RC time, using a Stop Watch, RC time is longer in time is the capacitor bad?

You want to measure the current passing through the capacitor when the voltage across it is X volts.

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Yes I get this, but you said:

The cap will charge to whatever voltage the supply has, regardless of leakage

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Which means the Time difference it takes the cap to charge will be the SAME regardless of leakage or not

*A Leaky Cap will take longer to charge up to the supply voltage, which causes the Time to increase higher*

Where did I say anything about time? I said that the cap will charge to the voltage of the supply, and also that this voltage doesn´t depend on the leakage current.
Nothing about time, time depends on capacitance, leakage and charging current. Charging current through a resistor depends on supply voltage. Leakage current depends on voltage across the cap.

Leakage current depends on voltage across the cap.

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So the Leakage current depends on the voltage across the cap NOT the supply voltage right?

The Charging current through a cap depends on the supply voltage or voltage across the cap?

I'm not sure if the charging current is the voltage from the supply or across the cap

Time depends on capacitance, leakage and charging current

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OK, Thanks for telling me this

So the Leakage current depends on the voltage across the cap NOT the supply voltage right? Yes, but usually when charged the cap has the same voltage as the supply has.

The Charging current through a cap depends on the supply voltage or voltage across the cap? It depends on both.

I'm not sure if the charging current is the voltage from the supply or across the cap Charging current is current, not voltage. In an RC circuit the charged current depends on the R, C, the supply voltage and the instanteneous voltage across the cap. More specifically Icap=(Vsupply-Vcap)/R at any given point.

Ok thanks

1.) There is Other Switches when TURN On takes 10 seconds to activate an Op-amp to switch it state, What is this timing RC networks called? they do this delay timing for 10 seconds so the input of the Op-amp comparators don't get false triggering's from noise or cause they want the Switch Turn ON state to stay on for 10 seconds before the Op-amp switches states or polarity on its output, What is this called when a designer puts RC delays charging like this?

2.) Also what I do at work is I have to measure the TURN ON and OFF points of Lights by using a Pot. I turn the Pot very slowly until the light turns ON or OFF and have to measure the voltage and current of how much voltage and current is I guess going into the Op-amp Comparator circuit.
What is this called what I am doing? I don't know the electronic name for what I'm doing for this

The Pot is either hooked up to an Op amp Comparator or to TTL or CMOS Logic GATES

Hello again,

The leakage current acts like a resistor in parallel to the cap. That means the charge time is shorter, but the level of voltage that can be reached across the capacitor is lower too so that requires attention.

For example, if you charge a 1uf cap up with a 1 Megohm resistor, the voltage will reach 63.2 percent of the supply voltage in one time constant, and one time constant is R*C which in this example is 1 second.
HOWEVER, if there is also a 1 Megohm resistor in parallel with the cap (the leakage) then it can never reach 63.2 percent of the supply voltage because of the voltage divider action. So there are two things at work when there is leakage current:
1. The charge time
2. The maximum voltage it can charge to

Now if the voltage is 10 volts and the leakage current is 10ua, that is equivalent to 1 Megohm so the circuit would behave similar to that above.

A less extreme example would be a leakage current of 1ua, where the parallel resistance would look like 10 Megohms so the time to charge to the original 6.32 volts would go from 1 second to about 0.91 seconds, a loss of about 10 percent. That isnt much so if the leakage is small then there's not too much difference.

Thanks Mr.Al

The leakage current acts like a resistor in parallel to the cap. That means the charge time is shorter

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Does this apply to all RC networks that are configured in these 3 types

RC timing networks , 3 types they use
1.) Resistor in series, capacitor is grounded
2.) Capacitor in series, Resistor is grounded
3.) Resistor is tied to VCC, capacitor is grounded, ( Not sure what this RC network is called tho) do you? The capacitor gets charged through the VCC through the resistor

There is an Op amp before the RC timing network and an Op amp After the RC timing network

Op amp#1 ----> RC timing network -----> Op amp#2

Also In an intergator circuit , if the capacitor in in a op amp intergator circuit has leakage current would the charge time be shorter?

Since the Charge time is shorts = leakage current

What will cause the Charge TIME to be Longer? since it's not leakage?

Please post a diagram of the three combinations along with the opamps, because I don´t see what purpose would number three serve.
Charge time will be longer when the capacitance differs, or the current differs, or the threshold differs, or the charging current differs.

Billy Mayo said:

Thanks Mr.Al



Does this apply to all RC networks that are configured in these 3 types

RC timing networks , 3 types they use
1.) Resistor in series, capacitor is grounded
2.) Capacitor in series, Resistor is grounded
3.) Resistor is tied to VCC, capacitor is grounded, ( Not sure what this RC network is called tho) do you? The capacitor gets charged through the VCC through the resistor

There is an Op amp before the RC timing network and an Op amp After the RC timing network

Op amp#1 ----> RC timing network -----> Op amp#2

Also In an intergator circuit , if the capacitor in in a op amp intergator circuit has leakage current would the charge time be shorter?

Since the Charge time is shorts = leakage current

What will cause the Charge TIME to be Longer? since it's not leakage?

Click to expand...

Hello again,

Well as i said, the time constant becomes shorter but the voltage that the cap charges to also changes so the operation of the circuit changes not only because of the time constant but also because the voltage changes. That means if the voltage trip point is fixed then the circuit could actually take longer to trip meaning the timing period will actually be longer even though the time constant decreased.

This can happen because with leakage current it actually takes longer to reach the SAME voltage level as it did without leakage current.

For example, with a 10k resistor and 1uf cap and a 1v charging supply voltage, it takes 10ms to reach 0.632 volts, which is one time constant which is t=R*C. So if your circuit is set to trip at 0.632 volts it will trip in 10ms. However, with an equivalent leakage current resistance of 50k in parallel with the cap, it takes nearly 12ms to reach that same level of 0.632 volts. So here we see an increase in the timing period even though the time constant came down to 8.33ms.

In the integrator we see a similar action. With a 10k input resistor and 1uf feedback cap, we see a time of about 6.32ms to reach 0.632 volts (with a -1v input) but with a 10k equivalent leakage current we see the time increase to 10ms to reach 0.632 volts which is again longer.

So in these cases the leakage actually caused an increase in the timing period.

For other circuits where the cap is still in series with the resistor, the action is the same when viewing the voltage ACROSS the cap, but the voltage referenced to ground may be different. For the cap connected to Vcc and the resistor to ground the voltage at the junction as measured to ground will appear to decrease with time.

It would be a good idea to post your exact circuit configuration so we can see exactly what you have there. We can determine what might be happening, although there can be several possibilities.

Caps tend to decrease in capacitance over time, and their ESR goes up. The ESR is another subject of interest as that can change the timing period as well. With much higher ESR the time to reach a certain voltage would decrease because the voltage across the ESR part of the cap appears almost immediately after the supply voltage is applied. How much difference it makes depends on how much the ESR has changed.

Thanks Mr. AL

That means if the voltage trip point is fixed then the circuit could actually take longer to trip meaning the timing period will actually be longer even though the time constant decreased.

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What is the timing period measured by in this case? is the timing period how long it takes the capacitor to charge to the fixed trip point?

RC has a time constant but I can't find in my electronic book about an RC timing period

With much higher ESR the time to reach a certain voltage would decrease because the voltage across the ESR part of the cap appears almost immediately after the supply voltage is applied

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If a capacitor has High ESR, the Charging time would get shorter or longer? and the discharge time would get short or longer?

Most of the voltage would be across the ESR?

Hi again,


Some very good questions.

The timing period is the time it takes the capacitor voltage to reach the trip voltage level from a known starting voltage which is usually zero. So it takes longer for the cap to reach a trip voltage of 2v than it does to reach 1.5v for example. So increasing or decreasing the trip voltage can change the time period too.

The RC time constant is just the resistance times the capacitance: T=R*C.

The ESR of a capacitor is likened to another smallish resistor in series with the cap, but this differs from a resistor you put in series with the cap on purpose for timing because the ESR is internal to the cap. What happens is that when the cap starts to charge the ESR drops some voltage so to the external world it looks just like a voltage across the cap.
In a timing application this would mean that as soon as the cap starts to charge the voltage jumps up to some level like 0.1v instead of starting at 0v. That means it reaches the trip point sooner and so the timing period is shortened.
If you have a large resistor like 10k as the charge resistor however it is unlikely that the ESR will affect it too much, but it's always a possibility.
During discharge, the ESR would make it appear to discharge faster too because the lower trip point would be reached sooner as well.
"Most" of the voltage should not be across the ESR because the ESR should be small, unless it increased a lot with aging.

Attached you'll find a graph of a capacitor charging through a resistor excited with a voltage source of 1v DC. The time constant of the circuit is 1 second because R*C is equal to 1000 times 1000uf which comes out to 1.
The first time constant tells us the time it takes the cap to charge to 63.2 percent of the maximum voltage and because the max is 1v during the first time constant the cap charges up to 0.632 volts here. That number is directly scalable so if the max voltage was 2 volts it would have charged to 2 times 0.632 volts (1.264v) after the first time constant.
The blue curve is the actual cap voltage, the red line from zero to the blue curve shows the approximation we often use to approximate the cap charge during the first time constant.

You can see that if you follow the blue line, as time goes on the voltage gets higher and higher and so a higher trip point means a longer time period.

The green line is just the 0.632 volt line which is 63.2 percent of the max voltage and the max voltage here is just 1 volt.

The RC time constant is just the resistance times the capacitance: T=R*C.

Yes I have this in my electronic Book , But I can't find the RC time period formula of how long it takes an RC time period to reach a Voltage Threshold/Trip Point to an IC input Pin, do you know the formula Mr. Al?

In a timing application this would mean that as soon as the cap starts to charge the voltage jumps up to some level like 0.1v instead of starting at 0v. That means it reaches the trip point sooner and so the timing period is shortened.

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High ESR capacitor will have a shorter charge time & shorter timing period and reach the Voltage threshold hold sooner
High ESR capacitor will have a shorter discharge time and shorter timing period

The ESR has a voltage drop across the capacitor that causes the shorter charge time, shorter discharge time and timing period to reach a voltage threshold/ trip point

Billy Mayo said:

The RC time constant is just the resistance times the capacitance: T=R*C.

Yes I have this in my electronic Book , But I can't find the RC time period formula of how long it takes an RC time period to reach a Voltage Threshold/Trip Point to an IC input Pin, do you know the formula Mr. Al?

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Hi,

Note that i included a graph of the cap charging in post #51 i think after you already read that post so you can look at that for more information too.

The formula for the cap charging from zero volts is:
Vc=Vmax*(1-e^(-t/RC))

and solving this for t we get:
t = -RC*ln(1-Vc/Vmax)

where
t is the time,
Vc is the capacitor voltage after time t,
Vmax is the max voltage usually the source voltage,
RC is the RC time constant R*C,
"ln" is the natural log function.

For the example i plotted in post #51:
Vmax=1
RC=1
Vc is plotted in blue.

RC is the time constant , is time constant and time period the same?

I thought the time constant was the TIME it took the capacitor to charge up to 100%

Time Period would the TIME the capacitors voltage would be at 0% to 100%

So they are the same right? Time constant and Time period

Hi,

Well, no, the time constant again is R*C and that is the time it takes the capacitor to charge from 0 to 63.2 percent of the max voltage. 63.2 percent is not 100 percent.
The capacitor is said to have charged to very close to the max voltage after 5 time constants which is 5*R*C. After 5 time constants the voltage reaches about 99.3 percent of the max charge voltage. After 6 time constants the voltage reaches to about 99.8 percent of max, which now is very close to 100 percent. After 10 time constants it is up to 99.995 percent of max which for most purposes could be called 100 percent.

The timer period is the time it takes the capacitor to charge from usually 0 to whatever the trip point is set at, which can be anything. So the trip point voltage can be 1v, 2v, 3v, or anything less than the max charging voltage.

If you look at the graph i posted you'll see how this works a little bit better.

So the timer period and the time constant are not the same in general, but in the case that the trip point is adjusted to 63.2 percent of the max charge voltage then they would be the same, but that's only for that one specific adjustment.

Look at the graph i posted and see if that makes more sense.

So the timer period and the time constant are not the same in general, but in the case that the trip point is adjusted to 63.2 percent of the max charge voltage then they would be the same, but that's only for that one specific adjustment.

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Yes , I get it , thats what i was thinking too

The timer period is the time it takes the capacitor to charge from usually 0 to whatever the trip point is set at

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Yes, I get it

the time constant again is R*C and that is the time it takes the capacitor to charge from 0 to 63.2 percent of the max voltage. 63.2 percent is not 100 percent.

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Why would anyone or a designer what to know the time constant of an RC network? What good is to know the time constant of an RC network when the time period is what you really need to know of an RC network

I know T=R*C , once you got the time constant , you can get the frequency value from it

To full charge the cap . it takes 10 time constants

1.) Cap with leakage, Time constant is shorter, time period is shorter, charging time is short, discharge time is short
2.) Cap with HIGH ESR, Time constant is shorter, time period is shorter, charging time is short, discharge time is short

At work

They use 741 op amps as a switch or comparators , but they use RC networks on the input pins to cause delay or slow the switching response

Any reasons why?

RC networks
1.) Resistor in series and cap in parallel grounded goes to input pin of 741 op amp ( as a switch or comparator )
2.) Cap in series and resistor in parallel goes to input pin of 741 op amp ( as a switch or comparator )
3.) Resistor tied to VCC and cap is parallel grounded goes to input pin of 741 op amp ( as a switch or comparator )

Hi,

Typical usage:
1. Output of 741 goes high after a delay.
2. Output of 741 goes high right away, then later goes low again (pulse output, one shot).
3. Output of 741 goes high (or low) after a delay after the circuit is turned on (power on reset).

The RC time constant plays into circuit analysis in a number of ways. By knowing the RC time constant you know that the cap charges near to 100 percent after 5 time constants for one thing. But there are many other ideas that come from the time constant. Knowing the time constant allows use to use the formula 1-e^(-t/T) where again you see the time constant T is just R*C. In an AC application it can also be used to calculate the cutoff frequency of the circuit.

In more complicated circuits the time constant is not always R*C but in the simple one or two resistor and one cap circuit it is easy to calculate so we try to take advantage of that and that in turn allows us to calculate other things almost as easy.

the time constant T is just R*C. In an AC application it can also be used to calculate the cutoff frequency of the circuit.

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Yes I have used this one if my electronics class

But what other things can you use the Time Constant for? any other explains u know about?