# Question about RC charging time

#### Knid

Joined Aug 28, 2020
20
Hi,

I have a circuit below. I am trying to find "How long does it take to get voltage at point A = 2.5 V after closing the switch? Both capacitors are completely uncharged at t = 0 Can anyone help to guide how to do hand calculation? It is DC voltage of 10 V.

Thank you!!!

#### Papabravo

Joined Feb 24, 2006
14,378
Where is point "A" on your diagram?

#### Knid

Joined Aug 28, 2020
20
So sorry. I for got to put point A in.

#### SamR

Joined Mar 19, 2019
2,529
Have you ever heard of an RC Time Constant?

#### Knid

Joined Aug 28, 2020
20
Have you ever heard of an RC Time Constant?
Yes, I am now wondering how do I apply in this equation.
Vc = Vs(1-e^(-t/RC))

#### MrChips

Joined Oct 2, 2009
21,815
Is this homework?
How exact an answer do you want?
By hand calculation I get 313μs.

#### Knid

Joined Aug 28, 2020
20
Is this homework?
How exact an answer do you want?
By hand calculation I get 313μs.
Nope it’s not a home work. It is circuit to measure capacitance(10470pF). Originally, there was no R and C(470pF). Some one add them later for there filter purpose. Point A is connectd to comparator to trig my FPGA to stop counting clock and back calculate to C. Could you guide me how did you calculate it? I am kinda rusty now.
Thank you!!!

#### MrChips

Joined Oct 2, 2009
21,815
If you want exact solutions you will have to solve two differential equations.

1) I made first order approximations by noticing that R1 is about 100 times larger than R2 and R3.

2) By assuming R1 is relatively large compared to C1 and C2, I made a second assumption that the voltage rise is relatively linear.

3) Hence I took the average of the charge current to take point-A from 0V to 2.5V.

The only three formulae I needed were

Ohm's Law I = V / R
charge Q on a capacitor Q = C x V
charge Q = I x t

#### MrChips

Joined Oct 2, 2009
21,815
Why not remove R2, R3, and C1 and make life simple?

#### Knid

Joined Aug 28, 2020
20
I couldn’t now. Someone add those in the circuit, not knowing that they will impact my circuit. What I can do now is to find its transfer function and it need to be accurate. Thank you.

#### WBahn

Joined Mar 31, 2012
26,139
I couldn’t now. Someone add those in the circuit, not knowing that they will impact my circuit. What I can do now is to find its transfer function and it need to be accurate. Thank you.
What does "it need to be accurate" mean?

How accurate is accurate enough?

What are the limits imposed by your other components? How accurate and precise is your 10 V supply? What are the tolerances on all of the components?

#### WBahn

Joined Mar 31, 2012
26,139
Nope it’s not a home work. It is circuit to measure capacitance(10470pF). Originally, there was no R and C(470pF). Some one add them later for there filter purpose. Point A is connectd to comparator to trig my FPGA to stop counting clock and back calculate to C. Could you guide me how did you calculate it? I am kinda rusty now.
Thank you!!!
If it's a circuit to measure the capacitance, then why do you want to know the specific time it takes Point A to reach 2.5 V for that specific value of capacitance? How does that help you measure that value when you don't know it?

#### crutschow

Joined Mar 14, 2008
25,367
Sure sounds like homework

#### MrChips

Joined Oct 2, 2009
21,815
In order to avoid confusion, relabel C1 to C2, and C2 to C3.

By applying KCL,
(1) i1 = i2 + i3

By apply KVL
(2) 10 = i1R1 + i2R2 + v2
(3) 10 = i1R1 + i3R3 + v3

where v2 and v3 are the voltages on capacitor C2 and C3 respectively.

Are you prepared to solve the above where v2 and v3 are differential equations?

Take the easy way out. Eliminate R2 and lump C2 with C3. Now that will simplify things.

#### Knid

Joined Aug 28, 2020
20
Sorry to make it looks like homework. Well, it is a test circuit to test my product. I could not share another part of the circuit since it is confidential. The whole circuit has so many features, dealing with high frequency stuffs. There is a part of a circuit used to measure capacitance of an element and this is an equivalent circuit. R1 is used as a bleeder resistance to protect the product from damaging by ESD. R2, C1 was tapped at the center to used as a low-pass filter to help eliminate switching regulator noise from other circuit. Point A in the circuit will connect to comparator, which has a threshold at 2.5 V. The output of comparator will trig FPGA to stop counting clock, while the switch is controlled by FPGA and start counting. After I get numbers of clock count. I can convert that back to capacitance by using RC time constant. I am kinda away from this kind of circuit for so many years. I know that I have to apply Vc=Vs(1-e^(-t/RC)) at some point, but can remember clearly how to calculate time constant from multiple RC circuit like this. That's why I am asking. And, yes, I will fix this in the next design.

#### MrChips

Joined Oct 2, 2009
21,815

How much current is flowing through R1?

#### Knid

Joined Aug 28, 2020
20

How much current is flowing through R1?
It will be:
i1 = i2 + i3
i1 = 10[e^(-t/(101.5k)(470p))] + 10[e^(-t/(101.5k)(10470p))]

#### MrChips

Joined Oct 2, 2009
21,815
On the left side you have amps. On the right side you have volts.

Actually, for the moment, I am looking for actual current values at a certain time.
Let us use t1 is the time taken for point-A to reach 2.5V.
t0 = time when switch is closed.

What is the actual (calculated numerical value) current at t0?
What is the actual (calculated numerical value) current at t1?

Furthermore, how much charge has accumulated in C2 and C3 at time t1?

#### Papabravo

Joined Feb 24, 2006
14,378
Doesn't the answer depend on the nature of the switch and how fast the voltage rises across the switch and if the switch bounces or not. With a sub microsecond risetime and no bounce the transition through 2.5 V is so fast you can't see it, and point A is at 10V pretty quick?

#### MrChips

Joined Oct 2, 2009
21,815
Doesn't the answer depend on the nature of the switch and how fast the voltage rises across the switch and if the switch bounces or not. With a sub microsecond risetime and no bounce the transition through 2.5 V is so fast you can't see it, and point A is at 10V pretty quick?
My first approximation pegs the time to be about 300μs to reach 2.5V.