Notes on Experiment #11

Be sure to print a copy of experiment #11 and bring it with you to your lab session. There will not be any copies available in the lab.

You should be able to finish this experiment very quickly. Use you extra time to practice with the equipment. Work on measuring techniques that are giving you trouble like voltage measurement with the scope (especially using the differential method) and current measurement using the DMM. The lab exam will be the last week of the term. We cannot provide you with practice time in the lab. So, use your extra time to practice.

This week we will do experiment 11 almost AS IS. Your data will be the graphical images on the display of the scope. So, BRING GRAPH PAPER! cm X cm is best since that is the actual scale of the scope display. You will be sketching the transient response of a RC circuit. We will also take a look at the capacitor as an integrating device.

Procedure

Part 1

Set up the circuit as shown in the lab manual with C = 0.01uF and R =27K. Set the amplitude of the square wave to 6 volts peak-to-peak (NOT the 20 volts indicated in the lab manual.). It is important that the frequency of the 6 volt (peak-to-peak) be exactly 200 Hz. Do not trust the scales on the function generator. The scope scales are much more accurate. So, do this:

1. Set the time base to 0.5msec/DIV.
2. Adjust the frequency control dial on the function gen. so that there is one complete cycle of the input and output in exactly the 10 horizontal divisions. (i.e. one cycle is 5msec and therefore f = 200Hz.)
3. Draw a large (half page at least) accurate sketch of the input V_S(t) and output V_C(t) (on the same sketch) just as you see it on the scope.
4. Repeat steps 1 to 3 with C = 0.02uF and C = 0.068uF.

Now we will be making an estimation of the values of the RC time constants (tau) for each sketch of V_C(t). To understand how to do this consider the following explanation.

For the part of V_C(t) that starts up at t = 0:

V_C(t) = V_CSS(t) + [V_C(0) - V_CSS(0) ]e^-t/RC where V_CSS(t) = -3 Volts (a constant for all time).

The slope of the tangent line to V_C(t) can be found by taking the derivative of V_C(t)

d{V_C(t)}/dt = (-1/RC)[V_C(0) - V_CSS(0) ]e^-t/RC

At t = 0 this becomes

d{V_C(t)}/dt|_t=0 = (1/RC)[ V_CSS(0) - V_C(0)]

So, for a vertical change of V_CSS(0) - V_C(0) the horizontal change is RC which is tau.

Note if V_CSS(t) is constant then V_CSS(0) = V_CSS(infinity) = V_C(infinity).

If you sketch the tangent line of V_C(t) from the point t = 0 to the -3 Volt line the the amount of horizontal change must be tau. Project that amount of change up to the t axis and you have graphically found the value of tau. See Figure 1.

Figure 1.

Part 2

Let R = 100K and C = 1uF.

At 200 Hz V_C will be very small compared to V_S as required. Use only one trial of V_S(t) for this part:

V_S(t) = 3cos(400(pi)t)

Is V_C(t) approximately 1/RC times the integral of 3cos(400(pi)t)?

Verify this by checking the amplitude and phase of V_C(t).

Do not use the square wave and triangle wave for input as suggested in the lab manual.

Circuit Analysis

Part 1

Use the model in the lab manual (with voltage sources at 3 volts and not 10 volts) to find the general expression for V_C(t). Calculate the expected value of tau for each capacitor.

Part 2

Show that if V_C(t) is very tiny compared to V_S(t) then V_C(t) approximately 1/RC times the integral of V_S(t). (Hint: if V_C(t) is very small then i_C(t) is approximately V_S(t) /R )

Read and know the setup of this experiment and

Have fun!