ELA
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Capacitor Discharge Equation
- Thread starter mull982
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gar
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- Ann Arbor, Michigan
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- EE
101208-1036 EST
ELA:
This appeared to me as a rather general question. Load is not defined and change is not defined.
But read in a narrower sense you are correct that his intent could have applied to a discharge curve only.
.
ELA:
This appeared to me as a rather general question. Load is not defined and change is not defined.
But read in a narrower sense you are correct that his intent could have applied to a discharge curve only.
.
gar
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- Ann Arbor, Michigan
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- EE
101208-2123 EST
What is the slope of the charging voltage curve relative to time at time 0 when you know the change in voltage from the starting point voltage to the final voltage at time infinity, and you know the RC time constant.
The charging voltage curve is
v = V*( 1-e^(t/RC) )
and the slope is
dv/dt = V/RC
Suppose RC = 10^6*10^-6 = 1, the time constant is 1 second for a 1 megohm resistor and a 1 mfd capacitor.
Make a very small change in time in the v equation, t=0.001 seconds, then the
change in v = V*(0 + 1 - 0.999) = 0.001*V volts change in 0.001 seconds which is 100 V/second and that is the same as you get from the
dv/dt equation.
Suppose you use the 1 second RC time constant circuit with a source voltage of 100 V, then when you reach a time of 1 time constant what is the slope of the curve? The remaining change in voltage is 100*e^-1 = 100*0.367879441. So the slope at this point is 100*0.3678---- = 36.78--- volts per second.
Average these two slopes (100+36.78)/2 = 100*0.683939721 or about 68.39 V/second. How does this slope compare with the slope at a time of 1/2 time constant.
At a time of 1/2 time constant the remaining voltage to change is 100 - 100 + 100*e^-0.5 = 100* 0.606530660 or about 60.65 volts/second. So the difference is about 7.74 V or 13 % above the actual slope.
For some purposes still useful. But with modern calculators it is easier to get the actual value.
These comments are partly to help you get an intuitive feeling of this type of curve.
For those unfamiliar with the term slope it is the rate of change of a straight line tangent to a curve at a point. At any min or max of a curve the slope is zero. Thus, at either the +/- peak of a sine wave the slope is zero.
I could have made mistakes above. If so, comment.
.
What is the slope of the charging voltage curve relative to time at time 0 when you know the change in voltage from the starting point voltage to the final voltage at time infinity, and you know the RC time constant.
The charging voltage curve is
v = V*( 1-e^(t/RC) )
and the slope is
dv/dt = V/RC
Suppose RC = 10^6*10^-6 = 1, the time constant is 1 second for a 1 megohm resistor and a 1 mfd capacitor.
Make a very small change in time in the v equation, t=0.001 seconds, then the
change in v = V*(0 + 1 - 0.999) = 0.001*V volts change in 0.001 seconds which is 100 V/second and that is the same as you get from the
dv/dt equation.
Suppose you use the 1 second RC time constant circuit with a source voltage of 100 V, then when you reach a time of 1 time constant what is the slope of the curve? The remaining change in voltage is 100*e^-1 = 100*0.367879441. So the slope at this point is 100*0.3678---- = 36.78--- volts per second.
Average these two slopes (100+36.78)/2 = 100*0.683939721 or about 68.39 V/second. How does this slope compare with the slope at a time of 1/2 time constant.
At a time of 1/2 time constant the remaining voltage to change is 100 - 100 + 100*e^-0.5 = 100* 0.606530660 or about 60.65 volts/second. So the difference is about 7.74 V or 13 % above the actual slope.
For some purposes still useful. But with modern calculators it is easier to get the actual value.
These comments are partly to help you get an intuitive feeling of this type of curve.
For those unfamiliar with the term slope it is the rate of change of a straight line tangent to a curve at a point. At any min or max of a curve the slope is zero. Thus, at either the +/- peak of a sine wave the slope is zero.
I could have made mistakes above. If so, comment.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101208-2331 EST
rattus:
It sure should be a - in the exponent. That is why I need proofreaders. Common mistakes like that I just can not catch in many cases. Thanks.
.
rattus:
It sure should be a - in the exponent. That is why I need proofreaders. Common mistakes like that I just can not catch in many cases. Thanks.
.
Open Neutral
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- Engineer
To really befuddle the OP; capacitors are not perfect.
They DO have an ESR, equivalent series resistance.
The ESR only really gets to be an issue in edge cases, but one of those is designing switch mode power supplies such as in computers. A SMPS takes the 120/240 60Hz, rectifies it into DC, charges a cap, has a higher frequency inverter, then transformer, rectifier, filter and regulator.
It does all that nonsense so that transformer is at {say..} 100 Khz, not 60 hz, and is a small fraction of the size and co$t. But the caps must have low ESR at the switching freq, or the design engineer shall soon go bonkers trying to keep it stable. (When you consider that such supplies are built in 100,000-1E6 piece production runs, function for years, and cost a pittance.... they are a modern miracle.)
They DO have an ESR, equivalent series resistance.
The ESR only really gets to be an issue in edge cases, but one of those is designing switch mode power supplies such as in computers. A SMPS takes the 120/240 60Hz, rectifies it into DC, charges a cap, has a higher frequency inverter, then transformer, rectifier, filter and regulator.
It does all that nonsense so that transformer is at {say..} 100 Khz, not 60 hz, and is a small fraction of the size and co$t. But the caps must have low ESR at the switching freq, or the design engineer shall soon go bonkers trying to keep it stable. (When you consider that such supplies are built in 100,000-1E6 piece production runs, function for years, and cost a pittance.... they are a modern miracle.)
Gar
The information and responses that you provided are exactly what I am looking for. I've been busy and have'nt had a chance to review your responses in depth however I will review them to see if I grasp everything as well as see if I have any follow up questions.
Thanks for the detailed explanation.
The information and responses that you provided are exactly what I am looking for. I've been busy and have'nt had a chance to review your responses in depth however I will review them to see if I grasp everything as well as see if I have any follow up questions.
Thanks for the detailed explanation.
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