Searching \ for '[EE] capcitor charge time' in subject line. ()
Make payments with PayPal - it's fast, free and secure! Help us get a faster server
FAQ page: www.piclist.com/techref/timers.htm?key=time
Search entire site for: 'capcitor charge time'.

Exact match. Not showing close matches.
PICList Thread
'[EE] capcitor charge time'
2005\11\07@124208 by talgo

flavicon
face
Hi all

Do you know if it is possible to know the time that take a capacitor to
be fully charged?

capacitor 1000u  is connected to the base of 2n3094 and the resistor is
3.6K
voltage in circuit is 5v.

                        +5v
                        |
                  R    load 120ohm
Vin ------+/\/\/\/\+---|/
              |       |\
              = C       |
              |       gnd
            gnd


Thanks

Tal

**** I rest my cake **** :)


2005\11\07@125740 by Hopkins

flavicon
face
R*C = time to charge to 77% (I think - can look it up if you want :) )

Note a capacitor in theory is never fully charged as the charging slows
as the voltage increases.

_______________________________________

Roy
Tauranga
New Zealand
_______________________________________

>
> Do you know if it is possible to know the time that take a capacitor
to
> be fully charged?
>
> capacitor 1000u  is connected to the base of 2n3094 and the resistor
is
{Quote hidden}

--
No virus found in this outgoing message.
Checked by AVG Free Edition.
Version: 7.1.362 / Virus Database: 267.12.8/162 - Release Date:
5/11/2005


2005\11\07@131219 by Mauricio Jancic

flavicon
face
hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html#c1

Mauricio Jancic
Janso Desarrollos
Microchip Consultant Program Member
spam_OUTinfoTakeThisOuTspamjanso.com.ar
http://www.janso.com.ar
+54 11 4542 3519

> {Original Message removed}

2005\11\07@133500 by olin piclist

face picon face
talgo wrote:
> Do you know if it is possible to know the time that take a capacitor to
> be fully charged?
>
> capacitor 1000u  is connected to the base of 2n3094 and the resistor is
> 3.6K
> voltage in circuit is 5v.
>
>                         +5v
>                         |
>                   R    load 120ohm
> Vin ------+/\/\/\/\+---|/
>               |       |\
>               = C       |
>               |       gnd
>             gnd

Do the math.  If the "fully charged" value in this circuit is 700mV and you
have a 5V supply, then you want it to get .7/5 = .14 of the way.
Ln(1/(1-.14)) = .15 time constants.  One time constant is 3.6Kohms x 1mF =
3.6 sec.  3.6 sec x .15 = 540mS.


******************************************************************
Embed Inc, Littleton Massachusetts, (978) 742-9014.  #1 PIC
consultant in 2004 program year.  http://www.embedinc.com/products

2005\11\07@142238 by Enrico Schuerrer

picon face
Charge to 63%

KR
Enrico
----- Original Message -----
From: "Hopkins" <.....rdhopkinsKILLspamspam@spam@ihug.co.nz>
To: "'Microcontroller discussion list - Public.'" <piclistspamKILLspammit.edu>
Sent: Monday, November 07, 2005 6:57 PM
Subject: RE: [EE] capcitor charge time


| R*C = time to charge to 77% (I think - can look it up if you want :) )
|
| Note a capacitor in theory is never fully charged as the charging slows
| as the voltage increases.
|
| _______________________________________
|
| Roy
| Tauranga
| New Zealand
| _______________________________________
|
| >
| > Do you know if it is possible to know the time that take a capacitor
| to
| > be fully charged?
| >
| > capacitor 1000u  is connected to the base of 2n3094 and the resistor
| is
| > 3.6K
| > voltage in circuit is 5v.
| >
| >                          +5v
| >                          |
| >                    R    load 120ohm
| > Vin ------+/\/\/\/\+---|/
| >                |       |\
| >                = C       |
| >                |       gnd
| >              gnd
| >
|
| --
| No virus found in this outgoing message.
| Checked by AVG Free Edition.
| Version: 7.1.362 / Virus Database: 267.12.8/162 - Release Date:
| 5/11/2005
|  
|
| --

2005\11\08@070810 by Bill & Pookie

picon face
First you need to find the RC time constant of the circuit.  This is simply
the resistor (measured in ohms) times the capacitor (measured in farads) and
the results is in seconds.  In your case it is 3.6 * (10 ^ 3) * 1 * (1000
^ -6) = 3.6 seconds.  That is the RC time for your circuit.

The capacitor will charge to 70.7% of the available voltage in 1 RC time
constant.  The capacitor is considered to be fully charged after 5 RC time
constants.   In your case, after 18 seconds.

A quick way to calculate the voltage is to use 2/3 or 66.7% of the voltage
in 1 RC time.

Easy to figure that it charges 2/3's V in 1 RC time. The second RC time it
will increase to 2/3 of the remaining voltage.  In your circuit you should
take into account the 0.6v or so dropped my the solid state  device.

A important number to remember is .707  just like the airplane.  If you have
a AC peak to peak voltage and you want to know the RMS of it (sort of the DC
equivalent), multiply the peak to peak by 9.707..  To go from RMS to peak to
peak, multiply AC RNS reading by 1.4.

1.4 is the square root of 2 and .707 is 1 / square root of 2

sqrt(2) = 1.41421356

1 / sqrt(2) = 0.707106781

Bill

{Original Message removed}

2005\11\08@083020 by olin piclist

face picon face
Bill & Pookie wrote:
> The capacitor will charge to 70.7% of the available voltage in 1 RC
> time constant.

No, the value is 1 - 1/e = 63.2%.

> The capacitor is considered to be fully charged after
> 5 RC time constants.

This is an arbitrary choice of what you consider close enough.  The
capacitor of an R/C filter is never fully charged to the input value.  It's
an exponential decay that never actually gets there, so the question is
really how close is good enough for the context.  Saying that it's
"considered fully charge after 5 RC time constants" is misleading at best
without context.

> In your case, after 18 seconds.

Actually in his case it will never get there by a long shot since the B-E
junction will start conducting around 500-700mV.  That's whay I answered how
long it would take to reach 700mV in a previous post.

2005\11\08@091734 by Enrico Schuerrer

picon face
To calculate the voltage across a capacitor for a given time t the exact
formula is

UC = UIN * (1-e-(t/Tau))  -    Uc = Uin * (1 - e^-(t/Tau))

e is around 2.71828, Tau = R * C [seconds, Ohm, Farad]

Maybe it helps you for calculating.

Kind regards

eNrico

{Original Message removed}

2005\11\08@111444 by talgo

flavicon
face
Hi 2 all

So i am confused more then before :-)

i got 6 replys and more than 1 way to solve my question, and some
different answers and calculations.

what went wrong?

thanks

Tal

**** I rest my cake **** :)





Bill & Pookie wrote:

{Quote hidden}

> {Original Message removed}

2005\11\08@121240 by William Chops Westfield

face picon face
On Nov 8, 2005, at 8:13 AM, talgo wrote:
>
> what went wrong?

Most of the answers were close.   "about 70%" or "about 2/3" or "exactly
1-1/e" are all close enough to each other for most engineers...  The
sqrt(2)/2 response was a bit worrisome, though, even though reasonably
correct numerically... :-)

The other point of confusion is the looseness of the definition of
"fully charged"; normally I'd take this to mean that the voltage on
the capacitor equalled the input voltage, but in the circuit diagram
you provided, the cap would never reach that voltage because the
transistor
Vbe junction would begin to conduct when the voltage on the cap reached
the
Vf of the junction.  So it looked to some of us as if the question
actually
was "how long after power is applied till the transistor turns on",
which
isn't really the same as "till the cap is fully charged."

BillW

2005\11\08@125640 by talgo

flavicon
face
Bill

If I have any error i apologize, but I take everybody answers seriously,
but I can't understand the time margins between Olin's calculation 540ms
and Bill 18 seconds.

I am not an engineer but all of the explanation look good to me, but the
results are very different.  540ms<>18 seconds isn't it? :-(

Tal

**** I rest my cake **** :)


William Chops Westfield wrote:

{Quote hidden}

2005\11\08@132425 by Harold Hallikainen

face picon face
The RC time constant uses the R "seen" by the C. In the circuit you
supplied, that R is is the Thevenin equivalent of the resistor and the
dynamic base-emitter junction resistance of the transistor. That can be
approximated as 25mV/Ie. It will generally be much less than the 3.6k
resistor you have. So, another way of looking at the circuit is to say
that the base-emitter junction is an open circuit until the base voltage
reaches about 700mV. Then the transistor conducts and the load is
energized. So, what you need to figure out is how long it takes for the
capacitor voltage to reach 700mV.

The explanation below is true of a general RC circuit. Here, however, the
transistor base-emitter junction limits the capacitor voltage to 700mV.

Another replay indicated that the voltage as a function of time was

V(t)=Vin*(1-e^(-t/T))

where t is timer, and T is the RC time constant. As t approaches infinity,
the we have V(t)=Vin(1-0). Applying a little algebra we get

T(v)=T*ln(1/(1-(v/Vin))

With your values (3.6k and 1,000uF), T=RC=3.6 seconds. We're interested in
knowing when v will be 700mV. Vin is 5V. Solving for this (on my trusty
HP15C), I get 543ms .

To get a longer time, you need to have a higher trip voltage (than the
700mV). The NE555 timer has a couple comparators that have their trip
points set to 1/3 and 2/3 Vcc. When operating in monostable mode, the
capacitor charges from 0V to 2/3Vcc. When operating in astable mode, the
capacitor charges and discharges between 1/3Vcc and 2/3Vcc.

One problem with getting long times with RC time constants is you need
large R and large C. Large C tend to have leakage current so you end up
with a voltage divider being formed by the R and the leakage current. This
voltage divider limits the maximum capacitor voltage, often to something
below your trip point, so the circuit never times out.

Harold




{Quote hidden}

>> {Original Message removed}

2005\11\08@140824 by talgo

flavicon
face
Thanks guys!
all of you who explain with patient...:-)


you have enrich me with your experience & wisdom and i thank you again.

Tal

**** I rest my cake **** :)


Harold Hallikainen wrote:

{Quote hidden}

>>>{Original Message removed}

2005\11\08@162311 by William Chops Westfield

face picon face

On Nov 8, 2005, at 9:55 AM, talgo wrote:

> but I can't understand the time margins between Olin's
> calculation 540ms and Bill 18 seconds.
>

The actual equation is:
   V(t)=Vin*(1-e^(-t/(R*C))

Olin solved:
    V(t) = 0.7           (approx when the transistor turns on.)

Others solved for:
       V(t) = 1 - 1/e       (one time constant; t = R*C   The usual answer
                          for "how long till charged?", but probably not
                          correct for this case.)

and others for:
    V(t) appox= Vin      (several time constants: "FULLY charged")

BillW

2005\11\08@164118 by olin piclist

face picon face
talgo wrote:
> i got 6 replys and more than 1 way to solve my question, and some
> different answers and calculations.
>
> what went wrong?

Some answers were bad, but in the end there is no substitute for knowing
what you're doing.  So go do the math yourself.  What's the problem!!?

2005\11\08@165305 by olin piclist

face picon face
talgo wrote:
> If I have any error i apologize, but I take everybody answers
> seriously, but I can't understand the time margins between Olin's
> calculation 540ms and Bill 18 seconds.

Right, so figure out why.  Different people were making different
assumptions, and there were a few errors too.  Everyone showed how they
arrived at their figures, so not it's up to you to decide which assumptions
were valid for your case.  Another good idea is to build the circuit and
measure the time delay, then let us all know what you found.


******************************************************************
Embed Inc, Littleton Massachusetts, (978) 742-9014.  #1 PIC
consultant in 2004 program year.  http://www.embedinc.com/products

2005\11\09@090433 by Bill & Pookie

picon face
As usual, Olin brings his quiet knowledge to the scene.

And I will ponder the question "Is the cup  full after 5 RC time constants
or not?"

Bill

{Original Message removed}

More... (looser matching)
- Last day of these posts
- In 2005 , 2006 only
- Today
- New search...