Power to the People

The questions below are due on Friday April 25, 2025; 05:00:00 PM.

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Today we're going to do something quite magical. We're going to light up a little string of lights. Well, maybe that doesn't seem so magical, but it is! For now you'll just have to trust me, but by the end of the day hopefully you'll feel the same way.

The circuitry we're going to use to accomplish this is something very cool, and it utilizes the energy storage capabilities of inductors and capacitors in a new way that's going to let us do things that we had previously been unable to do.

Start by grabbing yourself a little string of lights from up front, and a AA battery (1.5 Volts) and holder. Ultimately, our goal for the day is to power this string of lights using the battery.

Each of the bulbs in our little string of lights is an LED (a "light emitting diode"). We haven't covered diodes too much in class, but as a quick introduction: when a voltage of the proper polarity and magnitude is applied, current will flow, the LED will light up, and it will maintain a roughly-constant voltage drop across it. For the lights we're using, they'll turn on when the voltage drop across them is around 2.4V (though it depends on the color of the LED).

1) Lights

To start, let's hook up the lights to our battery. You can plug the battery leads into a breadboard, and then you can use alligator clips to connect up one end of the lights to a little piece of wire that you can jam into the board, like so:

Also note that the lights have a polarity (i.e., it matters which side is connected to + and which is connected to -). A little protrusion on the plastic casing of each LED marks the + side:

OK, so hook up your battery across the string of lights. Do they light up? No they don't. How sad. But has your day been ruined? Of course not! During the remainder of the lab, we're going to light up the darkness and push the bad feelings away through the magic of power electronics.

But before we get to fixing the problem: why didn't the bulbs turn on? Well, remember that in order to turn on, each bulb needs about 2ish Volts across it. So we need a much bigger voltage than the 1.5V we get from the battery in order to turn the lights on. Let's test a little bit more to figure out what we're aiming for.

Check Yourself 1:

With the benchtop power supply off, set it to 0V and 100mA (0.1A) current limit, and hook the lights up to it (again, be careful that the + side of the power supply is on the side of the lights with the little plastic protrusion).

Turn the power supply on and slowly, SLOWLY and carefully turn the voltage up. DON'T GO ABOVE LIKE 15V, the lights probably won't like that.

What voltage is needed to make the lights start to light up? What is needed to make them light up nice and bright?

Make a note of these voltages.

2) Boost!

OK, so yep, we need substantially more than 1.5V to light up the lights. What are we going to do? Luckily, we can use some of the cool features of the RLC circuits we've seen in class the past week or so in order to build a circuit called a Boost Converter, a circuit which can take a small voltage as input and produce a large voltage as output.

Here is a conceptual model of the circuit we're going to build:

The circuit generally operates in three stages:

In stage 1, S_1 is closed and S_2 is open.
In stage 2, S_1 is open and S_2 is closed.
In stage 3, both switches are open.

These stages are repeated in quick succession, forever; so after stage 3, we proceed back to stage 1. Importantly, the amount of time we stay in any stage is intentionally way smaller than the period of an oscillation at the natural frequency.

Let's take a look at how the voltages and currents in the circuit evolve during each of these stages, theoretically.

2.1) Analysis: Stage 1

During stage 1, switch S_1 is closed and switch S_2 is open; so the inductor has one side shorted to ground and the capacitor is completely disconnected from the circuit, like so:

Let's think about the case where, just before entering stage 1, the system was at rest (i.e., no current through the inductor and no voltage across the capacitor). Then we enter stage 1 at time 0.

Answer the following questions, assuming these initial conditions; use V_IN, T_1, C, L, and whatever mathematical constants you need in order to answer the question.

What is the current through the inductor, i_{\rm L}, just after we enter stage 1?

i_{\rm L}(0^+) =~

We stay in stage 1 for some arbitrary amount of time T_1, where T_1 is much smaller than the period of oscillation we would get from this LC combination.

What is the current through the inductor, i_{\rm L}, just before we exit stage 1?

i_{\rm L}(T_1) =~

If we were to sketch the shape of i_{\rm L} during stage 1, which of the following best describes the shape we would see?

If we were to sketch the shape of v_{\rm C} during stage 1, which of the following best describes the shape we would see?

Check Yourself 2:

Summarize, in your own words, what is happening to the inductor and capacitor in stage 1.

Sketch out how both i_{\rm L} and v_{\rm C} are changing as functions of time over the course of stage 1. We'll add more pieces to this sketch in a moment, and you should be prepared to discuss your sketch during the first checkoff.

2.2) Analysis: Stage 2

Next, we enter stage 2! During this stage, switch S_2 is closed and switch S_1 is open, meaning that the inductor and capacitor are connected to each other directly (during this stage, this looks like the driven LC oscillator we saw in lecture before!):

What is the current through the inductor, i_{\rm L}, just after we enter stage 2?

i_{\rm L}(T_1^+) =~

We stay in stage 2 until the inductor current reaches 0 (call this time T_1 + T_2). T_2 is also intentionally much smaller than the period of oscillation we would get from this LC combination.

What is the current through the inductor, i_{\rm L}, just before we exit stage 2?

i_{\rm L}(T_1+T_2) =~

If we were to sketch the shape of i_{\rm L} during stage 2, which of the following best describes the shape we would see?

If we were to sketch the shape of v_{\rm C} during stage 2, which of the following best describes the shape we would see?

Check Yourself 3:

Summarize, in your own words, what is happening to the inductor and capacitor in stage 2.

Sketch out how both i_{\rm L} and v_{\rm C} are changing as functions of time over the course of stage 2. We'll add more pieces to this sketch in a moment, and you should be prepared to discuss your sketch during the first checkoff.

2.3) Analysis: Stage 3

Finally, in stage 3, both switches are open.

What is the current through the inductor, i_{\rm L}, just after we enter stage 3?

i_{\rm L}((T_1+T_2)^+) =~

We remain in this state for some arbitrary amount of time, until t = T_1 + T_2 + T_3, where T_3 is again much smaller than the period of oscillation we would get from this LC combination.

What is the current through the inductor, i_{\rm L}, just before we exit stage 3?

i_{\rm L}(T_1+T_2+T_3) =~

If we were to sketch the shape of i_{\rm L} during stage 3, which of the following best describes the shape we would see?

If we were to sketch the shape of v_{\rm C} during stage 3, which of the following best describes the shape we would see?

Check Yourself 4:

Summarize, in your own words, what is happening to the inductor and capacitor in stage 3.

Sketch out both how i_{\rm L} and v_{\rm C} are changing as functions of time over the course of stage 3.

Now imagine repeating the whole pattern two more times. Say that the capacitor's voltage after 1, 2, and 3 full cycles, respectively, are v_{C1}, v_{C2}, and v_{C3}. How do these values compare against one another?

v_{C1} = v_{C2} = v_{C3}

v_{C1} < v_{C2} < v_{C3}

v_{C1} > v_{C2} > v_{C3}

v_{C1} < v_{C2} = v_{C3}

v_{C1} > v_{C2} = v_{C3}

something else

Check Yourself 5:

Now, continue your sketches of i_{\rm L} and v_{\rm C} for two more cycles (two more complete times through stages 1, 2, and 3). Be prepared to talk about your sketches during your checkoff!

Checkoff 1:
Discuss your results from above (and your sketches!) with a staff member.

3) Build It!

Phew! Now that we've done some analysis to know what to expect, we can carry on with actually building the circuit! In order to do so, we're going to need to make a few small changes/additions to our circuit. The main difference is that we'll need to use real components to create switches that can be controlled electronically (rather than mechanically).

We'll use a MOSFET for one of our switches. We've seen MOSFETs before (as a variable current source, way back in week 4), but today we'll use them in a different way: as a switch. When the voltage on its gate terminal is high enough, it will open a path for current to flow from the drain to the source (effectively closing the switch).

For the other switch, we'll use a diode. The diode will allow current to flow in one direction (from the side without the line on in its symbol toward the side with the line) but not in the other direction, which effectively makes it act like a switch (we can model it as being an open in some cases and a short in other cases).

We'll also add another diode right next to the voltage source, to keep current from flowing back in that direction (this will keep things closer to the idealized model we analyzed above).

All told, our circuit will look like this:

Let's go ahead and build this. As you're laying our your circuit, take note of the following:

The 1.5V source in the diagram is a stand-in for our AA battery.
Build things without the battery connected first, and only connect it when you need to.
Try to keep things relatively compact on your breadboard if you can.
The inductor for today is on the table near the front of the room, but the capacitors are in the cabinet as usual.
The MOSFETs we're using are ultra cute and should fit right into the breadboard. Along with lots of other useful information, their datasheet will tell you which pin is which (if you have trouble finding it, let us know).
- Note also that, like the op-amps we've used, these MOSFETs are perfectly sized to cross the gap in the middle of the breadboard; you shouldn't need to bend any pins to hook things up.
The diodes have a little silver line printed on them, which needs to be oriented the same way as the line in their little symbol.
We're going to grab V_{\rm sq}, which will control the switch in our circuit, from the wave gen on the scopes. Use the following settings:
- square wave
- 85kHz
- 5Vpp
- 2.5V offset
- 50% duty cycle (the fraction of the time that the value is held high, versus low)

If you want us to check your circuit before powering it on, let us know!

4) Measurements

Let's also connect up a load resistor, as a temporary stand-in for the lights we'll hook up in a second. For now, use a 4.7{\rm k}\Omega resistor, hooked up like so:

Before we move on to the lights, let's make some measurements to make sure that the behavior of the circuit agrees qualitatively with what we expect. For each experiment, it's a good idea to disconnect your battery before performing the experiment, and only to connect it when you're ready.

Please read each of these carefully, and actually do all the steps. We'll expect you to be ready to talk about all of these during your checkoff.

Experiment 1: Output Voltage

The goal of this circuit is to produce a constant output voltage. Let's hook up a multimeter to measure the voltage drop across the load, and then connect the battery up. What value does it reach?

Check Yourself 6:

Our theoretical results from above indicated that this voltage should have kept climbing forever, approaching infinity. But here, our measured output voltage is decidedly finite. Why is this?

Think about your sketches from above; what is different about this circuit, compared to the one we analyzed earlier, and how does that affect how i_{\rm L} and/or v_{\rm C} are changing in the various stages?

Experiment 2: Duty Cycle Effect on Output Voltage

Try adjusting the duty cycle of the square wave down from 50% (you can measure V_{\rm sq} with the scope probes to see the effect this has on the input wave if you want). What effect does this have on the voltage?

Try adjusting the duty cycle upwards as well; what effect does this have?

Make a note of all of these, and be ready to talk about them during your checkoff.

Before moving on, adjust your duty cycle so that the output voltage is sitting at around 15V (remember that this is around where the lights were nicely lit up earlier).

Experiment 3: Effect of Load Resistance

Swap out the 4.7k\Omega resistor for a 1k\Omega resistor. What happens to your output voltage?

Check Yourself 7:

Why did the voltage change in this way? Think about it in terms of your sketches from earlier on.

Experiment 4: The Approach

Next, let's do another experiment to see, on a macro scale, how our output voltage changes as it approaches its final value.

Put the 4.7k\Omega resistor back in (replacing the 1k\Omega resistor), and set up your scope as follows:

Connect channel 1 so that you're measuring the voltage across the capacitor. Connect the channel 1 probe directly to the legs of the capacitor, not via wires into the breadboard.
Set the voltage scale for channel 1 to be around 2V/div
Slide the "ground" symbol for channel 1 down so that it's near the bottom of the screen
Put the trigger at around 7.5V
Set the timescale to be around 5ms/div
Click "single" (which will cause the scope to wait for a single trigger event and then freeze)

Then, disconnect the battery, and then quickly reconnect it. This will show us how things are changing in the early moments.

Note that in order to make this work, the connection needs to be made quickly. What I would suggest is leaving the ground side of your battery connected and then, instead of plugging the + side into the breadboard, just touch it to the leg of the diode briefly, like so:

You don't need to leave it held there very long; just long enough to trigger (which should happen fast).

Check Yourself 8:

If we work out the theory (which you'll get a chance to do in this week's p-set for a similar circuit), we'll see that the output voltage should grow roughly proportionally to \sqrt{n}, where n is the number of full cycles we've gone through. Does that match what you see here?

Take a photo of your scope screen so that we can discuss this shape during your checkoff.

Experiment 4.1: A Different Scale

To get a better sense of how this is happening, we'll repeat that same experiment but on a different scale on the scope. Adjust your scope as follows:

Set the voltage scale for channel 1 to be around 500mV/div
Slide the "ground" symbol for channel 1 down so that it's near the bottom of the screen
Put the trigger at around 750mV
Set the timescale to be around 10 microseconds/div
Click "single"

Then repeat the experiment, quickly connecting the battery using the method described above. What you should see on screen is how the voltage across our capacitor is changing over the course of the first few cycles through stages 1, 2, and 3.

Check Yourself 9:

Does the shape of this curve match your expectations? How does it compare to your sketch of v_{\rm C} from earlier?

Take a photo of your scope screen so that we can discuss this shape during your checkoff.

Experiment 5: Inductor Current

Let's also make sure that the inductor current is changing in the way we expect it to, from our graphs. Our scopes can't measure current directly, but we can use multiple voltage measurements to take care of this. Let's (temporarily) adjust our circuit to make a measurement by adding a little resistor, like so:

If we measure the voltage on the left side of this resistor on channel 1 and the voltage on the right side on channel 2, then their difference (which we can compute using the math functions on the scope) will be proportional to i_{\rm L}!

Check Yourself 10:

Go ahead and do this, and fully connect the battery (don't just do a quick touch for this one).

You should be able to adjust your zoom levels to get a nice, clear view of the inductor current's general shape. Does it agree with your sketches from earlier? Can you pick out stage 1, stage 2, and stage 3 on it?

As before, take a photo of your scope screen so that we can discuss it during your checkoff.

5) Light Show

OK, now to (finally) put our circuit to use! What we'll do now is to use the circuit to power our light strip from earlier (there should be plenty of voltage now!), like so:

Disconnect the battery
Remove the 10\Omega resistor we just put in, so that the inductor is once again connected directly to the diode.
Remove the 4.7k\Omega load resistor and replace it with our light strip (remember that the side with the little plastic protrusion should go on the + side of our output, and the side without it should go on ground).
Connect the battery again.
Adjust the duty cycle so the lights are as bright as you can make them.

Do your lights light up? Hooray!

Checkoff 2:

Discuss the results of all of your experiments above with a staff member, and demonstrate your working lights.

When you're all set, clean up by doing the following:

Discharge your capacitor by disconnecting the battery while either the lights or one of the load resistors is connected
Return your scope to its default settings:
- Click the "Default Setup" button
- Under "Wave Gen," click "Setttings" and then "Wave Gen"
Turn off your scope and power supply
Throw away your resistors and any loose wires
Return the other components to the front of the room (diodes, MOSFET, inductor, capacitor, lights, alligator clips, battery, battery holder, breadboard, basically everything that isn't a resistor or a loose wire)