Post by thread:[EE] capcitor charge time

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

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 | | | --

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}

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.

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}

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

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 > 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

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}> 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}

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 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 > > > > > > >>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}