Electronics coursework

Fri. Mar 18, 12:33PM, 1 year and 6 months ago

Electronics coursework is finally finished, it took me around a week and a half to get it built, I’m now in the testing stage, finding ways to improve effeciency and just assessing the overall system performance.

Overall system function explanation:
The system function can be made by utilising several subsystems; in logical order these are: input subsystems, a logic system, a timer system, a frequency generator, a driver, and finally the output subsystem.

The first of these subsystems are the inputs, there are two inputs, the headlights and door opening. The headlight signal is generated by another, irrelevant subsystem, when the headlights are on, a ‘high’ signal is generated. The second input is a Push-to-make switch, when pressed the output, is ‘high’ for a very short period, however it can be held down to create a high signal for a longer period of time, however the process of pressing the PTM switch momentarily when the door is opens is an engineering, not electronic task to solve.

In order to trigger an AWD for a set amount of time, a 555 monostable must be created. This generates a ‘high’ signal for a custom amount of time, however, this IC is triggered only by a falling-edge pulse, from 1 to 0, high to low. In order to create this low-going pulse, a logic subsystem must output 0, to trigger the monostable only when; the headlights are on (logic 1) and the door is open (logic 1), it must not however, output 0 when the lights are open and the door is closed, and when the lights are off and the door is open. This can be explained the boolean table as shown:

0 0 1
1 0 1
0 1 1
1 1 0

Where A is the headlights, and B is the door.

As you can see from the above, Q is low, when headlights (A) AND the door (B) are high, this can be achieved easily using a NAND gate. The NAND gate takes the two inputs and AND’s and then NOT’s them, generating a 0 signal. If the headlight is off, and the door is open, the signal is high and will not make a falling-edge pulse and will be high. If the door is closed, and the headlight is on, the output signal is also high. Since the PTM switch is only triggered 1 for a millisecond or so, the output of the NAND gate will also be 0 for a millisecond or so, this logic system generates the falling-edge signal required to trigger a monostable.

When the input voltage signal to the monostable, generated by the output of the NAND gate, Vin, goes below 1/3 of the supply voltage, +Vs, the monostable output, Vout, becomes +Vs and the discharge terminal becomes an open circuit. This allows the capacitor, C, to charge via its series resistor. The output voltage will remain at +Vs until the voltage across the capacitor becomes greater than the threshold switching voltage, which is 2/3 of +Vs. When this happens, Vout will return to 0V and the discharge terminal will connect back to 0V, this causes the capacitor to discharge, creating Vout’s 1 signal. This state is stable and will remain so until Vin becomes less that 1/3 +Vs, or until the output of the NAND gate goes low.

The stable high output signal of the monostable can then be used to trigger a frequency generator, in this case a 555 Astable, via it’s reset terminal. When the astable is first switched high by the monostable, via the reset, the capacitor, C is discharged and so the voltage across this capacitor is less than the trigger voltage, so the astable’s Vout, becomes +Vs. The capacitor then charges through the two series resistors, until the voltage across it is greater than the threshold switching level, at this point Vout becomes 0V and the discharge terminal is connected to 0V. The capacitor now discharges through the 2nd series resistor, until the voltage across it becomes less than the switching voltage. When this happens, Vout now becomes +Vs, and the process repeats until the monostable time period is over, i.e no signal at reset is received.

Since in reality the fluctuating output voltage does not actually become +Vs, due to the internal construction of the 555 IC, the actual voltage will depend upon the current flowing through the output, which is typically only few mA, the actual output voltage is approximately +Vs -1 V. A few mA at +Vs -1 is unsuitable to drive a high power component such as a loudspeaker, which is what is going to be used for the Audible Warning Device. Therefore a driver is required to operate it. In such a case where switching high and low at 2.8kHz is required, an Enhancement Mode metal-oxide-semiconductor field-effect transistor (MOSFET) is the only available solution. When the output of the Astable is sent to the gate of the MOSFET, it begins to saturate high at 0.1V when the Vout of the astable is about 2V. Any further increase in Vin has no effect on Vout, the MOSFET is therefore operating as a switch in the same way as transistor, and can be used to drive the loudspeaker.


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