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four switch inputs, software-controlled backlight, custom characters, and buzzer output. The data output of the AM-HRR3 on pin 14 is fed directly to the serial ILM-216L display. The ILM-216L display unit provides 8 pins for interfacing. Pins 1 and 6 are tied to ground, and 5 Vdc is applied to pin 2 to power the display. Serial data input is applied to pin 3, while data output from the display is provided on pin 4. A piezo alarm buzzer can be connected to pin 5. The baud rate can be changed via pin 8, using the configuration/test input on pin 7. The receiver is coupled directly to serial input on pin 3 of the display. The receiver and display are both powered from the 5-V regulator at U1. A piezo buzzer is connected to pin 5 of the display unit to provide an audible indication that an alarm condition has occurred. In operation, once the alarm controller is triggered, the STAMP 2 sends out serial data through the transmitter to the receiver at RX. The STAMP 2 sends out a serial word to activate the buzzer three times to get your attention; you can modify the program to preset a longer alarm sound if desired. Once the buzzer has been activated, the STAMP 2 begins sending out a serial data message to the LCD display unit, which indicates which alarm channel has been activated. The transmitter and receiver combination will operate up to 3 kHz, so the serial data stream from the STAMP 2 to the transmitter was set up to 2400 baud. Note that by using the Abacom FM transmitter/receiver pair the system is capable of 9600-baud operation.

Problem 5-4

In the circuit of Fig. 5-4, what does the ammeter A say This involves finding the current in any given branch. The voltage is 12 V across every branch; Rn 120. Therefore In, the ammeter reading, is found by Ohm s Law: In E/Rn 12/120 0.10 A 100 mA. All of the In s should add to get the total current, I .There are 12 identical branches, each carrying 0.1 A; therefore the sum is 0.1 12 1.2 A. It checks out. And there aren t any problems this time with significant figures. Those two problems were designed to be easy. Here are two that are a little more involved.

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The STAMP 2 alarm controller and transmitter were mounted on a small PC board measuring 4 by 4 in. It is advisable to use an integrated circuit socket for the STAMP 2 controller, in the event of a failure. Be careful to correctly orient the STAMP 2 when installing in the IC socket. When assembling the transmitter/controller, be sure to observe the correct polarity of the capacitors, diodes, and LEDs. A 4-pin 0.010-in male header at P1 is soldered to the PC board. A mating 4-pin 0.010-in female header at J1 is cabled to a 9-pin DB-9 female RS-232 connector, for programming the STAMP 2 processor. If you elect to use the

Problem 5-5

DCF is more wallet oriented, hence more meaningful to you unless you happen to be an accountant The discounted free cash flow approach uses corporate performance measures that focus solely on real cash dollars flowing into and out of the company Corporate activities that produce additional net cash inflows to the company, such as increased revenue growth or increased net operating profit margins, have a positive effect on stock value These are good activities Corporate activities that produce net cash outflows from the company, such as higher income tax rates, higher capital investment or working capital requirements, or lower net operating profit margins due to increasing labor costs or other costs of production, have a negative effect on stock value These payments may be necessary from the corporation s perspective but they are not particularly good activities for the stock s value.

RT5* 1 2 6 7

Three resistors are in parallel across a battery that supplies E 12 V. The resistances are R1 22 , R2 47 , and R3 68 . These resistors carry currents I1, I2, and I3 respectively. What is the current, I3, through R3 This is done by means of Ohm s Law, as if R3 were the only resistor in the circuit. There s no need to worry about the parallel combination. The other branches do not affect I3. Thus I3 E/R3 12/68 0.18 A 180 mA. That problem wasn t hard at all. But it would have seemed that way, had you needlessly calculated the total parallel resistance of R1, R2, and R3.

Problem 5-6

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