| T O P I C R E V I E W |
| n/a |
Posted - Aug 24 2003 : 8:50:10 PM
I am using a circuit in my car that include some CMOS and 555 timers.
I know this chips work OK within 3 to 15 Volts.
The projects work perfectly in my bench with a 12 volt battery BUT when I conect it with the battery from the car it work ok until I actvate a switch in the car like the door lock.
Its look like my project is getting some type of spike or glitch from the car electrical circuit.
What I have to do in order to fix this??? Do I have to install some kind of debounce circuit???
Do I have to install a Voltage Regulator???
Can somebody give me detailed instrucctions or point me to a wep page were I can find more info about this or a circuit that I can copy.
Thank you for your time and help.
Edited by - Turbo Boss on Aug 29 2003 8:10:06 PM |
| 5 L A T E S T R E P L I E S (Newest First) |
| Aaron Cake |
Posted - Aug 29 2003 : 3:51:58 PM
The 1N4007 will work fine.
The capacitor won't discharge since there is a much more abundant source of power available. Even if it did discharge, it would then charge right back up after the sag finished. Remember, every electronic device in your car will have some decent sized filter capacitors at it's input.
|
| n/a |
Posted - Aug 28 2003 : 11:06:51 AM
quote:
The diode would go on the input side of your power supply. It is a kickback diode, which prevents negative-going spikes from hitting the circuit.
What diode you recomend me for this?? Do a 1N4007 will be ok??? Or do I have to use a bigguer one like a 3AMP?
Also, about the diode in series in the positive imput side + line ...... You says I dont need it. So what will prevent the capacitor to get suddenly discharge into the car electrical circuit when there is a load in the own car circuit?
THX |
| Aaron Cake |
Posted - Aug 28 2003 : 09:13:15 AM
quote:
I will going to use this schematic I find.........
www.eleinmec.com/figures/020_02.gif
I going to regulate to 5 volts and on the imput side I going to install a electrolitic cap 2,200uF/50V and a 1N4001 diode in series in the positive line so it will stop the capacitor suddenly discharging into the load...... What you think??
That will work fine, but you don't have to worry about the diode.
quote:
So you say to put also a paralel diode on the imput side or on the output side????
What this diode will do???
The diode would go on the input side of your power supply. It is a kickback diode, which prevents negative-going spikes from hitting the circuit.
quote:
Also, what you mean "ICs should be bypassed with a 0.1uF capacitor at the power supply pins" Let se if I understand: one leg of the 0.1uF capacitor goes to the + IC pin and the other capacitor leg goes to the regulated 5 volts OR one leg of the 0.1uF capacitor goes to the + IC pin and the other to ground.
One end of the capacitor goes to the power supply pin(s) of the IC, then other goes to ground.
|
| n/a |
Posted - Aug 27 2003 : 8:11:48 PM
Hey Aaron, THX for your reply.
I will going to use this schematic I find.........
www.eleinmec.com/figures/020_02.gif
I going to regulate to 5 volts and on the imput side I going to install a electrolitic cap 2,200uF/50V and a 1N4001 diode in series in the positive line so it will stop the capacitor suddenly discharging into the load...... What you think??
So you say to put also a paralel diode on the imput side or on the output side????
What this diode will do???
Also, what you mean "ICs should be bypassed with a 0.1uF capacitor at the power supply pins" Let se if I understand: one leg of the 0.1uF capacitor goes to the + IC pin and the other capacitor leg goes to the regulated 5 volts OR one leg of the 0.1uF capacitor goes to the + IC pin and the other to ground.
Can you clarify this Please
Thank you again for your time and help
Turbo Boss
Edited by - Turbo Boss on Aug 27 2003 8:13:49 PM |
| Aaron Cake |
Posted - Aug 26 2003 : 11:02:21 AM
You need to filter the supply voltage. The electrical system in a car is very noisy. A 100uF and 0.1uF filter capacitor at the power supply of your circuit will make a huge difference. Also, ICs should be bypassed with a 0.1uF capacitor at the power supply pins. Finally, connecting a diode in parallel with the power supply wires (anode to +, cathode to -) will make a big difference as well.
|
