Home / Exercises 11 / Lights Revisited

Lights Revisited

The questions below are due on Wednesday November 20, 2019; 11:59:00 PM.
 
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In Lab 8 we designed a circuit that allowed us to light up a bunch of LEDs (that together need a lot of voltage such as 40V) using a low initial voltage (3.3V). Let's think about this problem another way.

We know both from 8.02 as well as earlier in the semester that if you have a component in steady state (DC) and there is a voltage across it, the power for that component will be equal to:

P = V\cdot I

Using the sign notation shown in the diagram below for this equation, if the power is determined to be negative, it means that the component supplies power to the circuit, whereas if the power turns out to be positive, it means the component in question consumes power.

We can take this consuming and supplying of power further and come up with a rule that says the sum of all power in a circuit must be 0. We won't use that much here, but it always holds up!

We can come up with special cases of our power equations as needed. In the case of a resistor, for example, because of Ohm's Law (V=I\cdot R) this can then be turned into two other forms:

P = \frac{V^2}{R}

or

P = I^2 R

Again nothing new with this, but we can extend this thinking into the sinusoidal steady state domain! For starters we will say that in a sinusoidal steady state power associated with a component is based on the following equation:

S = V_{rms}I_{rms}^*

where S is the complex power consumed or supplied by a component. The voltage V_{rms} is the Root-Mean-Square (RMS) of the voltage accros the component v(t) (we'll come to that in a minute) in phasor form. The current I_{rms}^* is the complex conjugate of the RMS current through the component in phasor form.

The power mentioned, S, is complex and as its name implies, it can be both real and/or imaginary. Real power is actually power that is supplied or consumed by a component, while imaginary power can be thought of sort of like reflected power. This is very similar to complex impedance. If the real part of the complex power is positive, it means the component is consuming power. If the real part of the complex power is negative, it means it is supplying power. The sign of the imaginary term has implications about how it is reflecting power which we won't worry too much about.

What about RMS?

The RMS of a signal is a means of quantifying the energy in it by taking the square root, of the mean of the square of that signal over all time. More formally this looks like:

V_{rms}=\lim_{T\to \infty}\sqrt{\frac{1}{T}\int_0^T (v(t))^2dt}

However, it so happens that if a signal is periodic then you only need to do this for one period, avoiding limits with our integrals (thank goodness), so if T is the duration of one period of the signal.

V_{rms}=\sqrt{\frac{1}{T}\int_0^T (v(t))^2dt}

RMS voltage is not a time-varying term...rather assuming as sytem has a steady state periodic signal, it will have one value which can be complex.

For a sine wave of amplitude V_A we can find the RMS signal by doing the following:

V_{rms}=\sqrt{\frac{1}{1/f}\int_0^{1/f} V_A^2\sin^2\left(2\pi f t\right)dt}

When this is worked out you'll see that

V_{rms} \frac{1}{\sqrt{2}}V_A

And this holds true of any sinusoidal function. The RMS value a sinusoid is always \approx 0.707 of its amplitude!

With that in mind, if the voltage out of American power sockets (you know the ones in walls) is sinusoidal at 60Hz with a value of 120V_{rms}. What must the peak voltage of this waveform be approximately?

RMS signals can be treated just like DC signals in how we analyze and work with them. Ohm's Law for example works the same and since the RMS values can be complex, we can use them just like we do other voltage and currents that are complex (when we're doing phasor math):

Building on this, two other relationships (similar to the DC ones above can be generated), although you need to take care with complex conjugates:

The complex power associated with a device with impedance Z is shown below in two forms:

S = \frac{\left|V_{rms}\right|^2}{Z^*}

or

S = \left|I_{rms}\right|^2Z

You'll notice both of these equations have analogs in the regular DC power equations at the top with resistors. These have been generalized to cover arbitrary frequencies as well (though it should be noted that they will collapse back to the equations towards the top of the page at DC).

If you have 120V RMS at 60Hz driving a 100 Ohm impedance what is the complex power associated with the device?:

If you have 120V RMS (from the wall) driving a 5 microFarad capacitor at 60 Hz, what is complex the power associated with the device?. Use j for complex numbers.

You should see two things from thre previous answers. The resistor had a purely real complex power (meaning it consumes actual power!) whereas the capacitor had a purely imaginary complex power, which means it consumes nothing and merely reflects the power back at the source out with some phase offset.

1) Teensy Time

In lab we had our Teensy generating a square of 3.3V to 0V at 50% duty cycle at some variable frequency which we set. While not a sinewave, you can still calculate the RMS of this signal.

What is the V_{rms} of the Teensy when running a 50% duty cycle signal varying from 3.3V and 0.0V at an arbitrary frequency? Think of how you could find the RMS of this without integrals

So the RMS voltage of our Teensy's digital pin at 50% Duty Cycle will be useful below!

2) The Predicament

There are certain types of components in life that are just inherently high in resistance. The string of LEDs from Lab 7 can be roughly modeled as such a situation. This causes an issue in an electronic environment when only low voltages are present such as in our Teensy microcontroller (or any microcontroller really).

Let's remember our string of LEDs from lab which we can model as a resistor. Looking at them electrically they'll just look like the following:

If a string of LEDs, which can be modeled as a 50 kOhm resistor, is connected to the output of a Teensy digital pin operating in PWM mode with a 50% duty cycle what is the power S dissipated by those resistors (in Watts)?

This is somewhat unfortunate since the Teensy can actually provide a lot more power than what you found above. Because the Teensy can supply up to 25 mA (instantaneous) out of its digital pins, how much power could the Teensy provide if it was driving a load with a smaller effective resistance, assuming a 50% duty cycle wave of V_{pp} of 3.3V? Hint: Figure out the maximum I_{rms} for the Teensy and then use it with the V_{rms}

Since the Teensy can supply up to 25 mA out of its digital pins, how much power could the Teensy provide if it was driving a load with a smaller effective resistance? (answer in Watts)

Comparing the previous two answers we can see that the ability of the Teensy to transfer power into the high-resistance load is sadly very low.

What is the ratio of Actual vs. Possible power delivered to the resistor by the Teensy? In other words what fraction of power that the Teensy could provide actually gets provided to the resistor?

3) Moving On

There is no Thevenin equivalent voltage as seen from the perspective of the Teensy microcontroller since the LEDs are strictly passive components modeled as a resistor.

What does the equivalent impedance of just the resistor look like (not a trick question)?

You then proceeded in lab to generate an L and a C which you incorporated as shown below. The network seen by the Teensy will now look different. While it will still lack a non-zero Thevenin voltage source, the Thevenin equivalent impedance will now be notably different Z_{Th}. Generate an expression for the equivalent circuit

What does the equivalent impedance of the circuit look like (answer in terms of RL, L, C, omega (or w), as well as j? Do not put numbers in!

We proceeded to drive our circuit at resonance. When the circuit is at resonance, what happens to the equivalent impedance seen by the the Teensy?

If we proceeded to drive this circuit at resonance (remember what is resonance in this circuit?), what will the impedance simplify to? Answer in terms of RL, L, C, as well as j? Do not put numbers in!

Assuming we chose values of 15 nF for C and 47 mH for L (not saying those were the best), and the LED resistor model is still 50 kiloOhm, at resonance what approximately are the real and imaginary components of the impedance seen by the Teensy's pin?

Real component of Thevenin Impedance at resonance (in Ohms)

Imaginary component of Thevenin Impedance at resonance (in Ohms)

What the numbers above should indicate is that at resonance, with properly chosen values of L and C, the effective impedance of what the Teensy's digital pin will drop drastically...in other words, the resistor which previously looked huge (and therefore drew very little power) will now look much, much smaller! What this means is that when the Teensy drives this load now, it will be able to deliver much more power.

What is the complex power delivered into the LED load at resonance using the values for the components we listed above? (in Watts) (multiply by j for the indicating the imaginary term).

What is the ratio of Actual Real Power vs. Possible power now delivered to the system by the Teensy? In other words what fraction of power that the Teensy could provide actually gets provided to the resistor?

The number you get above should be much, much higher than what we got before...in fact the number should be larger than one which means with this particular set of values, we're now capable of drawing more power from teh Teensy than what it can provide1

Objectively speaking, this is really cool. This is because at resonance this circuit (the L and C) makes the resistor appear to be a much smaller magnitude impedance. The result is that the Teensy can provide more power to it, which then gets handed off to the resistor by inductor and capacitor via their oscillations.

This type of circuit (the role that the L and C carry out here) is known as an impedance matcher. What we've done with the L and C is carry out impedance matching, and this is a huge field of engineering...even outside of circuits. Anything that mediates the transfer of energy from one form or source into another form or source, where if left on their own they'd not be able to achieve much or anything, is called impedance matching.


 
Footnotes

1While not ideal to try to draw up to a system's power-delivery limits it isn't too dangerous here (click to return to text)