 50 50,0
May I suppose the number in the "Power" column has to be displayed ... or is it just your Index ???
In this case ... It would be much easier to use if the "Volts" were integers numbers and "Power" decimals ... I know: some more work for you !!!
Alain
See Also Here : http://curveexpert.webhop.net/
That Helps ... isn't it ???
Quadratic Fit: y=a+bx+cx^2 ( y: Power ; x :Voltage )
Coefficient Data:
a = 0.00076569488
b = -8.0880958e-005
c = 0.019999816
I display only the voltsQuote:
Originally Posted by Acetronics
read from previous posts :Quote:
Originally Posted by savnik
<< I want to measure the volts at the input of pic , and show on LCD the power >>
Could you give me a decent explanation ???
Alain
I have a linear for FM and i want to measure the power.Quote:
Originally Posted by Acetronics
From the power meter i take the indication which is in volt.This volt i want to show on LCD. But the power change in logarithm scale and this is my problem.
This volt i want to show on LCD.
> Where's the problem, then, If you only want to display the voltage ???
But the power change in logarithm scale and this is my problem.
> This needs some USEFUL enlightments ... if you do not need to display the Power !!!
> Do you want to display Voltage in dBv ???
> Do you want to display Voltage in some BARGRAPH mode ??? ...
those ONLY could give any sense to your question !!!
Alain
My english is not so good.Quote:
Originally Posted by Acetronics
I want the voltage which measure with the pic , to show on lcd in dBm or dBv or watt (the table which post in #6)
Savnik,
You need to get a book on basic electronics. Voltage Current Power and Resistance.
Power = voltage * current
In your case you can not see the current so: Power = Voltage * Voltage /Resistance (Voltage squared)
To prove that look at your table. Voltage of 22.4 volts = 10 watts and Voltage of 44.7 volts = 40 watts. Voltage increases 2:1 and power increases 4:1. That proves there is a voltage squared * constant.
Voltage * Voltage/50 = Power
To convert to db is very hard or very simple depending on how you look at it. You can fight BIG math for weeks. When you understand db it is almost as simple as finding Power. I will not tell you how now. You need to think about it for a while.
RonS.
Hi Savnik,
If I read your problem correctly:
You have a RF amplifier that has an analog power meter. The power meter reads a voltage and displays the power on a Logarithmic scale. The input, to move the meter, is voltage. The scale painted on the meter face is logarithmical.Quote:
Originally Posted by savnik
You need the PIC’s adc input to read the DC voltage running the meter and then using mathematics or a “look-up” table convert the voltage to the equivalent power in Watts. You have already calculated the table, posted in your previous post.
Is this close to the question? I am just trying to help with clarifying the question. Sorry I don’t have an answer for you.
-Adam-
Yes, it is my question.Quote:
Originally Posted by Pic_User
I am a electronic for 30 years.Quote:
Originally Posted by ronsimpson
Thank you for your help. You give me some idea how to calculate.
Savnik,
Your problem has been bugging me, I like math puzzles, and the post by Pic_User did clear things up.
Now if I understand your table correctly, POWER column is the reading you want and the VOLT column is the value you now have. If this is correct, then keep reading, if not than ...
This is the formula:
Step #1:
D = (VOLTS / 10) * 0.2 'D' is a dummy variable
Step #2:
POWER = VOLTS * D
This give an answer within .1 of the value in the power column.
Hope this helps.
I look back and see that ronsimpson said the same thing.
Now you have two ways to do it. :)
P=v*d, D=(v/10)*.2
P=v*((v/10)*.2)
P=v*v*.2/10
P=v*v/50
I was looking at a way to change one set of numbers into another. Did not pay attention to your post or think in electrical terms. Just MATH. NUMBERS DO NOT LIE. NOT OFTEN:)Quote:
Originally Posted by ronsimpson
You must be an Engineer not a Lawyer or Statistician. On subject; I need to get back to taxes.
Fuzzy logic (fuzzy math)
Just an old farm boy having fun.Quote:
Originally Posted by ronsimpson
I use this code to read the forwardpower
and this for reflected powerCode:ADCIN 0, adval ' Read channel 0 to adval
dummy = adval * 5 * 10 ' Multiply to load internal registers with 32-bit value
value = Div32 1023 ' Divide 32-bit value by word and store in word
forwardpower = (value*value)/50
and the type for SWR is:Code:ADCIN 1, adval1 ' Read channel 0 to adval
dummy1 = adval1 * 5 * 10 ' Multiply to load internal registers with 32-bit value
value1 = Div32 1023 ' Divide 32-bit value by word and store in word
reflectedpower = (value1*value1)/50
..........1 + square root(reflectedpower/forwardpower)
SWR = --------------------------------------------
..........1 - square root(reflectedpower/forwardpower)
Any help how to show the SWR on LCD;
The largest ADVAL1 can be is 1023
The largest dummy can be is 1023 * 5 * 10 = 51150 (which is 16 bits or less)
Why do you do the * 5 * 10, why not * 50? and why *50?
You do not need DIV32 since you never exceed 16 bits with dummy
The same goes for ADVAL calcs
In final form, your
RP = (ADVAL1*50/1023)^2/50
FP = (ADVAL*50/1023)^2/50
If you plug these into your SWR equation and reduce, you get
SWR = (ADVAL1+ADVAL)/(ADVAL1-ADVAL), which should be easy to deal with
However, with all this noted, I do not think you are going to get what you want ... or will you? I haven't yet tried to follow this thread in any depth.
I use adval * 5 * 10 , because adval * 5 is to show on lcdQuote:
Originally Posted by paul borgmeier
the volts from 0 to 5v and 10 is multiplier to show on lcd 0 to 50v