Basics: Capacitor charging

We are going to see what happens when we charge a capacitor and why it blocks DC.
I am not going to go into all the theory behind capacitor, but lets dig in to the experiment setup.

cap circuit

our circuit for this experiment.

In the beginning, when the switch S1 is open, lets assume the capacitor is discharged. (i.e. the plates of the capacitor have zero potential across them.)

We are probing the setup at
1. the input of the capacitor C1 and (channel 2)
2. across the resistor R1. (channel 1)
3. We are going to subtract the channel 1 from channel 2 to get the voltage across the capacitor in our Math channel of the scope.

cap_charging

So we know what our setup is. Lets wait no longer and press the switch, S1.
A surge of current equal to about V1/R1 starts to flow and we get v1 (here which is 5v) across the resistor R1; as if the capacitor is a short circuit! At this very point we can see in the Math channel that the voltage across the capacitor is indeed zero. (a signature of a short circuit.)

This current that is now flowing will decrease gradually as the capacitor gets charged, (i.e. the between the the two plates becomes equal to the source voltage (here that is 5v).

Now as the current decreases, charging the capacitor more and more, we see a the voltage across the resistor decrease at the same rate. The current this time becomes,

i(t) = {{(V1 - V_c(t))} \over{R1}}

Soon the capacitor will be fully charged, with the left plate at around 5v (we see it to be around 4.8v) and the right plate at zero, at this point very little current flows and therefore we get very little voltage across the resistor.

To find the time for this to happen, we use the capacitor charging equation.

V_c(t) = V_s(1-e^{-t/RC})

t = -ln((1 - {V_c(t) \over V_s})) \cdot RC

t = -ln((1 - {4.8 \over 5})) \cdot 1k \Omega \cdot 1 \mu F = 3.9{\mu}s.

Our experimental data checks out with our math. No further current will flow, until the capacitor gets discharged.

Hope you like this little, ‘back to the basics’ post. I am waiting to hear back from you.

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#back-to-basics, #capacitor, #charging, #electronics

Complex number in A/C circuit analysis

I am going to do a series of posts and try explaining what impedance is and how to analyse a passive Low and High Pass filters. But for that we need to know what a complex number is and how to handle it. A complex number is in the format of A + Bj, where B is an imaginary number and A was real. That was the entire story I knew. Until recently, I didn’t know why it is necessary or what it represented. I will be going to great depth on this subject and begin with the absolute basics.

Graphical representation of a Complex Number

Let’s say we have a graph as shown in figure 1. 

We have three points, A = (0, 15), B = (10, 0) and another point Z. We defined Z as the sum of A and B. So what is the value of Z.? What point does it represent? That is simple:

Z = A + B = (0 + 10, 15 + 0) = (10, 15) .We just add the X values and Y values together and that’s the answer.

A complex number can also be thought of just a point in a graph. The Y axis is represents the imaginary numbers and X real.

For example the graph in figure 1, the point A is represents the imaginary part and point B the real part of a complex number. Therefore, that complex number must be B + Aj, or, 10 + 15j. And this complex number is what Z represents.

Another way of thinking is that all points in a graph represent some complex number. A point that lies on the Y axis is purely imaginary, and, that which lies on the X axis is purely real.

Thus A is a complex number, with no real part and B with no imaginary part. To find A + B we simply do a complex number addition.

Z = B + A = ( 10 + 0j ) + ( 0 + 15j ) = 10 + 15j.

Polar and Cartesian form of a complex number

We can represent a point on a graph either in polar or Cartesian (or rectangular) format. And because a complex number is thought as nothing more than just a point on a graph, we can represent a complex number in either of these forms.

In Cartesian format, we represent a complex number by the point on the graph it represents. For example in the figure 1, the point Z represents the point (10, 15). So its Cartesian form is Z = 10 + 15j.

In the polar format, we represent a complex by its distance from the centre and an angle. The distance from the origin, is, the magnitude of the complex number. The complex number can also be thought as a vector; with a magnitude and a direction.

The magnitude r = \sqrt[2]{{y^2} + {x^2}}

And the angle \phi = tan^{-1} (\frac{y}{x})

In the polar format we can demote Z with its magnitude and it’s angle, so Z = \sqrt[2]{{y^2} + {x^2}} \angle tan^{-1} (\frac{y}{x})

  1. Say we want to represent Z = 10 + 15j in polar format. This is in Cartesian format, so we know, that it denotes a point of (10,15) on the graph.

Z = \sqrt[2]{{15^2} + {10^2}} \angle tan^{-1} (\frac{15}{10})

\Rightarrow Z = 18 \angle 56.3^{ \circ}

  1. Take another example. Convert P = 15.5 – 10.5j in polar format.

    Again, we know that this complex number denotes the point (15.5 , 10.5) in a graph. So

    Z = \sqrt[2]{{15.5^2} + {10.5^2}} \angle tan^{-1} (\frac{-10.5}{15.5})

    \Rightarrow Z = 18.72 \angle -34.1^{ \circ}

We can convert a complex number in polar format to Cartesian format in the following way.

We know that the Cartesian format of a complex number is in the form

Z = x + yj.

From the basic trigonometry, cos \phi = \frac{x}{r} and sin \phi = \frac{y}{r}

\Rightarrow x=r.cos\phi and y = r.sin\phi

thus the Cartesian form of the complex number is

Z = r.cos\phi + r.sin\phi j

For example: A the Cartesian form of the complex number

Z=5.385 \angle68.2^{\circ} is Z = 5.385 * cos(68.2) + 5.385 * sin(68.2)j = 2 + 5j.

Multiplication, Division and Reciprocal of a complex number

Multiplication and division is easier when the complex number is represented in the polar format. For example:

Multiplication

We just multiply the scalar and add up the angles.

For example:

1. (18\angle90^{ \circ})(10\angle-45^{ \circ}) = 180\angle45^{ \circ}

2. (176\angle-320^{ \circ})(5\angle-10^{ \circ}) = 880\angle-330^{ \circ} or 880\angle30^{\circ}

Division

We divide the magnitude as usual, then to find the angle, we subtract the denominator from the numerator.

For example:

1. \frac{18\angle90^\circ}{10\angle-45^\circ} =1.8\angle45^\circ

Reciplocal

Reciprocal of a complex number is also easy to do, when the complex number is in the polar format. Just follow the division procedure, except the angle in numerator is zero.

1. \frac{1}{23\angle45^\circ}=\frac{1\angle0^\circ}{23\angle45^\circ}=0.043\angle-45^\circ

2. \frac{1}{15\angle-19^\circ}=\frac{1\angle0^\circ}{15\angle-19^\circ}=0.0667\angle19^\circ

That’s that.

In the next post, I will design a passive high pass and a low pass filter, and then you will see how important complex numbers are. It makes our lives a lot easier.

PS: I am not an expert in this, so please inform me, if I have done anything wrong, or something that you didn’t understand.


#back-to-basics, #electronics, #mathematics