# AC/Capacitive Coupling

1. Jan 6, 2013

### d.arbitman

In the attached picture R1 and R2 are assumed to be equal so that they form a voltage divider. Without any signal into C1, the voltage of the base of the NPN BJT is (1/2)Vcc.

If a square wave with Vhigh = (1/2)Vcc and Vlow = 0V, and a duty cycle of 80% is fed into C1, what would be the resultant voltage ranges of the base of the transistor?

I cannot find the answer to my next question anywhere and I figure it has to do a lot with duty cycles and average signal values. If we use capacitive coupling, why must the average output signal = 0?

I hope I have provided enough information. Thank you.

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2. Jan 7, 2013

### Runei

Lets take the input thingy first.

The impedance of the input capacitor is $\frac{1}{j\omega C}$. The input signal can be found by superposition.

The voltage due to the powersupply is still $V_{1}=\frac{1}{2}V_{ss}$

The voltage due to the input signal is $V_{2}=v_{i} \frac{R_{2}}{\frac{1}{j\omega C} + R_{2}}$

Thus the total voltage is $V_{t}=V_{1} + V_{2}$

Now, normally we don't care about the DC term (V1), since the AC term is the only thing that will matter, and I will show you why.

But first - in order for the voltage waveform to make it somewhat correctly through the impedance voltage divider, the impedance of the capacitor should be low. Meaning we should have a high capacitance.

The voltage at the emitter of the transistor is given by $V_{e}=V_{b} - V_{be}$

Now Vbe of course controls the current running through the transistor, but we can approximate it to around 0.7 volts.

When no input is applied Vin = 0 , the voltage at the emitter is given by $V_{e} = \frac{1}{2} V_{cc} - 0.7$

Here comes the thing: The output voltage is the thing on the other side of C2, and right now it is floating. So we just need to add a resistance to ground (the load resistance). So imagine that we have added a resistance to ground at that floating node.

The output voltage at DC (when no input is applied) is 0. Why? Because no current is flowing through the capacitor.

Imagine now that we introduce the square wave you wanted to use.

This square wave will make the emitter voltage oscillate because the emitter voltage is now given by $V_{e} = \frac{1}{2} V_{cc} + V_{2} - 0.7$. Remember V2 is the voltage due to the input signal. If our input capacitor is chosen correctly, V2 is almost equal to Vin.

$V_{e} = \frac{1}{2} V_{cc} + V_{in} - 0.7$

Now you should see, that as the voltage goes up at the emitter. A current will begin to run through the capacitor.
You can use the same arguments as for the input capacitor. The output capacitor + the load resistance form a voltage divider in the z-domain (impedance).

The voltage division is the same as before. But notice also, that if the frequency of a signal is 0, then the impedance of the capacitor will be infinite (an open circuit). And since out voltage at the emitter is comprised of two components - a DC term (1/2 Vcc - 0.7) and an AC term (Vin) - The output voltage will only react to the AC term, which has a non-zero frequency.

Thus the output voltage will be approximately equal to the input voltage (if the capacitors are chosen correctly).

Why is the average output voltage 0?
Because if it wasn't, a constant current would be running through the capacitor, which would mean that it would have infinite capacitance - which is equal to a short circuit allowing also DC to pass.

Conclusion
So AC coupling removes any DC components from a voltage signal, and almost replicates the AC components (if the capacitance is chosen correctly).

3. Jan 7, 2013

### d.arbitman

That is a brilliant explanation, Thank you. I just have a few follow up questions.

I went ahead and built a similar circuit, except I didn't include a transistor nor C2. I just had the input signal (a square wave, 0V to Vcc) fed to a voltage divider network similar to R1 and R2 on the left hand side of C1 (see attachment). When I probed (with my scope) the left hand side of C1 with respect to ground, I expected a square wave oscillating between 0V and (1/2)Vcc, but what I saw was a square wave oscillating between a voltage > 0V and a voltage < (1/2)Vcc. I'm curious as to why the square wave on the left hand side of the capacitor didn't reach the values that I was expecting.

Thank you.

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4. Jan 7, 2013

### Runei

First we convert the circuit into the z-domain. Only the capacitor has something special about it here. $Z_{c} = \frac{1}{j\omega C}$

So the two resistors left of the capacitor we call R1 and R2. The two resistors to the right we call R3 and R4.

We simply use the node voltage method to find the voltage between the resistors R1 and R2 (lets call it V1). V2 is the voltage between R3 and R4.

Eq 1: $\frac{V_{1}-V_{in}}{R_{1}} + \frac{V_{1}}{R_{2}} + \frac{V_{1}-V_{2}}{Z_{c}} = 0$

Eq 2:$\frac{V_{2}-V_{cc}}{R_{3}} + \frac{V_{2}}{R_{4}} + \frac{V_{2}-V_{1}}{Z_{c}} = 0$

Solve for V1 and you would get an expression for V1 (where you were probing). But the thing becomes a complex garble I can see.
But clearly the voltage at V1 will not be the same as the input, since the input will be divided through a network of impedances (if you look at it using superposition - Turn off the Vcc and replace it by a short circuit to ground).

If you wanted something were you can see V1 rise and follow the input signal, you would need to replace R1 with a coupling resistor. As I see it then, the V1 should then oscillate between 0 and almost 1/2 Vcc (given a high enough capacitance).

Also - Remember the frequency of the signal also has something to say about what the impedance of the capacitor is.

5. Jan 7, 2013

### Runei

Oh yeah, and a square wave might be a bit problematic because of all the harmonics in it's spectrum... <--- Forget this

EDIT: Aha, the problem is of course that the square wave you are using is going from 0 to 1/2Vcc. However, that means that there is a DC component in it.
After some time, the DC component will be gone, and you will be left with the AC.

The DC component is the average value of the signal. The longer your duty cycle, the larger average value.

Last edited: Jan 7, 2013
6. Jan 7, 2013

### d.arbitman

Couple of things:
My resistors were all of equal values, ~1.5k.
Vcc = 5.4V
Input was between 0 and 5.4V.
The square wave had a duty cycle of about 77.8%; High for 2.8ms & Low for .8ms.

May I use the impedance formula for the capacitor despite the input being a square wave?

If so, then ω = 2$\pi$f , where f = 278Hz.

I am stuck here and I have no idea how to represent the square wave as an input voltage.

Are there any resources that anyone can recommend I read about capacitive coupling? I have checked other forums, youtube, my textbooks and not a single one talks about capacitive coupling.

How would I choose the correct capacitance?

What do you mean by "After some time"?

Last edited: Jan 7, 2013
7. Jan 8, 2013

### Runei

I will make some simulations and return stay tuned :)

8. Jan 8, 2013

### Runei

Choosing a capacitor
In general for a sinusoidal signal, we would like the impedance of the capacitor to be as small as possible. In that case, it looks like we should just choose as big a capacitance as possible.

This is also true in theory, but the true practical capacitor is not just capacitance, it is actually modelled by the following:

$Z_{c} = R_{c} + j\omega L + \frac{1}{j\omega C}$

Yes that is true, the capacitor also has an inductance (although very small), and a resistance (very big).

However, you can see, that as the frequency increases and the term from the capacitance becomes smaller, but the term containing inductance gets larger. So for every capacitor, there is a frequency where, instead of the impedance getting smaller and smaller, it begins getting bigger and bigger. This is called the self-resonant frequency.

Now the inductance L depends on how the capacitor is built, but in general, the bigger capacitor, the bigger inductance it will have. So higher capacitance will give you a higher inductance, which in turn will mean a lower self-resonant frequency.

So in conclusion, when choosing the capacitor you would of course look at the frequency or frequencies it's gonna work with, and then try to choose a high capacitance, but small enough so that the operating frequency is still lower than the self-resonant frequency.

How to do this? Look at the datasheet for the capacitor, it will have a graph of the impedance versus frequency, and there you can see the self resonant frequency.

But for starters: Just choose a high capacitance, since you most likely wont be using a high enough frequency (it is more in RF systems this has to be taken into account).

The square wave
Have you learned about Fourier Series? If you have, then you can decompose the square wave signal into a sum of normal sinusoids.

$v_{sq} = a_{0} + \sum{a_{n}cos(n\omega_{0} t)} + \sum{b_{n}sin(n\omega_{0} t})$

The sums are from 0 to ∞.

Now ω0 is the fundamental frequency and is the frequency of the periodic signal you are trying to model (the square wave).

a0 is the DC term. It is actually the frequencies where n = 0 (frequency 0).

What happens in the circuit
Now if the square wave signal is represented by all these different sinusoids, we can just apply superposition and and look at what happens for each and every one of them.

As you can see, all frequencies other than the fundamental one are greater than the fundamental frequency (because n is a positive integer). Only the DC term has a "lower" frequency (It's 0).

So if we have chosen the capacitance so that the fundamental frequency gives a small enough impedance, all other frequencies will have an even smaller impedance.

The DC term however, will have frequency 0, which means the impedance will be infinite.

The transient versus steady-state response
The impedance model of a circuit tells us about the steady-state response of the system. That is, how the voltages and currents will look after some time has gone by. What goes on before this is called the transient response.

You can find out the transient response by using LaPlace transforms and solving differential equations.

So what is the conclusion?
Well, since the DC term at steady state will meet an infinite impedance, it wont be visible at at the node left of the middle capacitor. However, all the other frequencies should be let through quite nicely. Here is a circuit that shows how it look: I have chosen the same values as you did, and I am probing the following places:

Blue: voltage of the signal generator
Green: voltage at left side of C1
Red: voltage at right side of C1

As you can see, after approximately 40ms, the "transient response" has died out, and we are left with the steady state response.

If you wanna know more, you should really study Fourier Series and Fourier transforms and Laplace Transforms and possibly Control theory. These things will give you several ways of understanding a system like this.

EDIT: Attached the image ;)

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• ###### Untitled (Recovered).bmp
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Last edited: Jan 8, 2013
9. Jan 8, 2013

### d.arbitman

Thank you very much for trying to help me. I built the circuit and simulated it in Multisim, but I really need to be able to do the math myself rather than through a simulation.
I am aware that a capacitor has inductance and resistance components of its impedance. I know the concept behind Fourier Series, although I haven't done many problems using them; I know the Laplace transforms, but I have zero knowledge about control theory.

Would you happen to know of a great textbook or resources that talks about capacitive coupling in depth, with worked examples?

I can't thank you enough.

10. Jan 8, 2013

### Runei

Very good that you want to do the math yourself! I know far too many engineers who rely on software to do the work for them! :D

I don't know about a text book I'm afraid, but I've been looking around a bit.

A video:

Otherwise, try doing some analysis using Fourier Series.

You can also view the circuit as a system and calculate its transferfunction for V1/Vin (where V1 is at the left side of the capacitor), and then see how it responds to different frequencies.

Draw a bode diagram of the magnitude response and see how different frequencies are attentuated. That should give you some ideas about the system. Then draw the spectrum of the input signal (using Fourier series) on top of the bode diagram and you can see what is passed and what is not.

Last edited by a moderator: Sep 25, 2014
11. Jan 8, 2013

### d.arbitman

I will try working some textbook examples using Fourier series. Thanks for the video, but I watched that one about 3 times now. I will resort to asking my physics and engineering professors for help.

Last edited by a moderator: Sep 25, 2014
12. Jan 8, 2013

### Runei

Good luck, hope I helped with some of it :)

13. Jan 8, 2013

### d.arbitman

You did, thank you much.