UChicago Instructional Physics Laboratories

Resistors: 1k, 2k, 2.2k, 100k

Capacitors: 560pF, 0.01  { $\mu$ F, 15 [Math Processing Error]μ { $\mu$ F  *

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Our { $0.01\mu$ F capacitors are suitable for this, you don't need something special This variety or RLC circuit's resonant frequency is  { $f_{resonance} = \dfrac{1}{2\pi \sqrt{LC}}$  (see page 114 of lab book)

Remember that -3dB point indicates that power is halved; voltage is reduced to   { $1/\sqrt{2} \approx 70\%$  of the peak value The signal frequency is displayed at the bottom right corner of the screen; alternatively you may use the “Measure” button to set up a frequency measurement.

You'll probably want to trigger the scope on rising square wave for this.

“While watching a square wave, remove the probe's ground clip” → disconnect the wire connecting your oscilloscope's ground to the rest of the circuit.

You aren't being asked to calculate the rms voltage here, however an example of where it comes from is included below.

“rms” stands for root mean square, often used for finding the power delivered by a time-varying signal.

Mathematically, this means taking the voltages from one period of your function, squaring them (to get all positive values), finding the mean of that over one period , and then taking the square  root  of the result (to get units of voltage again).

Consider a sine wave: { $V_{out}(t) = V_0\sin (2\pi f t )$ where [Math Processing Error]V0 { $V_0$ is the amplitude and [Math Processing Error]f { $f$ is the frequency (and [Math Processing Error]τ=1/f { $\tau = 1/f$ is the period) This function squared is { ${V_{out}(t)}^2 = V_0^2\sin^2 (2\pi f t )$ $ The mean of this function over one period is the integral of that function over one period divided by the time of a single period, as follows:

{ $Mean (V_{out}(t)^2) = \frac{1}{\tau}\int_{t=0}^{t=\tau} {V_{out}(t)}^2 dt = \frac{1}{\tau}\int_{t=0}^{t=\tau} V_0^2\sin^2 (2\pi f t )dt$ { $Mean = \frac{V_0^2}{\tau}\int_{t=0}^{t=\tau}\sin^2 (2\pi f t )dt$ using the half-angle formula { $\sin^2(\theta) = \frac{1}{2} - \frac{1}{2}\cos(\theta)$ , we find { $Mean = \frac{V_0^2}{\tau}\int_{t=0}^{t=\tau}\left(\frac{1}{2} - \frac{1}{2}\cos (2\pi f t )\right)dt$ We can split up the integral by pulling out the constant term:

{ $Mean = \frac{V_0^2}{\tau}\left(\frac{\tau}{2} - \int_{t=0}^{t=\tau}\left(\frac{1}{2}\cos (2\pi f t )\right)dt\right)$ Now, we take advantage of the fact that the integral of { $\cos (\theta)$ over a full period is zero, this simplifies to { $Mean = \frac{V_0^2}{2}$ remember that his is the mean of the voltage squared; to get the rms value we take the square root of this function:

{ $\sqrt{Mean} = \sqrt{\frac{V_0^2}{2}}$ This finally simplifies to:

{ $V_{rms} = \dfrac{V_0}{\sqrt{2}} \textrm{for a sine wave}$ where [Math Processing Error]V0 { $V_0$ is amplitude.

You can use 1N400X power diodes for this part if there aren't enough others, just stick to only a single variety in the circuit at once.

The book is being overly coy here, here's a translation of some parts:

Attaching a ground connection to one side of the secondary transformer coils will effectively result in one of the diodes being placed in parallel with the transformer (a voltage source) without any resistance.

What happens if you put  { $V>0.6 V$  across a diode without any resistance?  Look at section 3N.5 (pg 119) for a diode's IV curve Skip the paragraph asking about diode failure; you don't have a model that can answer that right now

Do the subsequent investigation of the 0V regions.

“Observe ripple, given C and load:”

That is a description of what you'll be doing, not a particular direction.

Our 15 { $\mu$ F capacitors are polar; see the picture below for how to use them

{ ${/download/attachments/205261933/Cap%20Guide.png?version=1&modificationDate=1555005676000&api=v2}$

Click to make larger

The “ripple” amplitude { $\Delta V_{ripple}$  is given by [Math Processing Error]ΔVripple=VpeakRΔTC { $\Delta V_{ripple} = \dfrac{ V_{peak}}{R} \dfrac{\Delta T}{C}$  where: { $V_{peak}$  is the maximum voltage of the output { $R$  is the load resistance { $\Delta T$  is half the period of the signal { $C$  is the capacitance of our attached capacitor This is assuming that the voltage drop at the output will be due to a current { $I = \dfrac{V_{peak}}{R}$ passing through a capacitor [Math Processing Error]C { $C$ for the time it takes to go from one maximum to the next ([Math Processing Error]ΔT { $\Delta T$ ) This (roughly) follows from section 3N.7.2 of your lab textbook, with the added assumption that the maximum current  { $I$  is equal to the maximum voltage [Math Processing Error]Vpeak { $V_{peak}$  over the resistance [Math Processing Error]R { $R$

You'll need to use the 5V power supply at the top of your board for this.

Remember to connect the function generator ground to the board's, otherwise strange things may occur.

Don't estimate the dynamic resistance here; you don't have a good model for how to do that right now.

If there aren't 2k resistors, use 2.2k and 1.1k for the voltage divider; the result will be similar enough for the effect we're looking for.

To get the Thévenin equivalent model of the voltage divider, we know that

{ $V_{out,\, no\, load} = 15 V \dfrac{1k\Omega}{1k\Omega + 2k\Omega} = 5V = V_{Thevenin}$ The short circuit current for this circuit is found from assuming that the 1k is shorted out (i.e. 15V is across the 2k resistor)

{ $I_{short} = \dfrac{15 V}{2k\Omega} = 7.5 mA$ The Thévenin resistance is thus

{ $R_{Thevenin} = \dfrac{V_{Thevenin}}{I_{short}} = 1.5 k\Omega$ Thus, the following circuit should behave the same as the one depicted in Figure 3L.8

{ ${/download/thumbnails/205261933/InterpretedClamp.png?version=1&modificationDate=1555013985000&api=v2}$

*Voltage clamp using Thévenin equivalent of voltage divider.** The representation has been changed to emphasize how components are connected You don't need to build this

This model might help you explain why the output of your 'clamp with voltage divider reference' output changes so much.