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RC time constant question

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shaneshane1

New Member
Hi im new to this whole time constant thing,i usually just ask for the answer, lol, but im at the stage were i should be working it out myself.

calculating how long it takes for a capacitor to reach 2/3 full

RC time constant = R X C (resistance x capacitor)

resistor = 68K (68,000 ohms)
capacitor = 470uF (0.000470 farads)

so the answer im getting = 31.96 (so 32 seconds)

so i put this to practice using approximatly 7V and timed it up to 32 seconds and the voltage was around 4.2V everytime and thats approximatly 2/3 of 7V

is this correct?????? and if so, how would i go about using that as a time delay to turn on a LED, so the LED comes on at approximatly 4.2V (32 seconds) what other components would i need?
 

mabauti

Member
Re:

4394-d73dd886b48875a2e7e542c8d6eba751.png
yields t = RC * ln(3) , for Vc = 7*(2/3) & V = 7, so your result is almost correct.

The problem is that you'll need a Op-amp in order to avoid the charge effect on the capacitor
 

Roff

Well-Known Member
The voltage at one time constant is 0.632*Vsupply. So, if you have a 7V supply, 68k, and 470uF, the time to reach 0.632*7V=4.425V will be 31.96 seconds. Your readings are within the typical tolerances of resistors and electrolytic capacitors. If you want to run the experiment you proposed, you will need a comparator IC and some resistors. You can also make a comparator out of a few transistors and resistors.
 

shaneshane1

New Member
i dont know anything about Op-amps :( , is that so the capacitor charges at the same rate(time)??? if so its not that critical, i just want to get the capacitor to act as a time delay switch to turn on a LED at a given time(Voltage),even if its 16seconds instead of 32(so half the time)eg: at 2.1V instead of 4.2

so like a switch that will only turn on after 2.1V ?????????
 

ClydeCrashKop

Well-Known Member
Most Helpful Member
If you use an LM339 comparator with a pot giving a 4.5 volt reference to the - pin, it will turn on when the input on the + pin rises to 4.5 volts. This is open collector out put so you need a pull up resistor. If you swap the inputs, you get inverted output that you can use to ground the LED at 4.5 volts so no pull up is required.
There are 4 comparators on that chip. Ground all unused inputs. With 4 different reference voltages you could use it to make a bar graph of 1 input.
 

Roff

Well-Known Member
shaneshane1 said:
how would i go about that? anyone
See below.

mabauit, i see what you mean,thanks for pointing that out.
Mabauti's circuit will give you about 80 microamps through the LED, which I don't think will be visible in daylight. You might see it in a dark closet after your eyes get accustomed to the dark.:rolleyes:
 

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audioguru

Well-Known Member
Most Helpful Member
You are just making a timer IC with transistors.
A 555 timer IC has an RC timing circuit and a comparator to detect when the voltage has charged high enough in the capacitor. Then its output becomes active with a pretty high available current that doesn't affect the weak charging current in the capacitor.
 

Roff

Well-Known Member
audioguru said:
You are just making a timer IC with transistors.
A 555 timer IC has an RC timing circuit and a comparator to detect when the voltage has charged high enough in the capacitor. Then its output becomes active with a pretty high available current that doesn't affect the weak charging current in the capacitor.
Except that the transistor circuit doesn't have any of the features of the 555 timer except for the comparator. I got the impression that he might have transistors on hand, but no timers or comparators. The comparator (integrated or discrete) would also allow him to change the voltage divider and see what happens to the turn-on time. He could even use a pot (gasp)!:D
 

audioguru

Well-Known Member
Most Helpful Member
He lives in Florida. Where will he buy a pot?
The alligators might eat him if he walks to RadioShack.
 

Roff

Well-Known Member
audioguru said:
He lives in Florida. Where will he buy a pot?
The alligators might eat him if he walks to RadioShack.
I was born in Florida. My daughter lived in Florida until recently. I have been there several times, and still have all of my limbs and most of my digital appendages.
 

audioguru

Well-Known Member
Most Helpful Member
I was also in Florida and didn't get eaten by alligators. I was very careful. i didn't walk to a RadioShack.
 

shaneshane1

New Member
ClydeCrashKop said:
If you use an LM339 comparator with a pot giving a 4.5 volt reference to the - pin, it will turn on when the input on the + pin rises to 4.5 volts. This is open collector out put so you need a pull up resistor. If you swap the inputs, you get inverted output that you can use to ground the LED at 4.5 volts so no pull up is required.
There are 4 comparators on that chip. Ground all unused inputs. With 4 different reference voltages you could use it to make a bar graph of 1 input.


iv went over what you have said and come up with a schematic for my application, i have swapped the pins over so i dont have to use a pullup resistor, is this schematic correct???

i have not purchased the LM339 yet to test this as im not sure if this schematic is right?
 

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Roff

Well-Known Member
Looks good, except for the LED details. See below. You should also connect a 0.1uF ceramic capacitor from +V to ground to prevent oscillation. You can also use LM393, which is the same comparator, but there are only two of them, in a 8 pin package.
 

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shaneshane1

New Member
so is the output a 0 or a 1????????

and whats the best way to get the input from 7 down to 4.5 for my input,eg: resistors...,im just a little unsure how to go about it, because iv never had to do it.
 

Roff

Well-Known Member
shaneshane1 said:
so is the output a 0 or a 1????????

and whats the best way to get the input from 7 down to 4.5 for my input,eg: resistors...,im just a little unsure how to go about it, because iv never had to do it.
The output of the comparator will go from off Open circuit) to on (zero volts) when the cap voltage exceeds 4.5V, or whatever voltage you set the noninverting (+) input to. Look at it like this: When the - input is higher than the + input, the output will go low, and vice-versa.
Use a resistor divider to get 4.5V from 7V. The resistor to ground will have 4.5V across it, and the resistor to +7V will have (7 - 4.5) 2.5V across it. So, the resistors have to be in the ratio of 4.5 to 2.5, which equals 1.8 to 1. I would use 180k from the + input to ground, and 100k from the + input to +7V.
 

shaneshane1

New Member
thank's for the help i really appreciate it, one last question before i go and buy the LM339, does the current coming out of the voltage divider need to be around a certain current (mA) to work properly on the input or is it just reliant on the volts???
 
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