Set the function generator to output a TTL compatible signal

Set the function generator to output a TTL compatible signal (as described in the previous sections of the lab), with a frequency of 1 Hz.

Using the counter we built last lab, disconnect pin 14 (INA) from the switch, and instead connect the output of the function generator to pin 14. Take note of the automatic counting.

Increase the frequency of the function generator to 16 Hz.

Use CH1 of the oscilloscope to take the following measurements:

INA

QA

QB

QC

QD

Frequency

Period

Increase the frequency to as close to 32768 as you can get. This is a common frequency for crystal oscillators to be set to for time purposes (e.g. for a quartz watch). Use CH1 of the oscilloscope to again take the following measurements:

INA

QA

QB

QC

QD

Frequency

Period

Q1) If we wanted to design a watch that ticked every one second, and we used a 32.768K quartz crystal, how many 73LS93 chips would we need?

Q2) Draw the schematic that could be used to generate a frequency of 1Hz from a 32.768K quartz crystal. Use only 74LS93 chips. You only need to indicate connections that occur on pins INA, QA, QB, QC, and QD. Be sure to label the input from the crystal and the 1 Hz output.

Search the internet for the datasheet for the SN74LS04N. A good reference website is:

http://www.alldatasheet.com

Open the data sheet, and use the pin out as well as the logic layout to set up the following circuit using 2-74LS04 chips:

Click AUTOSET to display both CH1 and CH2 signals. (Note: Clicking AUTOSET will undue the 20 MHz bandwidth limit so reset those using the CH1/CH2 Menu buttons)

Use the POSITION knob located over CH1 to center the CH1 signal vertically. Next, use the VOLTS/DIV knob to increase the size of CH1. You will need to re-position CH1 vertically. Continue this process until CH1 is centered vertically and as large as possible while still fitting entirely on the screen.

Repeat step 4 for CH2. Try to line up the bottom of CH1 with the bottom of CH2. Note the top of the wave might not line up perfectly.

Using the TIME/DIV knob and the HORIZONTAL POSITION KNOB, expand the time base and zoom into a LOW-TO-HIGH transition. Expand the transition horizontally as large as possible while making sure to keep the entire transition on screen.

Q3) Draw what you currently see on the screen.

Click the CURSOR button at the top of the oscilloscope. Change “Type” to “Time”. In the cursor menu you can use the KNOB at the top to compare the relative times of two positions on the screen. The cursor on-screen menu works as follows:

The top button, “Type” allows you to choose different measurements.

The second button, “Source” allows you to choose the source that you are moving the cursor across.

The third button is inactive, but displays the TIME DIFFERENCE and VOLTAGE DIFFERENCE between the two cursors.

Click the fourth button to move the first cursor.

Click the fifth button to move the fifth cursor.

Take the measurements below and fill out the table:

Input t_r

Output t_r

t_{pd_LH}

Input t_f

Output t_f

t_{pd_HL}

Q4) What would be the low-to-high propagation delay for 1 inverter? Look in the data sheet and see if you can find the typical t_PLH for SN74LS04. Does your chip work better/worse than it should?

	INA	QA	QB	QC	QD
Frequency
Period

CH1_Probe 5 V CH2_Probe

Solution

1 • A coil in an ac generator rotates at 60 Hz. How much time elapses between successive peak emf values of the coil? Determine the Concept Successive peaks are one-half period apart. Hence the elapsed time between the peaks is ( ) ms .338 s 602 1 2 1 2 1 1- === f T . 2 • If the rms voltage in an ac circuit is doubled, the peak voltage is (a) doubled, (b) halved, (c) increased by a factor of 2 , (d) not changed. Picture the Problem We can use the relationship between V and Vpeak to decide the effect of doubling the rms voltage on the peak voltage. Express the initial rms voltage in terms of the peak voltage: 2 peak rms V V = Express the doubled rms voltage in terms of the new peak voltage : V\'peak 2 2 peak rms V\' V = Divide the second of these equations by the first and simplify to obtain: 2 2 2 peak peak rms rms V V\' V V = peak peak 2 V V\' = Solving for yields: V\'peak peak = 2VV\' peak a)( is correct. 3 • [SSM] If the frequency in the circuit shown in Figure 29-27 is doubled, the inductance of the inductor will (a) double, (b) not change, (c) halve, (d) quadruple. Determine the Concept The inductance of an inductor is determined by the details of its construction and is independent of the frequency of the circuit. The inductive reactance, on the other hand, is frequency dependent. b)( is correct. 4 • If the frequency in the circuit shown in Figure 29-27 is doubled, the inductive reactance of the inductor will (a) double, (b) not change, (