My computer, which has previously been stolen, is one of the most valuable pieces of equipment in my home.
The text describes the specifications and an alternative solution for creating an alarm system. The system should sound and give a visual signal when an object, such as a computer, is lifted off its base. The specifications include the need for the alarm to sound immediately upon lifting, the ability to manually reset the circuit, independent reset options for the alarm and visual signal, a latch on feature for the visual signal, a 5-minute sounding period with an accuracy of ±30 seconds, and the ability to run off a low voltage supply like a battery pack.
The alternative solution involves using a motion sensor and a key switch. When the object is moved and the key switch is turned to “arm” the circuit, it would trigger a two-tone siren for maximum audible effect that would stay on for approximately 5 minutes (as per the specifications). Additionally, a flashing high-wattage filament lamp would latch on for the same duration.
Each would be able to be reset independently. I selected my solution over the previously described one due to its simplicity and affordability. The motion sensor requires complicated setup and configuration, ensuring it only responds to object movement and not other factors. On the other hand, implementing a two-tone siren would be more challenging and costlier, likely involving a pulse generator connected to a loudspeaker.
The cost of the flashing light would increase due to the need for an astable and a high voltage power supply to operate the high-wattage filament lamp. These added expenses could make the final product more cumbersome and limit its potential use. Project Development
NB: All output/input letters correspond with each other
Pressure Sensor (Input Sub-System)
Sub-System Testing
I used a multimeter (configured as a voltmeter) to measure the voltage at output A when the PTB switch was in both its normally closed and open states. The measurement was taken when the switch was open (i.
When the object is placed on it, the voltage level at A on the multimeter reads 0V. When the object is lifted and the switch is in its normally closed position, the display reads 4.
98V.I measured the voltage across the PSU terminals and found it to be 4.98V. The latch and reset for Processing Sub-System 1 is created using a D-Type Flip Flop, which is labeled as IC1.
When the clock input rises, the D-Type’s Q output will take on the state of the D input. Since the D input is tied high, the Q output will always become high on the first rising edge (when input A rises and the PTB switch closes). Additionally, the Q output will stay high for any subsequent clock input pulse.
The Q output is unconnected and the set pin is tied low. However, the reset pin is connected to a PTM switch (SW2) and a pull-up resistor (R2), allowing the Q output to be forced to 0 and resetting the latch until another falling edge is detected at the clock input. To test this subsystem, I measured the voltage at Q using a voltmeter while applying different logic levels to input A. When input A was set to logic level 1, Q was found to be at 0V.
I changed the logic level at the input to 0, and the output remained unchanged at 0V. I then raised the logic level at the clock input back up to 1, causing the D-Type flip flop to trigger on and the Q output to rise to 4.98V. The input state was then changed multiple times, but the output state remained unaffected by subsequent rising or falling edges.
When switch 2 was pressed and released, the Q output went back to 0V due to the momentarily high reset pin.
Visual Signal (Output Sub-System 1)
The visual signal consists of an LED (red, labeled D1) connected in series with a protective resistor (R2). It is driven by a transducer driver because the D-Type Flip Flop, which generates the signal for this output sub-system, is powered by the PSU but does not supply enough current to fully power the LED. The LED has a rating of 2V and 12mA.
Although my circuit is shown here running off of 5V, the finished product will run off of a 6V battery. If the LED has a voltage of 2V across it and draws a current of 12mA, the resistor must have a potential difference of 4V (as voltages in series add up to the total voltage, Vs). The value of the resistor (in ?) can be calculated using Ohm’s law: 4V ÷ 0.012A = 333?.
To ensure the LED’s protection, we utilize the E12 series to select the closest value above the calculated ideal value, which is 390?. R5 acts as a protective resistor and restricts the current to the transistor’s base.
During sub-system testing, when input B had a voltage level of 0, the LED remained unlit with a voltage across it of 0V. However, when input B had a high voltage (measured at 4.
96V) The LED was illuminated and had a voltage drop of 2.12V. This could be attributed to the allowable variation of the E12 series, which has a tolerance of 5%. The timer, known as Processing Sub-System 2, consists of a 555 timer configured as a monostable (IC2). It uses a 560KΩ resistor and a 470μF electrolytic capacitor to create the time delay. The input signal from A (pressure sensor) goes through a differentiator and then undergoes inversion using a Schmitt Inverter (IC3). This conversion is necessary since monostables are triggered by falling edges, whereas the pulse generated by the input sub-system when the object is lifted off its base is a rising edge.
On a falling edge, the C output will stay near the supply voltage for a specified period of time determined by the values of R4 and C1. The C output can be reset to 0 when pins 6 and 7 are set to logic 1, which charges the capacitor (C1) instantly. When switch 4 is pressed, it creates a path with negligible resistance, causing the capacitor to charge rapidly. The time delay of a monostable circuit can be calculated using the formula: “Time (s) = 1.”
The formula for determining the time delay produced by my monostable when triggered by a falling edge at pin 2 is “1 x resistance (MΩ) x capacitance (µF)”. By plugging in the values, we get “1.1 x 0.56MΩ x 470µF = 290 seconds”.
The duration of 290 seconds is 10 seconds less than 5 minutes, which meets the requirement I mentioned earlier – the time delay should be accurate within �30 seconds of 5 minutes. During the testing of the sub-system, the 555 monostable was triggered by a brief falling edge from the Schmitt Inverter. The average duration of the timed output pulse, based on 5 tests that ranged between 327 and 329 seconds, was found to be 328 seconds (with a voltage reading of 4.97V from the voltmeter used). It is worth noting that the duration I timed (328 seconds) may differ from the calculated duration of 290 seconds due to two main factors. Firstly, the E12 series of preferred values for gold band resistors allows for a tolerance range of 5%, with a maximum value of 588K? and a minimum value of 532K?.
Furthermore, human error may contribute to the time difference, which depends on the individual’s reaction time (in this case, mine). The measured value of R4 at 578 K? using an ohmmeter and the capacitance of C1 at 459�F using a capacitance meter. The Audible Alarm (Output Sub-System 2) includes components such as a transistor (labeled Q1), its protective resistor (R5), a buzzer (BZ1), and a protective diode (D2). The transistor is safeguarded by R6, which limits the base current.
The purpose of the transistor is to ensure that the load can be operated because the 555 monostable does not have enough current to power the buzzer. The buzzer generates the audible alarm for my circuit, while the diode safeguards the transistor from high voltages generated by inductive loads, like a buzzer. When input C, which is the output from the 555 monostable, becomes high at approximately 4.98V, the voltage at the transistor’s base will be 0.
The transistor turns on at 7V, resulting in a collector-emitter voltage of 0. The potential difference across the load matches the supply voltage. During testing, the buzzer had 0V across it when the voltage at C was low. However, when the voltage was high, the buzzer had the full 4.98V (supply voltage) across it and a current of 6.8mA was passing through.
The sound was caused by the high level of C during the monostable time period (refer to previous Sub-System Testing section). The complete circuit diagram provides a brief description of how it works. Note that for detailed descriptions, refer to each individual sub-system section. Switch 1 is closed when the computer or other object is placed on top of it. When the object is lifted off the switch, the logic level at the clock input of the D-Type Flip Flop (IC1) changes from 0 to 1. This change causes the Q output to take on the state of the D input, which is latched high to the supply voltage as it is rising edge triggered.
The D-Types output is raised to a high state, causing the LED (D1) to illuminate until the reset switch (SW3) is pressed. When the object is lifted off the switch, a rising edge is generated and passed through a differentiator (or spike producer). The resulting signal is then inverted using the Schmitt (IC3) to create a brief falling edge. This is necessary because without inversion, the input will remain low, causing the output of the monostable to remain high until another object is placed on Switch 1. This brief falling edge triggers the 555 monostable (IC2), causing its output to go high for a specific time period determined by the values of R4 and C1 (which was measured at 328 seconds, for more information refer to the separate Sub-System Testing section).
This causes the voltage at the base of the transistor to be 0.7, which results in the voltage across the buzzer (BZ1) being equal to the supply voltage. As a result, the buzzer will produce a sound for either 328 seconds or until Switch 4 is pressed. When Switch 4 is pressed, it provides a low resistance path and the capacitor instantly charges. This causes pins 6 and 7 to become logic 1 and the output to become logic 0. The diode is used to protect the buzzer from high voltages generated by inductive loads.
Simply put, when an object is removed from the pressure sensor (SW1), the LED stays illuminated until reset, while the buzzer remains on for around 328 seconds or until reset. During circuit testing, when Switch 1 is pressed (with the object on it), its resistance is essentially infinite, which indicates that it has the supply voltage across it (measured at 4 volts).
When the object (i.e. stolen) is removed, the switch closes with virtually no resistance, meaning 4.
The voltage across resistor R1, 98V, causes the clock input of the D-Type flip flop to be pulled to 4.98V (logic 1). This voltage level serves as the rising edge trigger that leads to the Q output becoming high, measured at 4.97V.
The transistor’s base voltage was set to 0.7V, turning it on. The LED had a voltage of 2.12V and a current of 12mA passing through it, resulting in full brightness.
The rising edge generated by the input (Switch 1 closing) was transmitted through the differentiator and a Schmitt inverter, resulting in a momentary decrease (from 4.98V to 0V) at pin 2, the input of the 555. Although this decrease was too short to be detected by the voltmeter and/or my eyes, I am aware that it happened because it caused the output (pin 3) of the 555 to become high, measured at 4.
97V, and this only happens when a falling edge is triggered. The base emitter voltage of the transistor was 0.7V, indicating conductivity. To determine if it was saturated, I measured the voltage between the collector emitter, which was 0V, proving saturation.
After conducting 5 tests, I observed that the buzzer took an average of 328 seconds to stop sounding when not reset. Please note that the LED remained on until reset. Additionally, I triggered the circuit again and tested the reset switches. The LED reset immediately upon pressing switch 3, and I measured a voltage of 4 at the reset pin.
The text describes the functioning of a monostable reset circuit. The circuit includes a switch 4 which charges a capacitor when pressed, causing the voltage across it to rise to 4.97V. This voltage change causes pins 6 and 7 to return to logic 1, resulting in the output returning to 0V and causing the buzzer to stop sounding. As part of a test, the voltage across the collector and emitter legs of the transistor was measured and found to be 4.
97V is the supply voltage that proves the transistor is off. The circuit has undergone thorough testing and meets all specification points. The alarm is triggered immediately upon lifting the object, with any time delay being very brief and insignificant to humans. The circuit can be reset manually by pressing switch 3 and/or switch 4 (see complete circuit diagram, page 6).
The visual signal can be reset independently by pressing either switch 3 or 4. The alarm will sound for an accurate period of time within �30 seconds of 5 minutes before automatically switching off. The alarm successfully met its accuracy target in all 5 tests, indicating that it meets the specification demands. Additionally, the circuit operated flawlessly and as expected when powered by a low voltage supply of 5V during testing.
I believe that it fulfills all requirements stated in my specification. The following is a list of costed components and the sources of information that were used:
- List of “Inductive Output Transducers” from Mr. Woolley’s class notes
- Pin out for logic gates obtained from doctronics.co.uk
- E12 series of preferred value resistors found on Wikipedia.org
- Costed component list created in Circuit Wizard, with prices from Circuit Wizard’s Rapid Electronics database
- All circuit diagrams were drawn in Circuit Wizard
- Schmitt Inverter IC # found on electronicsinschools.org
