Capacitor Discharge Equation

[mull982]

Can someone please provide me and explain to me an equation for capacitor discharge, or how to calculate the time it will take a capacitor to discharge given a particular load.

Can this same equation also be used for determining the time it takes for a capacitor to change from one voltage level to another when connected to a given load? For example lets say a capacitor was charged to 600V then how long would it take for the capacitor to discharge to 500V with a given load?

 

[bob]

check here http://www.tpub.com/neets/book2/3d.htm

 

[mull982]

I've also heard that capacitors or capacitive devices can be damaging on switching devices because there can be a great deal of current to charge a capacitor only limited by the system source impedance.

Is the large amount of charging current for a capacitor due to the fact that there is no impedance in the capacitor while it is charging?

 

[mivey]

I've also heard that capacitors or capacitive devices can be damaging on switching devices because there can be a great deal of current to charge a capacitor only limited by the system source impedance.

Is the large amount of charging current for a capacitor due to the fact that there is no impedance in the capacitor while it is charging?

Don't mix up the terms impedance and resistance.

The capacitor initially looks like a short circuit with very very little resistance.

 

[steve66]

Yes, like Mivey said, if the capacitor is initally discharged, the capacitor initially looks like a short circuit with very little resistance. The only real limiting parameters are the source impedence, and the small series resistance present in the diode bridge. The current spike will be larger if you just happen to apply power during an AC voltage peak.

On many smaller power supplies, its fairly common to include something like a 0.1 ohm resistor in series with the diodes to limit the current.

High current charging spikes can also damage the capacitors. The thin insulating layer produced in electrolytic caps can easily be damaged by these spikes with the result being a shorted capacitor.

 

[gar]

101205-1253 EST

mull982:

You should not think of a capacitor as a resistance or reactance when viewing it from a transient point of view.

For an ideal capacitor, no internal resistance or inductance, view it as a device defined by v = q/C.

For a simple series RC circuit with a switch and DC voltage source write the differential equation for the circuit, solve for i, apply any desired initial conditions and see what happens.

R*(dq/dt) + (1/C)*q = V

Ultimately you can determine for an initial charge of 0 on the capacitor that
q = -V*C*e^(-t/RC) + V*C
i = (V/R)*e^(-t/RC)

Thus, the source voltage divided by the loop resistance defines the initial current. More complex if you add inductance into the series circuit.

Were you to try to define the capacitor as some sort of time varying resistor, then what happens when you open the circuit and no current flows? The voltage across the device that is a capacitor but you want to call a time varying resistor drops to 0. Whereas actually a voltage remains on the capacitor when the switched is opened. The voltage on the capacitor remains at the voltage determined by the amount of charge on the capacitor at the time of opening the switch.

I do not think it is useful to try to view the capacitor as a resistor except at only the condition that zero charge is on the capacitor. Then just for convenience you call it a short.

You really need to think in terms of the differential equation and the solutions and what happens to the current vs time.

For the discharge of of a capacitor in a series RC circuit you are really dealing with the same differential equation except that there is no source voltage, it is zero, and there is an initial charge on the capacitor.

So the resulting equations become:
R*(dq/dt) + (1/C)*q = 0

Ultimately you can determine for an initial charge of V*C on the capacitor that
q = V*C*e^(-t/RC)
i = -(V/R)*e^(-t/RC)
Now the current flows out instead of in.


Now consider a 100 mfd capacitor with 400 V on it in a radio receiver when you turn the power off. Suppose there is the equivalent of 1000 ohms shunting the capacitor. The voltage on the capacitor will be
v = Vint*e^(-t/RC) = 400*e^(-t/0.1) where t is in seconds.
After 0.1 seconds the voltage is 400*0.368 = 147 V, after 1 second v = 0.02 . The value RC is typically called one time constant.

In an actual tube type radio the filaments cool quickly and the load on the power supply capacitors is actually a time varying load. Thus, the capacitor voltage may not drop this quickly.

After the first time constant the voltage v is 0.367879441*Vint.
After the second time constant it is v = 0.367879441*0.367879441*Vint = about 0.135*Vint.
This continues forever for as many time constants as you want.
After 6 time constants the voltage is down to about 0.0025 of the initial voltage. Safe in many applications.

.

 

[ELA]

101205-1253 EST

mull982:

You should not think of a capacitor as a resistance or reactance when viewing it from a transient point of view.

I do not think it is useful to try to view the capacitor as a resistor except at only the condition that zero charge is on the capacitor. Then just for convenience you call it a short.

You really need to think in terms of the differential equation and the solutions and what happens to the current vs time.

.... and a lot of extras

.

Gar,
All very prescriptive but why not just provide the equation requested by mul982 in his second paragraph?
Givens = (capacitance, resistance, V1 and V2) find t . Now that would be impressive yet helpful.

 

[mivey]

Gar,
All very prescriptive but why not just provide the equation requested by mul982 in his second paragraph?

Because if you teach someone to fish...

 

[gar]

101205-1420 EST

ELA:

Too many people just want an answer to a specific set of conditions. So what happens with a new set of conditions? They ask the same question again. If they know the theory, then possibly they can answer a similar question on their own in the future.

This is like the story about the hungry person. If you give them a fish, then they can live for a day before another handout. But if you teach them to fish, then until the fish become extinct, they can feed themselves forever without a handout.

Discharging the voltage on a capacitor will never theoretically reach zero. But after 10 time constants for most conditions it could be considered 0. Fewer time constants if the starting voltage is in the 100 V range and a couple volts is adequate for an acceptable end point. Note: a voltage of 3 V on 1000 mfd capacitor placed across a 1N4148 would probably destroy the 1N4148. In this situation this would not be considered an acceptable value for the discharged condition.

From 100 V a discharge time of 4 time constants reduces the voltage to 1.83 V.

I did provide the equation to determine the capacitor voltage after a period of time.
i = -(V/R)*e^(-t/RC)
So when you multiply i by R the result is
v = V*e^(-t/RC)
I also provided an actual example of the voltage vs time for discharge of a 100 mfd capacitor with an initial 400 V charge shunted by 1000 ohms.

Post #2 by bob provided useful information although I do not believe it provided the exponential equation. With the given graphs of the exponential shape of the curve one can get a reasonable answer of discharge time.

I chose to provide the basic information on how to derive the curve. Most on this forum will not understand how to solve the differential equation. But it illustrates summing the of the voltage around a closed loop. Indirectly it illustrates the relationship of current to charge.
So i = dq/dt.
In simple terms current is the rate of change of charge vs time. A constant current is thus a constant rate of change of charge, constant velocity. This is not for you ELA. Just expanding the discussion for others. Charge is like miles driven and current is like MPH, speed.

If the reader understands that the sum of the voltages around a closed loop is zero, then what I have in my basic differential equation is the voltage across the resistor, R*(dq/dt), and the voltage across the capacitor, q/C, added together equals the applied DC voltage. Capital V implied an invariant voltage. The equation as written was meant to include 0 and any positive number for time. (Note: dq/dt is substituted for i so the differential equation is all based on q rather than two variables. One of the tricks you use in solving equations. This in turn forced the equation to become a differential equation.)

.

 

[gar]

101205-1848 EST

Suppose we use an RC circuit to perform a timing function, time delay for example, and an FET is used as the threshold detector. Make it a 2N7000 FET. See http://www.fairchildsemi.com/ds/2N/2N7000.pdf . The gate threshold voltage is approximately 2.1 V at 1 MA drain current. Figure 6 shows the threshold variation with temperature.

The 2N7000 source is connected to the power supply common, you may want to call it ground. Its drain is connected to +5 thru a 5 K resistor, 1 MA when turned on. This resistor is for the circuit to work but otherwise does not relate to the discussion.

The connection point between a timing resistor and capacitor goes to the 2N7000 gate. First, connect the free end of the timing capacitor to common and the free end of the resistor to a normally open switch and then the other end of the switch to +5 V.

Make the resistor 1 megohm and the capacitor 1 mfd. Assuming zero charge on the capacitor initially, then at -50, 25, 150 deg C, what are the delays after the switch is closed until the drain voltage drops to near 0 volts?

At 25 deg C start with a -10 V charge on the capacitor, then what is the time delay?

Another way to control the start time is to connect the timing resistor directly to +5, and move the switch so that initially either 0, or -10 V is applied to the capacitor before the timing interval is initiated.

.

 

[ELA]

Can someone please provide me and explain to me an equation for capacitor discharge, or how to calculate the time it will take a capacitor to discharge given a particular load.

Can this same equation also be used for determining the time it takes for a capacitor to change from one voltage level to another when connected to a given load? For example lets say a capacitor was charged to 600V then how long would it take for the capacitor to discharge to 500V with a given load?

Mull982,

Here is a solution for the time:

t = RC * ln (Vi/Vf)
I do not know if you feel comfortable with natural logs but you can plug it into excel and let it do the calculation for you.

Because if you teach someone to fish...

101205-1420 EST

This is like the story about the hungry person. If you give them a fish, then they can live for a day before another handout. But if you teach them to fish, then until the fish become extinct, they can feed themselves forever without a handout.
.....
Most on this forum will not understand how to solve the differential equation.
.

???
I smell a carp ...

 

[gar]

101206-1321 EST

This post is not meant to imply anything wrong with ELA's equation for discharge. But to expand on it.

A form of the equation for either charging or discharging is

t = -R*C*ln(1 - dv/DV)

where
dv is the absolute value of the change of capacitor voltage from the initial voltage to the voltage point of interest
DV is the absolute value of the change of capacitor voltage from the initial voltage to the final voltage at t = infinity

So for the 600 to 500 example we have

t = -R*C*ln(1 - (100/600) ) = (-R*C)*(-0.18232---)

ELA's equation gives you

t = R*C*ln(600/500) = R*C*(0.18232---)

Thus, we agree for this comparison. Note: (-X)*(-X) = +X.


But if we use the intuitive meaning of Vi and Vf and apply these to a charging curve the results are a problem.

Next consider the charging curve. Let Vi be 0 and Vf = 100 V. Now ELA's equation blows up. But, nothing wrong with his equation for discharge. The problem arises from the choice of variables and their implied meaning.

If you make Vi = my DV, and Vf = (my DV - my dv), then ELA's equation works for both charge and discharge. As my dv approaches 0, then Vi/Vf approaches 1 and ln Vi/Vf or ln (DV/(DV-dv)) approaches 0. But these associations with Vi and Vf are not intuitive.



A major point to understand is that the charging and discharging curves for an RC circuit are essentially an inversion of each other and may have a different reference starting voltage and final steady-state voltage. But fundamentally the same basic differential equation applies.

.

 

[jghrist]

Give a man a fish, he'll eat for a day. Teach a man to fish and he'll be out on the #@*& boat every day drinking beer.

 

[mivey]

???
I smell a carp ...

Sorry I did not get back to you sooner as you may have already eaten, but fresh fish for consumption usually should not have a fishy smell as it is starting to spoil.

Bright or pink gills is an indicator of fresh fish. Dark or grey gills can be an indicator of an old catch (also a good flag for fishing tournaments).

The fish meat should be firm and not looked dried out.

FWIW, most of the other seafoods will not have a strong smell either when fresh.

Hope that helps.

 

[mivey]

Give a man a fish, he'll eat for a day. Teach a man to fish and he'll be out on the #@*& boat every day drinking beer.

I've chickened out since it turned cold, but you might could talk me into it.

 

[ELA]

Sorry I did not get back to you sooner as you may have already eaten, but fresh fish for consumption usually should not have a fishy smell as it is starting to spoil.

Bright or pink gills is an indicator of fresh fish. Dark or grey gills can be an indicator of an old catch (also a good flag for fishing tournaments).

The fish meat should be firm and not looked dried out.

FWIW, most of the other seafoods will not have a strong smell either when fresh.

Hope that helps.

Not in the least but good for incrementing one's post count I guess :roll:

101206-1321 EST

This post is not meant to imply anything wrong with ELA's equation for discharge. But to expand on it.


Next consider the charging curve. Let Vi be 0 and Vf = 100 V. Now ELA's equation blows up.
.

Not "my" equation but a discharge equation.
You seem to have ignored my point. I was giving an exact answer for a particular set of conditions. ...Trying to be concise as many here may not be interested in being lectured to -or have the time to fish. Understood that some do...

The question asked was discharging -not charging and thus it would be incorrect to apply the equation I presented to that set of conditions.

 

[mivey]

Not in the least but good for incrementing one's post count I guess :roll:

Yeah! One more. :grin:

 

[mivey]

Trying to be concise as many here may not be interested in being lectured to -or have the time to fish

If you want to be responsive to the OP, some explanation is in order. Gar only gave the OP what he asked for:

Can someone please provide me and explain to me an equation for capacitor discharge, or how to calculate the time it will take a capacitor to discharge given a particular load.

 

[skeshesh]

Behavior of Capacitive elements in the time domain is usually covered in the first circuit analysis course you could take at any school. Alternatively I suggest you grab a PDF of a circuit analysis book and read through the first few chapters. It's really not difficult but you need to see several cases and also do the calculations by hand to understand. There's been some decent explanations provided here but I think you'll be better off spending some time wrapping your head around it. As gar suggested, you may need some higher math to understand the depths of the material but you'll get a good enough clue reading the textbooks.

 

[gar]

101207-2143 EST

It should be noted that the original post ask about both charge and discharge.

I will try to add some more information that might assist one without ln or e tables, slide rules, or calculators.

As a rough approximation the change of voltage or current in one time constant is 63% of the difference between the final steady state and the initial value. This continues to repeat forever for each new time constant period. Within this 63% range you can approximate the value based on the assumption of a linear curve. Not real close, but if you want to estimate an RC value for a single shot multivibrator it should do. You can evaluate the maximum error if you desire.

.