Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

my explaination

Status
Not open for further replies.
555

Hi,
If i don't connect the pin 5 (Voltage Control) with a capacitor to GND, what will happen? I've seen a circuit which uses that pin with a variable resistor to control the voltage, then what it the purpose of connecting it with a cap to GND? The 555 timer will still work well without it right?
 
Look at the datasheet for the 555. Every circuit has a capacitor from pin 5 to ground if pin 5 isn't used for sweeping the frequency.
Without a capacitor, pin 5 can pickup interference, even from the power supply, that can vary the frequency.
 
From the datasheet, the maximum power dissipation is 600mW, so the output shouldn't be more than that, P=VI, right? How if i short pin5 to GND without a cap? Then the 555 won't work right?
 
bananasiong said:
From the datasheet, the maximum power dissipation is 600mW
No, that is the rating for a tiny surface-mount package. A regular-size LM555 has a max dissipation of 1180mW.

so the output shouldn't be more than that, P=VI, right?
No. The power dissipation in a load can be much more than the power dissipation in the 555.
15V supply, 200mA load. The output of the 555 when driving the load will be about 2.5V or 12.5V so it dissipates 500mW. The load dissipates 2.5W.

How if i short pin5 to GND without a cap? Then the 555 won't work right?
Of course it won't work.
 
No. The power dissipation in a load can be much more than the power dissipation in the 555.
15V supply, 200mA load. The output of the 555 when driving the load will be about 2.5V or 12.5V so it dissipates 500mW. The load dissipates 2.5W.
U mean, let's say i supply 5v to the 555, the output is 3.3v (from the datasheet). 600mW is dissipated by the 555, and the rest of it by the load. Let's say the load is 330Ohm, 0.5A*5V=2.5W, 600mW dissipated by 555, 1.9W by the load.
But the resistor used normally is only 1/4 watt, so we should increase the resistance??
 
bananasiong said:
U mean, let's say i supply 5v to the 555, the output is 3.3v (from the datasheet). 600mW is dissipated by the 555.
No. A 555 barely works with only a 5V supply. Its max current is about 100mA and its output voltage is 3.3V. Therefore it would dissipate 170mW and a 33 ohm load would dissipate 330mW.

Let's say the load is 330Ohm, 0.5A*5V=2.5W, 600mW dissipated by 555, 1.9W by the load.
No. The output voltage would be about 3.5V and the current in the 330 ohm resistor would be only 10.6mA. Therefore the 555 dissipates 24mW and the resistor dissipates 56mW.
 
??
No. The output voltage would be about 3.5V and the current in the 330 ohm resistor would be only 10.6mA. Therefore the 555 dissipates 24mW and the resistor dissipates 56mW.
I know the current is the division of the output and the resistor, 3.5/330. But how do u get the 555 dissipates 24mW and the resistor dissipates 56mW?
P=VI=VV/R=IIR
but i still can't get the 24mW and 56mW..

the other question (look my attachment if u've forgotten the circuit):
I'm doing a robot mower, i've tested the navigation system, it works well. When it goes to the multiple loops of wire (cat5), from the output of the 555, it will detect the electromagnetic field and turns to other direction.
Without running the blades (powered by 2 DC motors), it works well. I didn't connect them to the supply because i thought that it is just blades, nothing more, and it is quite noisy when running, so i test the operation without turn it on.
But when the time i try to turn on them, maybe because of the electromagnetic field produced by the DC motor, the LM393 circuit keeps response to it. Maybe is interference. The wires placed quite near.
Can i use the alluminium foil to cover the wires from the tank circuit to avoid the interference (red color from the diagram)? Then one end of the foil connected to GND. Is that called shielding interference?
 

Attachments

  • Rx.GIF
    Rx.GIF
    3.2 KB · Views: 131
Last edited:
The 555 and the 330 ohm load resistor are in series.
I goofed on calculating the current.
The 555 dissipates about 15.9mW and the 330 ohm resistor dissipates 37.1mW.
 
You need to keep the motors away from the detector coil.
If the motors are permanent magnet ones with brushes, then the brushes make sparks which transmits interference. A 0.1uF ceramic disc capacitor with very short wires across the motor will help reduce the interference.

The supply also needs a 0.1uF ceramic disc capacitor across it.
 
The 555 and the 330 ohm load resistor are in series.
I goofed on calculating the current.
The 555 dissipates about 15.9mW and the 330 ohm resistor dissipates 37.1mW.
Yes!! I got it!!

You need to keep the motors away from the detector coil.
But their place are fixed..
If the motors are permanent magnet ones with brushes, then the brushes make sparks which transmits interference. A 0.1uF ceramic disc capacitor with very short wires across the motor will help reduce the interference.

The supply also needs a 0.1uF ceramic disc capacitor across it.
the capacitor with short wire connect with the motor in parallel? What about the supply?
 
bananasiong said:
But their place are fixed..
Since the detector coil picks up the magnetic field from the motors then your project doesn't work. Can you mount the detector coil out on a stick in front of the robot?

the capacitor with short wire connect with the motor in parallel?
Yes.

What about the supply?
Connect a 0.1uF ceramic disc capacitor at the supply pins of the LM393.
 
Can you mount the detector coil out on a stick in front of the robot.
The tank circuit is already sticked in front of it, but still affected, i'm worried that, the wire connected from the tank to the input of the 393 will pick up the field.

Connect a 0.1uF ceramic disc capacitor at the supply pins of the LM393.
one end to the +ve and the other end to the -ve? means that in parallel?

*Can i shield the wires or shield the motors with the alluminium foil to avoid the interference as i mentioned before?
 
bananasiong said:
*Can i shield the wires or shield the motors with the alluminium foil to avoid the interference as i mentioned before?
Connect the metal covers on the motors to 0V with a wire.
Use shielded cable like is used to connect stereo components together (cut off the RCA connectors) to connect to the tank.
 
Connect the metal covers on the motors to 0V with a wire.
Yes.. Thanks.. The metal body of the motor won't cause any short circuit if connected to GND, will it?
Then do i still need to include the bypass capacitor?

Use shielded cable like is used to connect stereo components together (cut off the RCA connectors) to connect to the tank.
What is shielded cable? How to get it? Can i make it myself? What is the RCA connector?

*I need to use a diode to make to current to flow only one way, it is connected series with 2 DC motors. I was using 1N34A, but after it was turned on and off for once, the diode burned (i think) and cause open circuit. So i replace it with a 1N4001, now it works well.
My question is, if i replace the 4001 woth a LED, does it use more current?
 
bananasiong said:
The metal body of the motor won't cause any short circuit if connected to GND, will it?
Probably not.

Then do i still need to include the bypass capacitor?
Of course it is needed.

What is shielded cable? How to get it? Can i make it myself? What is the RCA connector?
In my country, the stereo tuner, CD player and amplifier are connected together with shielded cables with RCA plugs. I cut off the plugs and use the shielded wire.

I need to use a diode to make to current to flow only one way, it is connected series with 2 DC motors. I was using 1N34A, but after it was turned on and off for once, the diode burned (i think) and cause open circuit. So i replace it with a 1N4001, now it works well.
My question is, if i replace the 4001 woth a LED, does it use more current?
I don't know why current in your circuit doesn't flow one way without a diode. Is it AC?
Look at how much current the motors use when they work hard. Much more current than the max current ratings for the 1N34A or an LED.
Don't use an LED in series, connect a current-limiting resistor in series with the LED and connect the combination to the power source of the motors to indicate when they run.
 
I've shielded the wires and the motor. I've tested using batteries which are fully charge and not fully charged (not full). When i use the lower power, my robot mower operates well, no interference. When i use the fully charged batteries, of course, the motor run faster and interference occurs.
The motor should run as fast as possible for good mowing system, but the electromagnetic field produced by the motor is bigger.
Can i shield the wires and the motor thicker (turn more round) to solve this? I use alluminium foil (for cooking) to shield them.
 
Shielded cable or aluminum foil will stop the sparking of the motors from radiating electrical interference. But will do nothing for the magnetic interference.
 
!!?!???!?!?!??
What should i do besides moving them far away to each other?
Is there anything to remove or direct the unwanted electromagnetic field to somewhere else?
 
bananasiong said:
!!?!???!?!?!??
What should i do ?
I guess using a coil for a sensor is not the correct thing to do in a machine that is powered by electric motors.
Hee, hee. Diesel engines? Hee, hee.
 
Oh no... you're kidding right? I mean the diesel engines.. haha.. i use only AA's.. I'll try to solve it..
Anyway, you've helped me a lot, thank you very much.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top