# Understanding this simple circuit

I am basically having trouble understanding this circuit from a question asked on here.

Would be nice if someone could confirm that my understanding of it is right.

Here is what I have understood:

So without the cap, the voltage at the voltage divider output is 1 V . With the cap , On switching on the power supply, I would expect 0 V at Vout, as the cap is initially discharged , and slowly as it charges, The voltage would keep increasing based on the R1C time constant. After it has charged, I would see a stable 1V at Vout .

And finally the graph will be like this :

is it the correct understanding for the circuit ?

Thanks

The final voltage will be 1 V, yes, but you've got the time constant wrong. The time constant is Rp $\cdot$ C, where Rp = R1 and R2 in parallel. I know this is counterintuitive, but it's all Thévenin's fault. Look up "Thévenin equivalent".

One way to get some understanding why R2 also plays a role: suppose R2 is 1 ohm. Would the capacitor be charged more quickly? (The answer is yes!)

• The capacitor would be charged more quickly if R2 was 1Ω because it would only be charging to 2 microvolts. But you're right about the time constant. – Samuel Apr 27 '15 at 19:24
• @Samuel Actually it would be charging to a higher voltage, 1.99999V – ACD Apr 27 '15 at 19:57
• @ADC Right, sorry, the resistors numbers are reverse from normal ordering. Joris was, I believe, also referring to the lower resistor; R1 in this schematic when he wrote about changing R2. – Samuel Apr 27 '15 at 20:04
• @JorisGroosman Are you actually referring to R1 in your final two sentences? – Samuel Apr 27 '15 at 21:37
• Intuitively, charge time would depend R2, the resistor to the source. Trying to explain the non-intuitive part would be explaining R1, but apparently you're not trying to do.. that... – Samuel Apr 28 '15 at 5:35

A more systematic approach would be to find the transfer function of the system.

Let's say that the input voltage is given as $V_{in}$. Your question now is what happens to $V_{out}$ if $V_{in}$ goes from 0 to 2V?

To do that, find the transfer function, that is:

$$\frac{V_{out}}{V_{in}} = \frac{R_1||\frac{1}{j\omega C_1}}{R_2 + R_1||\frac{1}{j\omega C_1}}$$

As you said, it's a (complex) voltage divider, expanding the parallel term gives $$R_1||\frac{1}{j\omega C_1} = \frac{R_1 \frac{1}{j\omega C_1}}{R_1 + \frac{1}{j\omega C_1}} = \frac{R_1}{j\omega R_1C_1 +1}$$

$$\frac{V_{out}}{V_{in}} = \frac{R_1||\frac{1}{j\omega C_1}}{R_2 + R_1||\frac{1}{j\omega C_1}}=\frac{R_1}{j\omega R_1C_1 +1} \div( R_2+\frac{R_1}{j\omega R_1C_1 +1}) = \frac{R_1}{j\omega R_1R_2C_1 + R_2 + R_1}$$

This formula is a first order lag element. The general formula of this element is

$$G(j\omega) = \frac{K}{j\omega T + 1}$$

As you can see, the numerator is a polynomial of order 0 and the denominator is a first order polynomial. The coefficients and their values are not that important. That's why this is a first order lag element. The important thing is that many systems have a transfer function like your system. They are all behaving in a similar way, only varying due to values of T, K, etc. They all have a step response of the shape that yours has.

The transfer function is a way to express the behaviour of the system in an abstract mathematical way. It doesn't matter if it is a mechanical or electrical system. If I told you that the flow of heat through an object has the same transfer function, you now know quite a bit about what that means in terms of its behaviour without necessarily knowing anything about thermodynamics. You don't even have to know what heat actually is, because you have an abstract description of its behaviour.

Knowing that is helpful because the step responses for all the basic elements are known. It looks indeed the way you posted it.

The point here is that for more complex circuits, it is often not trivial to figure out what each element doe and how it affects the overall circuit. Analysing the circuit as a circuit with complex elements yields the transfer function that can be used to determine the step response (like in your case), the impulse response or generally speaking the response to any input signal.

• Thanks, I had a question regarding what you posted. So After getting the transfer function, how do I know its the first order lag element ? – Dallas Carter Apr 28 '15 at 10:59
• @DallasCarter I edited my answer =) – Magic Smoke Apr 28 '15 at 13:24

Mostly correct.

As pointed out by Joris Groosman you need to look at the thevenin equivalent for this circuit. It looks like this:

simulate this circuit – Schematic created using CircuitLab

I often would tell students to redraw their schematics to help them understand. For you, it may help looking at the initial circuit like this:

simulate this circuit

For me at least, this makes it more obvious that not only will the capacitor be seeing half the voltage, but it will also be seeing half the resistance when looking at the voltage source.

The rest remains the same, when the power supply it turned off the capacitor would be discharging through the lower 1MΩ resistor, following the dotted blue line shown here:

• No, it's not exactly correct. It's exactly wrong. See my answer. – Joris Groosman Apr 27 '15 at 18:54
• @JorisGroosman It's not exactly wrong. But you're right, it wasn't exactly correct. I've added the additional information and +1 to your answer. – Samuel Apr 27 '15 at 19:40
• Why the downvote? – Samuel Apr 27 '15 at 20:05
• Well, OP's reasoning wasn't that bad, but if you use the words "exactly right" then it has to be exact :-). It's as "exactly" right as it is "exactly" wrong. – Joris Groosman Apr 28 '15 at 5:49
• BTW, the downvote isn't mine. Maybe you thought so because of my comment, but I don't downvote that easily. Thing is that I got 3 downvotes (revenge?) in a row now on different answers. They're not yours, are they? – Joris Groosman Apr 28 '15 at 5:51

You are correct.

at t=0-, Vout=0, and there is no current, because there no voltage supply. At t=0+ (juuuuuust after you turn the power on), the capacitor acts like a short, so all of the current passes into the capacitor as it starts charging. As the voltage level in the capacitor builds up, a potential across the R1 is formed, and you get a current through the resistor at I=V/R1. Eventually, once the capacitor is fully charges, no current passes to the capacitor and it behaves as an open circuit, so you can ignore it and calculate Vout as if it is a simple voltage divider.