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'[EE] capcitor charge time'
2005\11\07@124208
by
talgo
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
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}> 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\07@131219
by
Mauricio Jancic
2005\11\07@133500
by
olin piclist
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) 7429014. #1 PIC
consultant in 2004 program year. http://www.embedinc.com/products
2005\11\07@142238
by
Enrico Schuerrer

Charge to 63%
KR
Enrico
 Original Message 
From: "Hopkins" <.....rdhopkinsKILLspam@spam@ihug.co.nz>
To: "'Microcontroller discussion list  Public.'" <piclistKILLspammit.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

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
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 BE
junction will start conducting around 500700mV. That's whay I answered how
long it would take to reach 700mV in a previous post.
2005\11\08@091734
by
Enrico Schuerrer
To calculate the voltage across a capacitor for a given time t the exact
formula is
UC = UIN * (1e(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

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}> 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@121240
by
William Chops Westfield
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
11/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

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}> 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
> 11/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@132425
by
Harold Hallikainen
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 baseemitter 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 baseemitter 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 baseemitter junction limits the capacitor voltage to 700mV.
Another replay indicated that the voltage as a function of time was
V(t)=Vin*(1e^(t/T))
where t is timer, and T is the RC time constant. As t approaches infinity,
the we have V(t)=Vin(10). 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}> 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:
>
>> 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@140824
by
talgo

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}>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 baseemitter 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 baseemitter 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 baseemitter junction limits the capacitor voltage to 700mV.
>
>Another replay indicated that the voltage as a function of time was
>
>V(t)=Vin*(1e^(t/T))
>
>where t is timer, and T is the RC time constant. As t approaches infinity,
>the we have V(t)=Vin(10). 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
>
>
>
>
>
>
>>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:
>>
>>
>>
>>>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@162311
by
William Chops Westfield
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*(1e^(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
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
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) 7429014. #1 PIC
consultant in 2004 program year. http://www.embedinc.com/products
2005\11\09@090433
by
Bill & Pookie
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}
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