circuit explanation

8/7/2019 Circuit Explanation

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Circuit explanationfor the ultrasonic detection unit(1)

The transmitter circuit, the receiver circuit of the ultrasonic are the circuit which is the same as the ultrasonic

range meter.

Ultrasonic pulse oscillator

IC1 is the oscillation circuit to control the sending-out time of the

ultrasonic pulse.

The circuit is the same as the ultrasonic range meter but the value of

the resistors and the capacitors are changed. The oscillationfrequency is the same.

The time of the oscillation pulse can be calculated by the following

formula. Actually, with the error of the parts, it is different from the

calculation a little.

The condition : RA = 1M-ohm, RB = 15K-ohm, C = 0.1F

TL= 0.69 x RB x C

= 0.69 x 15 x 103

x 0.1 x 10-6

= 1 x 10-3

= 1 msec

TH= 0.69 x ( RA + RB ) x C

= 0.69 x 1015 x 103

x 0.1 x 10-6

= 70.0 x 10-3

= 70 msec


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Ultrasonic oscillator

IC2 is the circuit to make oscillate the ultrasonic frequency of

40KHz.Oscillation's operation is same as IC1 and makes oscillate at

the frequency of about 40 KHz. It makes RB>RA to bring theduty(Ratio of ON/OFF) of the oscillation wave close to 50%.

The frequency of the ultrasonic must be adjusted to the

resonant frequency of the ultrasonic sensor. Therefore, I am

made to be able to adjust the oscillation frequency by making

the RB the variable resistor (VR1).The output of IC1 is connected with the reset terminal of IC2

through the inverter. When the reset terminal is the H level,

IC2 works in the oscillation. The ultrasonic of 40KHz is sent

out for the 1 millisecond and pauses for the 68 milliseconds.

The calculation example of the frequency is shown below.The condition : RA = 1.5K-ohm, RB = 15K-ohm. C = 1000pF

TL= 0.69 x RB x C

= 0.69 x 15 x 103

x 1000 x 10-12

= 10.35 x 10-6

= 10 sec

TH= 0.69 x ( RA + RB ) x C

= 0.69 x 16.5 x 103

x 1000 x 10-12

= 11.39 x 10-

=11 sec

f = 1 / ( TL + TH )

= 1 / (( 10.35 + 11.39 ) x 10-6

)

= 46.0 x 103

= 46.0 KHz


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Ultrasonic sensor drive circuit

The inverter is used for the drive of the ultrasonic

sensor. The two inverters are connected in parallelbecause of the transmission electric power increase.The phase with the voltage to apply to the positive

terminal and the negative terminal of the sensor has

been 180 degrees shifted. Because it is cutting the

direct current with the capacitor, about twice of

voltage of the inverter output are appied to the

sensor.

Signal amplification circuit

The ultrasonic signal which was

received with the reception sensor is

amplified by 1000 times(60dB) of

voltage with the operational amplifier

with two stages. It is 100 times at the

first stage (40dB) and 10 times (20dB)

at the next stage.As for the dB (decibel), refer to

"Logarithm Table".Generally, the positive and the negative

power supply are used for the

operational amplifier. The circuit this

time works with the single power supply of +9 V. Therefore, for the positive input of the operational

amplifiers, the half of the power supply voltage is appied as the bias voltage and it is made 4.5 V in the central

voltage of the amplified alternating current signal. When using the operational amplifier with the negative

feedback, the voltage of the positive input terminal and the voltage of the negative input terminal become

equal approximately. So, by this bias voltage, the side of the positive and the side of the negative of the

alternating current signal can be equally amplified. When not using this bias voltage, the distortion causes the

alternating current signal. When the alternating current signal is amplified, this way is used when working the

operational amplifier for the 2 power supply with the single power supply.As for the operation of the operational amplifier, refer to "Operation explanation of the triangular wave

oscillator".

Detection circuit
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The detection is done to detect the received ultrasonic signal. It is the

half-wave rectification circuit which used the Shottky barrier diodes.

The DC voltage according to the level of the detection signal is gotten

by the capacitor behind the diode. the Shottky barrier diodes are used

because the high frequency characteristic is good.As for the Shottky barrier diode, refer to "Diodes".

Signal detector

This circuit is the circuit which detects the ultrasonic which

returned from the object. The output of the detection circuit is

detected using the comparator. At the circuit this time, the

operational amplifier of the single power supply is used instead ofthe comparator. The operational amplifier amplifies and outputs the

difference between the positive input and the negative input.In case of the operational amplifier which doesn't have the negative

feedback, at a little input voltage, the output becomes the saturation

state. Generally, the operational amplifier has tens of thousands of

times of mu factors. So, when the positive input becomes higher a

little than the negative input, the difference is tens of thousands of

times amplified and the output becomes the same as the power

supply almost.(It is the saturation state) Oppositely, when the

positive input becomes lower a little than the negative input, the

difference is tens of thousands of times amplified and the output

becomes 0 V almost.(It is in the OFF condition) This operation is

the same as the operation of the comparator. However, because the

inner circuit is different about the comparator and the operational amplifier, the comparator can not be used as

the operational amplifier.At the circuit this time, it connects the output of the detection circuit with the negative input of the signal

detector and it makes the voltage of the positive input constant.

Vrf= ( Rb x Vcc )/( Ra + Rb )

= ( 47K-ohm x 9V )/( 1M-ohm + 47K-ohm )

= 0.4V

So, when the rectified ultrasonic signal becomes more than 0.4

V, the output of the signal detector becomes the L

level(Approximately 0 V).

There is another device in this circuit. It is the diode (D) which

connects with the side of the positive input.The pulse sending-out timing signal of the transmitter is appied

to this diode. So, it makes not detect the transmission signal

which was crowded when sending out the ultrasonic signal from

the transmitter and going around to the reception sensor, making

the voltage of the positive input of the signal detector rise in the
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pulse sending-out timing signal.The transmission signal has the remaining signal even if it stops the transmission timing pulse. So, it makes

the falling of the transmission timing pulse gentle with the capacitor (C) and it is preventing from the mis-

detection by the remaining signal.The value of this capacitor is the one point which decides the efficiency of the equipment. The detection start

time becomes late when the value of this capacitor is big and can not do the short distance. The equipment this

time makes the transmission pulse long(About 1 millisecond) to make detect possible to the about 10-mdistance and makes the capacitor of the detector big a little. Therefore, the shortest distance becomes about 40

cm.To detect the short distance, making TL in IC1 short, the value of the capacitor of the signal detector must be

made small. Way, in the time that the ultrasonic goes and returns in the 30-cm distance at 20C, it is the 1.75

milliseconds.

Time measurement gate circuit

This circuit is the gate circuit to measure the time which is

reflected with the object and returns after sending out the

ultrasonic. It is using the SR (the set and the reset) flip-flop. For

the details of SR-FF, refer to "The operation explanation of the

D-type flip-flop".The set condition is the time which begins to let out the

ultrasonic with the transmitter. It uses the transmission timing

pulse.The reset condition is the time which detected the signal with the

signal detector of the receiver circuit.That is, the time that the output of SR-FF (D) is in the ON

condition becomes the time which returns after letting out the

ultrasonic.

Alarm detector
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This circuit is the circuit to judge

whether or not the reaching time of

the reception signal is shorter or

longer than the setup time.It sets the time using the 555 timercircuit.The transmission timing pulse is used

for the trigger of the timer. The 555

timers begin the operation when the

input trigger becomes the L level. So,

it is inputting the transmission timing

pulse through the inverter(It uses the

2 input NAND).The alarm output(E) becomes the L

level if the output of the timer(D)

becomes the H level and the output

of the measurement gate circuit(C)

becomes the H level. When the timer does in the time-out, the output of the timer(D) becomes the L level. So,

after that, even if the output of the measurement gate circuit(C) becomes the H level, the alarm output (E) is as

the H level.If the D point becomes the H level before the C point becomes the L level, the wrong pulse with the alarm

output (the E point) is output. In the actual operation, because that the C point becomes the L level is earlier

than the operation beginning of the timer, I don't put the delay circuit.

It made the range of the alarm from about 40 cm to 10 m. The most short distance is limited by the

transmission pulse mis-detection prevention circuit. The most long distance is limited by the transmission

level of the ultrasonic, receiver sensitivity and then the interval of the transmission of the transmission pulse.The time that the sound wave goes and returns in the 40-cm distance

When the ambient temperature is 20C, the propagation speed of the sound wave is 343.5

m/second.In the time to be propagated by 80 cm (the going and returning), it is as follows.

TS = 0.8/343.5

= 0.00233

= 2.33 milliseconds


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The time that the sound wave goes and returns in the 10-m distance

In the time to be propagated by 20 m (the going and returning), it is as follows.

TL = 20/343.5

= 0.05822

= 58.2 milliseconds

In the time of the alarm detection timer, it makes change with the variable resistor.It is possible to calculate as follows in the minimum time and the maximum time.

The time of the 555 timer = 1.1CR

( C=F R=ohm )

That is, it becomes R = T/1.1 C.At the circuit this time, 1F of the capacitor is used.

The resistance value to get the minimum time (The variable resistor=0 ohm)

RS = (2.33 x 10-3

) / (1.1 x 10-

)

= 2.12 x 103

= 2.12 K-ohm

At the actual circuit, 2K-ohm is used.

The resistance value to get the maximum time (The variable resistor=the maximum)

RL = (58.2 x 10-3

) / (1.1 x 10-6

)

= 53.0 x 103

= 53 K-ohm

As the value of the variable resistor, it becomes the value which deducted RS from RL.At the circuit this time, 50K-ohm is used.


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Alarm output circuit

The output of the alarm detector isn't as long as it

works a relay. The alarm output circuit makesconstant time output last making the output of the

alarm detector a trigger.It works a relay for about 1 second using 555

timer.

T = 1.1CR= 1.1 x 10 x 10

-6x 100 x 10

3

= 1.1 seconds

The output of the timer (the H level in case of

operation) is inputted to the transistor and a relay

is driven. The diode (D5) which is connected withthe coil of the relay in parallel is to protect a

transistor from the opposite electromotive force which occurs with the relay.

Also, the LED(D4) lights up at the same time as the relay works. Making it be possible to see this LED from

outside by installing it in the case. It is to make it easy to set the distance of the alarm detector.

The capacitor(C24) to be putting between the base of the transistor and the ground is put to delay the release

(the non-operation) of the relay when the timer does in the time-out. After the timer does in the time-out, the

following trigger signal is inputted a maximum of 70 milliseconds later. It is because the interval of the

transmission of the ultrasonic is the about 70 milliseconds. When detecting the obstacle continuously, the

output of the timer becomes the L level condition for the about 70 milliseconds every second. Even if the

output of the timer becomes the L level in short time, by the electric charge to have stored the capacitor(C24),

the relay continues to work. The value of the capacitor is due to the release electric current of the relay. In caseof the relay which was used this time, it is doing the continuous operation at 220 F.

Power circuit

The circuit this time is using the various oscillators. The

frequency of those oscillators is related with the measurement

precision. So, the power supply voltage to use must be stable.

Because the circuit is using the CMOS, the power supplyvoltage is OK rather than +5 V. The inner power supply

voltage is made +9 V by the 3 terminal regulator.

The input voltage is using about +12 V voltage in the relation

with relay drive voltage.


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circuit explanation

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