Tutorial: RC Circuits

The questions below are due on Monday October 07, 2024; 10:00:00 PM.

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In this problem, we'll step through the process of solving a couple of different circuits involving a capacitor and a resistor ("RC" circuits), and we'll see some common patterns in the solutions that we can bring to bear on future problems of this sort that we encounter, without needing to rederive them.

1) Capacitors

In lecture this week, we introduced capacitors. Introducing these components was a big leap forward for us, because they differ from the other components we've seen so far in at least one key way: they are stateful. That is to say, if I want to know the voltage drop across a capacitor, it's not enough to know the current flowing through the capacitor just at that moment; rather, I need to know the whole history of currents flowing through the device.

In our lumped element abstraction, we'll draw capacitors like so:

And, as with the other components, the capacitor is defined by its constitutive equation (i.e., by the way it constrains i and v). There are a number of (equivalent) ways that we could write this equation for the capacitor:

v(t) = {1\over C}\int_{-\infty}^t i(\lambda){\rm d}\lambda

v(t) = v(t_0) + {1\over C}\int_{t_0}^t i(\lambda){\rm d}\lambda

i(t) = C{{\rm d}\over {\rm d}t}v(t)

All of these are equally valid, though we'll probably primarily use the last form i = C{{\rm d}v\over {\rm d}t}.

It's important to note that now that we have capacitors in the mix, voltages are currents are functions of time (they won't necessarily be the same from one moment to the next). For simplicity of notation, though, we'll often skip writing the (t) part and just write the variable name; but it's important to remember that we're talking about functions of time.

2) "Natural" Response

Let's start by considering the following circuit, with a resistor and capacitor connected up to each other:

Applying KVL and KCL, we can see that i_{\rm R} = -i_{\rm C} and v_{\rm R} = v_{\rm C}. We'll use that as the basis for solving out this circuit. Let's try to solve for v_{\rm C}(t). We can start with our KCL equation, rearranged slightly:

i_{\rm R}(t) + i_{\rm C}(t) = 0

Then we can substitute in values for each current based on the (shared) voltage:

{v_{\rm C}(t)\over R} + C{{\rm d}\over {\rm d}t}v_{\rm C}(t) = 0

and multiply through by R to get to a more comfortable form:

v_{\rm C}(t) + RC{{\rm d}\over {\rm d}t}v_{\rm C}(t) = 0

Our next step is to take an educated guess at the form of v_{\rm C}(t). In order for that sum to work out to zero, we need v_{\rm C}(t) to look a lot like its time derivative. As such, we'll assume that v_{\rm C}(t) is of the form Ae^{st} for some values of A and s.

Plugging in our assumed answer v_{\rm C}(t) = Ae^{st} to our equation from above, we get:

Ae^{st} + RCsAe^{st} = 0

This equation alone is enough to solve for s:

s = -{1\over RC}

The value RC (which has units of seconds) is important enough that we will give it a name (it's the "time constant" of this circuit) and a symbol (the Greek letter tau: \tau).

We're going to need to do a little more work to completely characterize our equation, though (KCL isn't enough!). Because of the stateful nature of the capacitor, we'll need to know the value of v_{\rm C}(t) at some point in time in order to solve for values of v_{\rm C} after that point in time (that is, we'll need to know some initial condition in order to fully solve). For now, let's assume that v_{\rm C}(0) is a known value. We can use this to find A:

v_{\rm C}(0) = Ae^{-0/\tau} = A

And we've done it! If we note that this will only hold after our initial condition, we can say that:

v_{\rm C}(t > 0) = v_{\rm C}(0)\times e^{-t/\tau}

Or, slightly more generally, if we know the value of v_{\rm C} at an arbitrary time t_o, we can say that:

v_{\rm C}(t > t_o) = v_{\rm C}(t_o)\times e^{-(t - t_o)/\tau}

2.1) Key Times

If we plot this function starting at time t_o, we see a very characteristic shape:

There are a few key values on this curve that are worth knowing (at least approximately). Use the formula we derived above to find the following values, and enter each one as an exact expression (you can use e to represent e, pi to represent \pi, and ln(x) to represent \ln(x)).

After one time constant, to what fraction of the original value has the voltage fallen? We'll often approximate this value as 0.368, or sometimes as 1/3.

{\displaystyle {v_{\rm C}(t_o+\tau)\over v_{\rm C}(t_o)}} =~

After two time constants, to what fraction of the original value has the voltage fallen? We'll often approximate this value as 0.135, or sometimes as 1/9.

{\displaystyle {v_{\rm C}(t_o+2\tau)\over v_{\rm C}(t_o)}} =~

After five time constants, to what fraction of the original value has the voltage fallen? We'll often approximate this value as 1% (.01).

{\displaystyle {v_{\rm C}(t_o+5\tau)\over v_{\rm C}(t_o)}} =~

After around 5 time constants, notice that the voltage doesn't change much at all; we'll often assume that things have converged and are sitting steady after around 5 time constants.

3) "Step" Response

Now let's consider a slight variant of the circuit from above, now with a voltage source introduced in series with the components:

Let's assume, as an initial condition, that v_{\rm C}(t_o) = 0 and think about how this circuit behaves.

At that instant, what must the current i_{\rm C}(t_o) be? Enter your answer as an expression in terms of V_s, R, C, and/or any other constants you need.

i_{\rm C}(t_o) =~

Note that the derivative of the voltage is proportional to this current. So v_{\rm C} will start to increase from 0. But as v_{\rm C} increases, the voltage drop across the resistor decreases, causing the current to decrease, causing the derivative of v_{\rm C} to decrease, and so on and so on, until, in the limit, things settle out to a point where v_{\rm C} is a constant.

In this limiting case, what is the current i_{\rm C}?
i_{\rm C}(\infty) =~

In this limiting case, what is the voltage v_{\rm C}?
v_{\rm C}(\infty) =~

In class, we solved out for v_{\rm C}(t) in a more careful way, but here we'll just jump to the conclusion: v_{\rm C} starts at some initial value v_{\rm C}(t_o) and follows a familiar exponential curve that asymptotes out at some final v_{\rm C}(\infty), with the same time constant as our original circuit. So we can say that, for t > t_o:

v_{\rm C}(t > t_o) = \underbrace{v_{\rm C}(t_o)}_\text{initial value} + \left(\underbrace{\left(v_{\rm C}(\infty) - v_{\rm C}(t_o)\right)}_\text{difference between initial and final values}\times \underbrace{\left(1 - e^{-(t - t_o)/\tau}\right)}_\text{upside-down exponential curve}\right)

Plotting this curve, we see a similar kind of shape:

After one time constant, what fraction of the way between its original and final values has the voltage gone? We'll often approximate this value as 0.632, or sometimes as 2/3.

{\displaystyle {v_{\rm C}(t_o+\tau)-v_{\rm C}(t_o)\over v_{\rm C}(\infty) - v_{\rm C}(t_o)}} =~

After two time constants, to what fraction of the original value has the voltage fallen? We'll often approximate this value as 0.865, or sometimes as 8/9.

{\displaystyle {v_{\rm C}(t_o+2\tau)-v_{\rm C}(t_o)\over v_{\rm C}(\infty) - v_{\rm C}(t_o)}} =~

After five time constants, to what fraction of the original value has the voltage fallen? We'll often approximate this value as 99% (.99).

{\displaystyle {v_{\rm C}(t_o+5\tau)-v_{\rm C}(t_o)\over v_{\rm C}(\infty) - v_{\rm C}(t_o)}} =~

As before, after about 5 time constants, we often assume that the system has fully converged.

We'll see these values come up quite a bit over the next couple of weeks as we continue working with these circuits, so it might even be worth committing some of these ratios (or at least their approximations) to memory!