another basic math question
japer
Posts: 105
hello,
trying to make sure that my math is right before i go on to end of chapter 6 basic analog and digital
______________
C= .01uf
Ra = 3.5 k ohm
Rb = 2 kohm
F = 2 hz
____________
using the formula to predict the 555 Timer's Frequecy
Do i drop the multipiler of the resistors, then add them to the final answers ?
as
F = 1.45/[noparse][[/noparse]C*(Ra + 2Rb) this seeems to come up right if do
same on Pulse width calculation
a) 0.69 x C x (Ra + Rb)
and on
Time off
b) 0.69 x C x Ra
while using a scope "not real time" this come up right every time +-10%
why do i have to drop the resistors multiplier ?
trying to make sure that my math is right before i go on to end of chapter 6 basic analog and digital
______________
C= .01uf
Ra = 3.5 k ohm
Rb = 2 kohm
F = 2 hz
____________
using the formula to predict the 555 Timer's Frequecy
Do i drop the multipiler of the resistors, then add them to the final answers ?
as
F = 1.45/[noparse][[/noparse]C*(Ra + 2Rb) this seeems to come up right if do
same on Pulse width calculation
a) 0.69 x C x (Ra + Rb)
and on
Time off
b) 0.69 x C x Ra
while using a scope "not real time" this come up right every time +-10%
why do i have to drop the resistors multiplier ?
Comments
and:
····· C· =· 0.01uf
····· Ra = 3.5 k ohm [noparse][[/noparse]3,500]
····· Rb = 2 kohm···· [noparse][[/noparse]2,000]
f·= 1.44 / C (Ra + 2Rb)
f =·1.44 / C (Ra + 4,000)
f = 1.44 / C (3,500 + 4,000)
f = 1.44 / C (7,500)
f = 1.44 / 0.00000001 (7,500)
f = 1.44 /·0.000075
f = 19,200 Hz
The period of 19,200 Hz is 52 usec (0.000052 sec)
Timea·(time HIGH) =·0.69 x C x (Ra + Rb)
························ =·0.69 * 0.00000001 * 5500
························ = 0.69 * 0.000055
························ = 0.00003795 sec
Timeb·(time LOW)·= 0.69 x C x Rb·· [noparse][[/noparse]it's Rb, NOT Ra]
························ = 0.69 * 0.00000001 * 2000
························ = 0.69 * 0.000035
························ = 0.0000138 sec
Timea + Timeb = period
0.000038 sec + 0.0000138 sec = period
···························0.0000518 = period
So, clearly, it's not "2 Hz", but nearly 20 kHz.
Please take a look at the Picture that I've attached.· Is it at all like your output?
Although this doesn't answer your question directly, you may find the attached 555 Timer Calculator handy.
Regards,
Bruce Bates
obviosly i am looking at my cap rating wrong
everything you calculate is comparitible to my caculations
except the "cap valve"
i surely see where my weakness is in this formula
as far as time low
You are right, wrote the formula down wrong in the post
Thanks for Straighting me out
japer
I want to thank bruce again for the time and effort put into his post.
the detail are incredible ,I feel i did not properly thank bruce for his effort
also i am an idiot
first the cap was 0.1 mf,and that's how i calculated it ,duh
i should have calculated at .000001
so i bought a book from radio shack and with the stamp program books
i should be able to cut down on these stupid questions
thank's again bruce and Parallax forum for your patience
japer
You rang?
In all fairness, and giving credit where it IS DUE, YOU were certainly the one who offered all the math assistance and examples. I was merely the guy who offered a 555 timer calculator program.
I fear Jasper mixed up our replies to him. No harm or slight intended, I'm sure.
Regards,
Bruce Bates
I want to thank PJ Allen ,I feel i did not properly thank PJ Allen for his effort
also i am an idiot
first the cap was 0.1 mf,and that's how i calculated it ,duh
i should have calculated at .000001
so i bought a book from radio shack and with the stamp program books
i should be able to cut down on these stupid questions
thank's again PJ Allen and Parallax forum for your patience
japer the idiot
· By the way, the Calculator that comes with "Windows" has·scientific notation·(View, *Scientific).· So, instead of typing in 0.000001 for 0.1 uF, you'd type 1, Exp, 6, and then the·+/- (change sign) key.· And with that you can also "translate" numbers from hex, binary, octal, and decimal.
· 1000pF is a lot easier to handle as "1000 e-12" than 0.000000001
thanks PJ Allen
so these calculators can translate binary numbers. this was what i was looking for,
also i have another basic math question coming up. i am tying to work it myself
but so far the example in the book shows an ending digit of 3 and i'm dealing with
5 and 1
you want a crack at it ?
japer
Let's have all the details, no omissions.·
(BTW, just to clarify further, some scientific calculators may have the binary-hex-decimal conversion feature/s, but the "Windows" calculator definitely does.· So, if you go shopping for one, make sure.)
Ryan
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Ryan Clarke
Parallax Tech Support
RClarke@Parallax.com
Mr, PJ Allen
Sorry I have not Responded to you Lately
Ok here's the fomula unfortunaltly i do not know how to lower case or upper case the text as you did
by naming the formula you will proably know it well
"Exponetial Decay"
the Problem Iam Having is when the formula is Broken Down into Properties of logarithms were
t
= 1n( Vdd) what is the " 1n" i found the "E" as = to 2.718 i know this is not the exact formula but having trouble with "1n"
R X C
Vinput
thanks
japer
· http://www.reference.com/browse/wiki/Natural_logarithm
My take on the natural log is that it answers the question, "what value do I have to raise e (2.718...) to the power of·to get this value".
For example, e2 = e X e ≈ 2.718 X 2.718 ≈ 7.389.·
So, if you want to know the natural log·of 7.389,·the answer is 2.· In equation form, that's:
· ln(7.839)·≈ 2
Another way to say it is, "the natural log gives you the exponent you had to raise e to to get the value."· In the case of 7.839, we have to raise e to a power of 2 to get that value.· The exponent is 2, so that's the answer of ln(7.839).
If you have a scientific calculator, there should be an eX button on it to raise e to various powers, and a ln (or sometimes LN) button to take the natural log of your result, which should get you back to the value of X.
If you're concerned about Basic Analog and Digital v1.3, pages 150 to 151, open up the attached PDF, and then continue reading.· The attached PDF breaks those equations into smaller steps, each one numbered to the right.·
We started with equation 1.·
Equation 2 takes the natural log of both sides of the equation.· That's fair game so long as everything you take the natural log of is greater than or equal to zero.·
Getting from equation 2 to equation 3 uses a property of logs that goes like this: log(a X b) = log(a) + log(b).· The same applies for ln(a X b), it's equal to ln(a) + ln(b)
Getting from equation 3 to equation 4 uses the property of logs at the beginning of this post.· If you take the natural log of e raised to some exponent, the answer is the number that was e's exponent.
Getting from equation 4 to 5 is the rule that adding a negative value is the same as subtracting that value.
Next, add t/RC - ln(VP15) to both sides of equation 5, and you'll get equation 6.
To get from equation 6 to 7, there's a property of logs that's similar to the step 2 to 3 property: log(a/b) - log(a) - log(b).·
Post Edited (Andy Lindsay (Parallax)) : 11/16/2005 5:31:41 AM GMT
Thanks for breaking down the formula for me.
japer