obstacle detector

[meowth08]

We have a new project. It's a obstacle detector that uses ultrasonic sensor/s with the proper circuitry mounted on a toy car. The project should respond like this. Once it senses an obstacle, it would either turn left or right so that it would not hit the obstacle. Our professor noted that if we can make the toy car move backwards first before turning left or right, she would give a credit for that. I have already googled for this project and I found this link.



I just don't know if these circuits are correct for the transmitter and receiver. However, the transmitter is being drove by a 555 timer operated in the astable mode. I think I can do that one since I can always refer to the datasheet of NE555. For the receiver, I honestly can't understand how the circuit works.

Am I also to use Hbridge for the steer?

I would appreciate any useful information you can give.

 

[ericgibbs]

hi me8.
Which device are you using for the controller.?? eg: PIC or logic IC's

 

[alec_t]

I just don't know if these circuits are correct for the transmitter and receiver

In the receiver R15 seems too low in value and one of the two transistors T5,T6 seems superfluous. Nevertheless the Rx should work. The Tx draws quite a lot of current so a PP3 battery won't power it for long.

For the receiver, I honestly can't understand how the circuit works.

The signal from the RX transducer is amplified by two transistor stages T3,T4 and the amplified signal is rectified to give a DC voltage. This is compared by IC2 with a threshold voltage. If the threshold is exceeded transistors T5,T6 turn on to energise relay RL1.

 

[meowth08]

Sir eric,

We are going to use logic IC's for this project.

alec_t,

Thank you for that information sir. I was able to follow your explanation.

 

[meowth08]

I am studying the circuit again. I was not able to understand the use of T1, T2, D1, D2, R1, and R5 on the transmitter part.

On the receiver part, could I change the two transistors to a darlington pair since their function is just to amplify the received signal and tap the output to inverting input of the op-amp?

I will also remove T5 or T6.

And I have never used a relay before. I have read that it is used as a switch, will it control my h-bridge? What I am saying is, with no signal received by the receiver, means no obstacle. The toy car would just go straight. With signal detected, the relay commands my wheels to turn right/or left.

 

[alec_t]

I was not able to understand the use of T1, T2, D1, D2, R1, and R5 on the transmitter part.

IC1 outputs a square wave. When the wave is 'high', current through R3 and D1 turns on T1. When the wave is 'low', current through R3 and D2 turns on T2. The top terminal of the transducer is thus pulled towards the positive supply and ground alternately. D1 and D2 ensure that both T1 and T2 can't be turned on at the same time (which would short the supply).

On the receiver part, could I change the two transistors to a darlington pair

Just replace T5 with a BC807 or BC327 or similar, dump T6, replace R15 (33 Ohm) with 3k3.

will it control my h-bridge?

If you want it to. Or you could do away with the relay and use T5 to control the H-bridge directly.

no signal received by the receiver, means no obstacle

Are you sure? If you use a reflection system then the ultrasound is only received when there is an ostacle in front.

 

[Boncuk]

Hi meowth08,

you might consider using this transmitter circuit to preserve battery power.

The timer IC produces a 20% duty cycle pulsing the oscillator circuit built around U1:A and U1:B.

U1:'D gates the oscillator signal for an ON-time of 1ms while the pause is 5ms between pulse bursts.

The circuit uses 10.5mA of current.

Here are schematic and an oscilloscope screenshot. Channel A shows the NE555 timer output signal, channel B the 40KHz continous 40KHz wave signal of the oscillator circuit and channel C shows the output signal at the US-transducer.

Boncuk

 

[meowth08]

alec_t:

with no signal received by the receiver, means no obstacle.

Is this not true for reflection systems?

Boncuk, I'll still see if the price of those IC's are not very expensive.
Thank you for the information sir.

 

[Boncuk]

alec_t:



Is this not true for reflection systems?

If there is no signal received there can't be a valid output signal.

Boncuk, I'll still see if the price of those IC's are not very expensive.
Thank you for the information sir.

An NE555 costs $.22, a CD4011 is $0.32 and a CD4049 is $0.35.

Total cost for all ICs = $0.89.

I hope this is not beyond your budget.

Boncuk

 

[alec_t]

"with no signal received by the receiver, means no obstacle."
Is this not true for reflection systems?

Yes. Brain not engaged when I commented

 

[meowth08]

Boncuk,

Sir, Im here in the Philippines. As of January 27, 2012, exchange rate is 42.713 Philippine peso per US dollar. But, that's not where my point is. Only four electronic shops owned by Chinese are available near me. They get their supplies from Manila which is the capital of my country. When the components arrive here in my place, the price is already twice or three times the original price. What I can remember is NE555 is worth 35 Philippine peso. That is already about .82 dollars.

Added later:

I forgot. I already have CD4011 and and NE555. I'm lucky!

 

[Boncuk]

Hi meowth08,

if I remember correctly the project is a school assignment and I expect your school to pay for the necessary parts.

Thinking of all the features your lecturer wants I guess he might as well participate in sharing expenses if school won't pay.

Boncuk

 

[meowth08]

That is not the case here in the Philippines sir. This is a third world country. I go to school. I pay for my tuition fee. The school pays the professor. It ends there.

Sir, I can't clearly see the capacitor values.

One question sir, is C5 connected to ground?

 

[Boncuk]

Hi meowth08,

please forgive me my ignorance.

Thailand is somewhat also a third world country and schools provide everything necessary.

Colleges and universities are not free of charge, but an annual fee of Thai Baht 3,000 (US$100) is affordable.

The technical college here has the finest equipment one can imagine, starting from PCs for everyone via design software up to Gerber photo plotters and electrochemical plating machines to make professional PCBs.

If all the burden for parts is on students in the Philippines you should take care of making your circuit designs anti copy proof by building in some faulty connections to hand out to your lecturer.

He might use your design and sell it to an industrial company for further development.

Here is the reworked circuit design using Eagle. The PCB measures 1.55X2.20".

Parts should now be easily to identify. C5 through C8 are decoupling caps which should not be omitted for proper function.

Boncuk

 

[meowth08]

Thank you sir.

If all the burden for parts is on students in the Philippines you should take care of making your circuit designs anti copy proof by building in some faulty connections to hand out to your lecturer.

I am always doing it sir. Actually this is a group work. Since I'm the one who is giving more effort in our group, I don't give the correct copy of the design to my group mates. They might sell it to other students. But sometimes, my conscience tells me not to do so since I get free help/information from ETO.

Question sir, US3C is an unused not gate?

 

[Boncuk]

Yes, it is an unused gate.

CMOS inputs must always be connected to either ground or VDD.

Boncuk

 

[meowth08]

CMOS inputs must always be connected to either ground or VDD.

They never mentioned this in our lecture. Thanks again sir. I didn't expect that my question will lead me to knowing more than what I asked.

 

[meowth08]

Please check on the circuit appearing on the pdf.

For the receiver, R15 will be changed to 3K3. T6, RL1, and D5 will be dumped and T5 will be changed to BC327. The additional circuit I have is to be connected on the collector of T5 (BC327). By the way, the aim of the circuit is to make the vehicle move backwards first, then steer the wheels and go forward.

Boncuk gave me this suggestion:

Boncuk: The moment an obstacle is sensed the first action should be reversing the drive motor for a certain time period. After the period has ended the steering motor has to be energized for some time and when that timing ends the drive motor will have the normal forward signal again.

Boncuk: All you'll need is two different monostable circuits which you can make using NE555s.

Unfortunately, I was not able to understand why I need two different monostable circuits. However, considering the aim, I came up with the circuit which appears on the PDF.

 

[meowth08]

I'll try to explain the circuit sir. Kindly verify.

On the receiver circuit:

T6, RL1, and D5 will be dumped.

On the circuit you gave me:

Q1 turns on when the PBNO closes. The PBNO represents T5(PNP transistor) on the receiver. T5 turns on when the output of the comparator is low. C2 net charge is still zero when Q1 is not turned on. Hence, no current flows to the trigger(pin 2) of U1. When Q1 turns on, the other side of the capacitor(left side) will be effectively grounded. Current flows through R3 and C2 will be charged for a very short period and then it discharges allowing current to flow to the trigger (pin 2) of U1. U1 is a monostable multivibrator. The on time is determined by the values of R1 and C3. When output of U1 is high, it turns on Q2. RL1 is then energized so the motor reverses. When U1 output is low, RL1 which is normally open switches back to the original position. The on time of U1 is also the time the motor reverses. Triggering of U2 is the same with the triggering in U1. It happens when output of U1 is already low. The on time of U2 is determined by R7 and C5. When output of U2 is high, it energizes RL2 and steering takes place. When U2 output is low, the vehicle is moving straight again.

I have a question sir. Isn't it that even Q2 is off, RL1 is still energized through the path going to the LED and R5?

 

[Boncuk]

I'll try to explain the circuit sir. Kindly verify.

On the receiver circuit:

I have a question sir. Isn't it that even Q2 is off, RL1 is still energized through the path going to the LED and R5?

For the LED the resistance of the relay coil is negigable compared to the additional 470Ω (R5) resistor. So the relay can only be activated by connecting its bottom pin to the collector of Q2.

When Q2 is turned off the relay would have to take ground via D2 and R5, hence the relay is deactivated.

Boncuk