Pull-up, Pull-down
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In the hardware problem from week 2, where we attempted to use a potentiometer to create a variable voltage to create a lightbumb "dimmer" switch (but failed...). In this problem, we'll look at some related circuits that may help explain why that demo behaved the way it did, and in future weeks we'll explore some related circuits (and try to fix the issue with that demo!).
Note that these questions allow a limited number of submissions.
1) Reminder: Potentiometers
A potentiometer is a three-terminal device whose electrical properties depend on the angle of a mechanical knob. We can think of a potentiometer as being made up of two resistors, whose resistances sum to some value R_P, and whose resistances vary with the angle of the knob, which we will quantify using a number \alpha: one resistor has resistance \alpha R_P, and the other has resistance (1-\alpha)R_P. Schematically, we can represent pots in the following two ways:
The quantity \alpha corresponds to the angle of the knob, normalized to be in the range [0, 1]. \alpha=0 corresponds to the knob being turned all the way in one direction, and \alpha increases as the angle increases, until reaching \alpha=1 when the knob is turned all the way in the other direction.
As the angle of the knob increases, the resistance between the bottom and middle terminals increases and the resistance between the middle and top terminal decreases. These changes in resistance occur such that the sum of the top and bottom resistors is constant.
The illustrations below show how these resistors are actually connected inside the potentiometer:
The way this is implemented is with a single band of resistive material (total resistance R_P) and a "wiper" that moves with the knob. As the angle varies between from \alpha=0 to \alpha=1 and the wiper arm moves, the resistance between the middle terminal and each of the two outside terminals will change. The internals look something like the drawing below:
2) Reminder: Approximating Series/Parallel Combinations
Consider the series combination of a 500\Omega resistor and a 5\Omega resistor. Which of the following is closest to the equivalent resistance of this combination?
- 5\Omega
- 250\Omega
- 500\Omega
- 1{\rm k}\Omega
Since v = iR, the 5\Omega resistor accounts for less than 1% of the voltage drop across the combination. Thus, it can be largely ignored.
The exact equivalent resistance, of course, is 505\Omega, but we can approximate the combination as just the 500\Omega resistor since the smaller resistor contributes so little.
In the last "check yourself" question, we noticed that we can sometimes approximate the series combination of two resistors with resistances R_1 and R_2 as \text{max}(R_1, R_2). When does this approximation break down?
And when R_1 = R_2, it isn't very good at all, since the series combination of R_1 and R_2 is, in that case, 2R_1.
Consider the parallel combination of a 500\Omega resistor and a 5\Omega resistor. Which of the following is closest to the equivalent resistance of this combination?
- 5\Omega
- 250\Omega
- 500\Omega
- 1{\rm k}\Omega
Since i = v / R, the 500\Omega resistor accounts for less than 1% of the current flow through the combination. Thus, it can be largely ignored.
The exact equivalent resistance is slightly tedious to solve, but we can approximate the combination as just the 5\Omega resistor since so little current flows through the bigger resistor.
In the last "check yourself" question, we noticed that we can sometimes approximate the parallel combination of two resistors with resistances R_1 and R_2 as \text{min}(R_1, R_2). When does this approximation break down?
And when R_1 = R_2, it isn't very good at all, since the parallel combination of R_1 and R_2 is, in that case, R_1 / 2.
3) Problem
Each of the graphs below shows the output voltage v_o of a certain circuit as a 5{\rm k}\Omega potentiometer is turned linearly from \alpha = 0 to \alpha = 1.
Response A:
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Response B:
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Response C:
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Response D:
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Response E:
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Response F:
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Response G:
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3.1) Configuration 1
In this configuration, we will consider a circuit with a single 5{\rm k}\Omega potentiometer (R_P = 5{\rm k}\Omega), and a load resistor with resistance R, connected as shown below. The intention is not to solve exactly, but to use the estimation techniques from above to estimate the answers.
3.2) Configuration 2
In this configuration, we will consider a circuit with a single 5{\rm k}\Omega potentiometer (R_P = 5{\rm k}\Omega), and a load resistor with resistance R connected as shown below. The intention is not to solve exactly, but to use the estimation techniques from above to estimate the answers.
3.3) Configuration 3
In this configuration, we will consider a circuit with a single 5{\rm k}\Omega potentiometer (R_P = 5{\rm k}\Omega), and two load resistors with resistance R, connected as shown below. The intention is not to solve exactly, but to use the estimation techniques from above to estimate the answers.
