Member - Posts: 39
Joined: March 25, 2006
Location: United States
Posted: October 30, 2006 at 3:02 PM / IP Logged
I am wondering whether I should put a 11 Ohm 1/4 Watt resistor, or maybe a 12 Ohm 1/2 Watt resistor..... I just want it to slow the rate of the discharge so it would last approximately 4 seconds instead of just a rapid discharge (Which would completely eliminate the 300,000 capacitor's reason for being there!)
KPierson
Platinum - Posts: 3,526
Joined: April 14, 2005
Location: Ohio, United States
Posted: October 30, 2006 at 3:28 PM / IP Logged

You lost me on your last post.  I think you are trying to slow the discharge rate to the coil of the relay.  However, like the filament in the lamp, the coil in the relay is your 'resistor'.  Once you select a relay you can measure the resistance of the relay and calculate the capacitance.  However, one thing to look out for is the drop out voltage of the coil.  Most 12vdc relays pull in around 12vdc and drop out around 6 (I'm guessing again).  So, once your cap discharges to 6vdc your relay will drop out (although it will continue to drain the cap).

I would shoot for the 30 second delay, but I would throw in an extra relay that will kill the relay delay off part of the ciruit if the ignition is turned on.  That would be sweet.

Kevin Pierson
KPierson
Platinum - Posts: 3,526
Joined: April 14, 2005
Location: Ohio, United States
Posted: October 30, 2006 at 3:31 PM / IP Logged
Oh, one more thing.  The wattage of the resistor tells you how much heat it can dissipate, not how much power it will create.  A 12 ohm resistor at 12 volts will create 1A, therefore a 1/2 watt resistor would not be big enough (watts = voltage x current).  In electronics you generally double all your heat ratings, so you would want a 24 watt (12vdc x 2A) 12 ohm resistor if you were to use that size resistor at 12vdc (also consider your max voltage in a car to be closer to 16vdc in a worst case situation).
Kevin Pierson
master5
Silver - Posts: 1,123
Joined: October 10, 2006
Location: United States
Posted: October 30, 2006 at 4:36 PM / IP Logged

The power rating of a resistor is simply that...a rating. Just like a lightbulb. That rating never changes BUT, if you exceed it's rating it will get quite hot and probably burn out. Put as with wire gauge, it will never hurt anything to go extra large, it can only help.

Ok next, I know you have simplified what you wish to do but now I am still curious since I put some thought into your initial plan.

Does anyone know if using a cap and resistor in parallel with the light allow the light to fade on. As I stated I base this theory on the princible of a cap. It will block DC. So it would seem logical that the discharged cap would shunt through the resistor, This shunting or short will not allow power to reach the bulb. As the capacitor charges the light should fade on. One the cap is fully charged it should have no other effect on the circuit or light.

This sound like it would work anyone?

master5
Silver - Posts: 1,123
Joined: October 10, 2006
Location: United States
Posted: October 30, 2006 at 4:54 PM / IP Logged

Oh wait. To answer my own question I realize this won't work. With a resistor it will never see a short and the current would simply branch through the resistor and light.

Now if we put a cap by itself in parallel I imagine this would have no effect as well as it would charge instantly hence no fade.

Out of curiosity does anyone know a simple RC circuit that would case a light to fade on?

haemphyst
Platinum - Posts: 5,052
Joined: January 19, 2003
Location: Michigan, Bouvet Island
Posted: October 30, 2006 at 6:09 PM / IP Logged
There really isn't one... POSSIBLY you could use a resistor divider network, but it's gonna be kludgy, and not terribly glamourous. Also, your light will never run at full brightness...
Actually, belay that... Let me draw something up. Got it happening in my brain right now. I'll get back with you!
It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice.﻿ You should write it down."
master5
Silver - Posts: 1,123
Joined: October 10, 2006
Location: United States
Posted: October 30, 2006 at 6:43 PM / IP Logged

Kudos to you if you can figure this out haemphyst, but if it's possible, I know you will.

I am severly out of practice with circuit design and really haven't had much use for it the past several years until joining here. But I have a feeling over time it will come back ,however so slowly and with plenty of mistakes (re-learning/learning process) along the way.

I racked my brain for a while with the "fade-on" and after several failed attempts have given up. Other then some kind of IC or module I can't for the life of me determine how a simple RC can do this.

Thanks and good luck.

master5
Silver - Posts: 1,123
Joined: October 10, 2006
Location: United States
Posted: October 30, 2006 at 6:51 PM / IP Logged

There must be a way, although I don't know how simple, to build an electronically controlled variable resistor (potentiometer). It would need the input voltage to decrease the output resistance over time, until zero resistance (or very close to it) is accomplished and the light fades on.

The question remains...how to do design this using nothing but resistors, caps and perhaps diodes/relays. hmmm

Member - Posts: 39
Joined: March 25, 2006
Location: United States
Posted: October 31, 2006 at 7:21 AM / IP Logged

I think I got it guys...... brb

Member - Posts: 39
Joined: March 25, 2006
Location: United States
Posted: October 31, 2006 at 9:23 AM / IP Logged

Okay guys, this is what I come up with......

Now I have heard of the time delay relays, if these relays were .2 second delay then it would take 2 seconds for the full fade effect. The reason those resistors are in that order and size is because....

1. Total of 150 Ohm resistance, Use regular relay.

2. Total of 135 Ohm resistance, Use a  .2 second delay relay

3. Total of 120 Ohm resistance, Use a .2 second delay relay

4. Total of 105 Ohm resistance, Use a .2 second delay relay

5. Total of 90 Ohm resistance, Use a .2 second delay relay

6. Total of 75 Ohm resistance, Use a.2 second delay relay

7. Total of 60 Ohm resistance, Use a .2 second delay relay

8. Total of 45 Ohm resistance, Use a .2 second delay relay

9. Total of 30 Ohm resistance, Use a .2 second delay relay

10. Total of 15 Ohm resistance, Use a .2 second delay relay

11. There is no resistor, it should allow light to be on full brightness. Use a .2 second delay relay

The regular dome light I have is 14 ohms, so this setup will have the affect of having 10 lights in a series circuit, and every .2 seconds 1 bulb is removed.... in turn making it brighter! The capacitor in for the sudden surges of the next relay turning on and to cause a fade effect to take placewhen it is shut off, it will last approximately 3 seconds .

Page of 7

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot delete your posts in this forumYou cannot edit your posts in this forumYou cannot create polls in this forumYou cannot vote in polls in this forum

•
 Search the12volt.com