Since we are dealing with reactive components we can no longer expect the voltage and current to be in phase. Last time we left off asking, “What is the phase of the current through this circuit, and how do we get the voltages for each circuit component to add up to 1 Volt rms (the source voltage)?” and we were looking at these graph of the impedance, current and voltage:
To calculate the current and voltages requires a bit of math using complex numbers. Don’t get intimidated by “complex” numbers; solving complex number problems works just like algebra in most cases. Just remember this: “i” (or “j” as engineers call it) is just a way to account for the “phase” or “rotation” of something.
When we multiply a number by -1, we say it’s the opposite of what it was before. If it was up, now it’s down, if it was left, now it’s right, if it was a dirt pile, now it’s a hole in the ground. (Sometimes I wish I were a math teacher).
Now stay with me for a minute… If I gave you directions to Klipsch and said, “Go east five miles” I could also logically say, “Go west negative five miles,” or “Go five miles in the direction opposite of west.” That is the concept behind multiplying by -1.
Now imagine that to get to Klipsch I say, “Go 3 miles east, turn left, then go 4 miles north.” If you were a bird and didn’t have to follow the roads, I could say, “Face east, turn 53.13° to the left and go five miles” and you would end up in the same place; (53.13°=arctan(4/3) and 5=√(32+42).
When we multiply something by “i” we rotate it 90° counter-clockwise (with polar coordinates zero degrees points to the right and 90°is straight up). We could also say “turn left”. If we multiply by “i” a second time, we rotate it another 90°, for a total rotation of 180°, and so on. If we divide by “i” it’s the same as rotating something by 90° clockwise, or -90°, or we could say “turn right.”
Now it gets fun… If we rotate something by 90° twice (or A*i2), it’s pointed the opposite direction from where it started, just like when we multiply something by -1 (A*-1 = -A ). Guess what? i2 = -1 so A*i2 = -A. Fascinating! If we multiply by -1 twice (-1*-1=1) it’s pointing back the way it started. If we multiply by “i” four times (i4) it turns a complete 360° and again it’s pointed the same direction it started.
Complex numbers add, subtract, multiply and divide just as you did in algebra except the “x” is a “j”.
Anyway, back to the point of this blog… To get the phase response we must now concern ourselves with the complex impedance of the circuit components (i.e. their reactive nature), not just the magnitude of their impedance. Since we left off with 1kHz, we’ll pick back up there:
Z(L)=0+j(2πfL); at 1kHz Z(L)=0+j62.83
Z(C)=0-j/(2πfC); at 1kHz Z(C)=0-j15.92
Z(R)=R+j0; at 1kHz Z(R)=10
Ztotal=R+j(2πfL)-j/(2πfC)=10+j62.83-j15.92; so at 1kHz Ztotal=10+j46.91
Knowing the complex impedance and we can calculate the current. We will almost always reference the phase of the source, and its phase will be zero degrees, in other words no rotation or phase shift, and the “j” term will be zero, or V=1+j0.
From Ohm’s Law we know that V=I*Z, but since we want current I=V/Z so the current at 1kHz is:
To solve the above expression we multiply top and bottom by the conjugate, or in this case the complex conjugate of the denominator, which is just a fancy way of saying its “reflection” on the “x” or “real” axis. The conjugate of 10+j46.91 is 10-j46.91 (see picture here http://en.wikipedia.org/wiki/Complex_plane).
I’ll take my time here so you can follow each step:
Intuitively we don’t really know what this means yet, but let’s take a look and see what we can see. The current seems rather small, somewhere around 20mA, and since the “-j” term is present we expect a clockwise or negative rotation (i.e. phase shift), meaning the current is lagging behind the voltage. Also since the coefficient of the ”-j” term is about 5 times the real coefficient the angle will be much closer to -90° than zero.
The magnitude of the current using the Pythgorean is:
|I|=√[0.004352+(-0.0204)2] = 0.02086
|I| = 20.86mA
Now the phase:
Phase = arctan(complex coefficient/real coefficient)
Phase = arctan(-0.0204/0.00435)
Phase = -77.96°
“Holy Cow, Andy! Isn’t there an easier way?”
Well if you have a cool scientific calculator (or an Excel spreadsheet) that does complex numbers, no! You just plug in (1,0) divide it by (10,46.91), then hit the button that converts it to polar form to get magnitude and phase. Man, I miss my HP 28S! :-(
If your scientific calculator is not so cool then, yes! There is an easier way. We start where we started before, but before we do anything we convert to polar coordinates, which is represented by the form (r,θ)
We recognize that j0 means no rotation or zero degrees, so the numerator becomes:
The denominator requires the Pythagorean to find the magnitude and an arctan to find the angle.
Magnitude = √(102+46.912) = √2300.5481 (Now where have I seen that number before?)
Magnitude = 47.964 Ohms
Angle = arctan(46.91/10) = 77.97° (Hmmmm…)
So altogether the denominator is:
I = (1,0°)/(47.964,77.97°)
Now we just divide 1 by 47.964 and subtract 77.97 from zero… drum roll….
I = (0.02085,-77.97°)
Or 20.85mA with a phase of -77.97 degrees.
Now we’ll use the computer to do the same thing for the rest of the frequencies. Here’s the graph:
Now go back to the impedance graph at the top. When the total impedance is close to the capacitive impedance (we say the capacitor “dominates”) the phase will be close to +90°. When the total impedance is dominated by the inductor the phase will be close to -90°. At resonance (the frequency where the capacitive and inductive impedance are equal but opposite in phase) the phase is zero and the load as seen by the voltage source is purely resistive.
Resonance is the place on the impedance curve with the sharp dip, and it corresponds to the peak in the current and voltage curves. At resonance the complex impedance of the inductor and capacitor are complex conjugates. (i.e. 0+jX and 0-jX).
In this circuit resonance occurs at 503.3Hz.
Well that was a pretty big chunk, so tune in next time and we’ll talk a bit about resonance and find out where all the “extra” voltage comes from!