# Simple Capacitor Charging Question

Right, I have a very simple circuit that I aim to use as a delay circuit, everything was going fine until I started to over think it all and now I have totally confused myself and it is slightly embarrassing.

This is my circuit: simulate this circuit – Schematic created using CircuitLab

As you can see, nice and simple. I know the time constant of the capacitor is RC to reach 63% of the voltage, but my issue is what is the voltage that it is reaching 63% of? I initially thought the 2V supply (capped at 1V though) that I am using at R2, but then it would charge to 1V in less than RC (not what I want).

I take the time constant as $1M\Omega \times 1 \mu F = 1$s but would that be the time to charge to 63% of 2V (but cap out at 1V) or 63% 1V (i.e. take 5 times longer to get to 1V)?

• If you removed the capacitor from your circuit what would be the (maximum) voltage you could get at the junction of R1 & R2? – JIm Dearden Apr 27 '15 at 13:31
• @JImDearden As mentioned, 1V – MrPhooky Apr 27 '15 at 13:31
• and 63% of 2V is? – JIm Dearden Apr 27 '15 at 13:35
• 2V across 2 series 1Mohm resistors produces 1V and this is equivalent to having a 1V supply in series with two parallel 1Mohm resistors OR a 1V supply in series with 500kohm – Andy aka Apr 27 '15 at 13:39
• @JImDearden 1.26V... Are you being purposefully patronizing? – MrPhooky Apr 27 '15 at 13:44

Well... Let's calculate it using Laplace transform.

$$T(s)=\frac{\frac{\frac{1}{sC}\cdot R_{2}}{\frac{1}{sC}+ R_{2}}}{\frac{\frac{1}{sC}\cdot R_{2}}{\frac{1}{sC}+ R_{2}}+R_{1}}=\frac{R_{2}}{R_{1}+R_{2}+sC\cdot R_{1}R_{2}}=\frac{R_{2}}{R_{1}+R_{2}}\cdot \frac{1}{1+sC\cdot \frac{R_{1}R_{2}}{R_{1}+R_{2}}}$$ You can see, that we have a resistor divider and RC filter with time constant equal:

$$\tau = C\cdot \frac{R_{1}R_{2}}{R_{1}+R_{2}}=0.5s$$

So in 0.5s time, you will have a 0.63% of input voltage multiplied by resistor divider - 2V * 0.5 * 63% = 0.63V

Continuing from where @przeski left off, I was curious to see what the transient response of Vout would look like. The following code was written:

t = 0:0.001:100;
Vin = 2;
R1 = 1e6;
R2 = 1e6;
C = 1e-6;

den = R1+R2+(j*(R1*R2*C*2*pi)./t);
Vout = (Vin*R2)./den;
Vout = abs(Vout);
y = 0.63.*(t>0);
plot(t,Vout);
hold on;
plot(t,y);


The output resulted in the following response: To answer your question yes it is 63 % of 1V this is as a result of the voltage divider present.

An interesting finding.

I know we shouldn't pose questions in the answers section, but i noticed this in my results.

Looking the the graph we can see that the 63% of the steady state value is achieved at 2.54 sec.

I continued to solve the equation given by:

$$\frac{V_{in}R_{2}}{(R_{1}R_{2})+jR_{1}R_{2}\omega C}$$

This was obtained by applying the voltage divider rule treating the capacitor as a reactance.

$$0.63 = \left |\frac{V_{in}R_{2}}{(R_{1}R_{2})+jR_{1}R_{2}\omega C} \right |$$

$$0.63 = \frac{V_{in}R_{2}}{\sqrt{(R_{1}R_{2})^{2}+(R_{1}R_{2}\omega C)^{2}}}$$

$$0.63 = \frac{2\times 10^{6}}{\sqrt{(2\times 10^{6})^{2}+\frac{(2\times\pi \times10^{6} )^{2}}{t}}}$$

Solving for t we obtain 2.5485 seconds.

The Question is, is this the time constant, 2.5485sec?

You are right ,the time constant is t=RC=1000000ohms*0,000001F=1second.
Your supply voltage is 2V,so that's the voltage the capacitor will have when it's fully charged;every capacitor will reach 63% of the supply voltage in one time constant.These are confirmed by this site:http://www.electronics-tutorials.ws/rc/rc_1.html

In conclusion,after one second,the capacitor will have 1,26 volts across it.