Hi everyone. I don't know if this is the correct place to ask this, but i am gonna give it a go.
I have my finals exam tomorrow and will have to explain a circuit which was handed out a couple of days ago. I understand most of it, but i will have to present it for about 20 minutes and i don't feel that i got enough information to do that. So i was wondering if any of you could help me explain the attached circuit more specifically?
Regards, Adam.
Here is the text from the paper which was handed out:
"This circuit (Fig. 1), used in conjunction
with a thin piezoelectric plate,
senses the vibration generated on
knocking a surface (such as a door or a
table) to activate the alarm. It uses readily available,
low-cost components and can
also be used to safeguard motor vehicles.
The piezoelectric plate is used as the sensor.
It is the same as used in ordinary
circuit. When someone knocks on the
door, the piezoelectric sensor generates
an electrical signal, which is amplified
by transistors T1 through T3.
The amplified signal is rectified
and filtered to produce a low-level
DC voltage, which is further amplified
by the remaining transistors. The
final output from the collector of pnp
transistor T6 is applied to reset pin 4
of 555 (IC1) that is wired as an
piezobuzzers and is easily available in the
market.
The piezoelectric plate can convert any
mechanical vibration into electrical variation.
As it doesnt sense sound from a distance
like a microphone, it avoids false
triggering. The plate can be fixed on a door,
cash box, cupboard, etc using adhesive. A 1-
1.5m long, shielded wire is connected between
the sensor plate and the input of the
astable multivibrator. Whenever the collector
of transistor T6 goes high, the astable
multivibrator activates to sound an alarm
through the speaker. The value of resistor
R12 is chosen between 220 and 680 ohms
such that IC1 remains inactive in the absence
of any perceptible knock.
When the circuit receives an input signal
due to knocking, the alarm gets activated
for about 10 seconds. This is the
time that capacitor C5 connected between
the emitter of transistor T4 and ground
takes to discharge after a knock. The time
delay can be changed by changing the
value of capacitor C5. After about 10 seconds,
the alarm is automatically reset.
The circuit operates off a 9V or a 12V
battery eliminator."
I have my finals exam tomorrow and will have to explain a circuit which was handed out a couple of days ago. I understand most of it, but i will have to present it for about 20 minutes and i don't feel that i got enough information to do that. So i was wondering if any of you could help me explain the attached circuit more specifically?
Regards, Adam.
Here is the text from the paper which was handed out:
"This circuit (Fig. 1), used in conjunction
with a thin piezoelectric plate,
senses the vibration generated on
knocking a surface (such as a door or a
table) to activate the alarm. It uses readily available,
low-cost components and can
also be used to safeguard motor vehicles.
The piezoelectric plate is used as the sensor.
It is the same as used in ordinary
circuit. When someone knocks on the
door, the piezoelectric sensor generates
an electrical signal, which is amplified
by transistors T1 through T3.
The amplified signal is rectified
and filtered to produce a low-level
DC voltage, which is further amplified
by the remaining transistors. The
final output from the collector of pnp
transistor T6 is applied to reset pin 4
of 555 (IC1) that is wired as an
piezobuzzers and is easily available in the
market.
The piezoelectric plate can convert any
mechanical vibration into electrical variation.
As it doesnt sense sound from a distance
like a microphone, it avoids false
triggering. The plate can be fixed on a door,
cash box, cupboard, etc using adhesive. A 1-
1.5m long, shielded wire is connected between
the sensor plate and the input of the
astable multivibrator. Whenever the collector
of transistor T6 goes high, the astable
multivibrator activates to sound an alarm
through the speaker. The value of resistor
R12 is chosen between 220 and 680 ohms
such that IC1 remains inactive in the absence
of any perceptible knock.
When the circuit receives an input signal
due to knocking, the alarm gets activated
for about 10 seconds. This is the
time that capacitor C5 connected between
the emitter of transistor T4 and ground
takes to discharge after a knock. The time
delay can be changed by changing the
value of capacitor C5. After about 10 seconds,
the alarm is automatically reset.
The circuit operates off a 9V or a 12V
battery eliminator."