Very simple capacitor question!

Discussion in 'The Projects Forum' started by circa27, Jun 19, 2011.

  1. circa27

    Thread Starter Member

    Feb 22, 2011
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    So I have a lamp that has 2 settings, High and Low

    High is regular 12v power at 55 watts, I assumed "Low" was 6 volts or somehow just a drop in power. After rewiring to use a relay to switch the lights on, I found that the "Low" is actually short and rapid pulses of 12v to give the halogen filament a dimmer look since its going too fast to see a fluctuation in brightness of the filament

    Now for the question

    How do I determine which capacitor to use to attempt to even out the charge from pulsing to steady flow of current? I ask because now my relay is switched on by the original power supply to the light, so on high the relay just turns on but on "Low" the relay rapidly clicks and whether that is okay or not, its too loud for my application

    Help please :/
     
  2. WellGrounded

    Member

    Jun 19, 2011
    32
    2
    circa27,

    I am a new member and this my first reply. What is the voltage supply to the lamp? Is it like a desk reading lamp with 120 VAC that is stepped down to 12 VAC, like track lighting. If so, it easier to accomplish what you want by putting a high amp diode (10 amps or more) in series with one of the secondary wires to your light. When the diode is working you only get dim light (half power) because you are only getting a pulsed 1/2 wave cycle. To get full power flick a switch that is wired NORMALLY OPEN across the diode and it will shunt the diode, allowing the current to flow through the light on both parts of the AC cycle.

    If you have DC supply then you can look at a 555 Timer circuit with a power transistor on a heat sink. That will give you a range of brightness since it will act as a dimmer switch. That will require a proper schematic.

    Hope this helps,

    WellGrounded
     
  3. circa27

    Thread Starter Member

    Feb 22, 2011
    38
    0
    Thanks for the quick reply, its actually just straight from a 12v power source, no step down. It is also DC as far as I know, and since it is not a ccfl style light it would not need to be AC so lets just assume DC for now.

    I dont need anything that specific, I just need HIGH only. no dimming choice as the power is going to signal the relay now, not to a light. That's why the relay is clicking, its getting the pulsed power that was originally for LOW on the halogen bulb. So I just want to smooth out the LOW setting, and since the relay is sensitive to ~4 to 24 volts It doesnt matter what comes out of the capacitor, as long as the current is not pulsing anymore so the relay will not click rapidly
     
  4. Pencil

    Active Member

    Dec 8, 2009
    271
    38
    For a better understanding search for "Pulse Width Modulation" or "PWM"

    Post up some sort of drawing/schematic of what you have done.

    It may be possible to replace relay with a transistor (BJT) or MOSFET
    commonly available at any electronics supply or Radioshack for literally
    just pennies (well maybe not pennies lower quantities).
     
  5. wayneh

    Expert

    Sep 9, 2010
    12,100
    3,034
    You want your relay to be the "R" in an RC tank. You need a "C" big enough to hold voltage high enough to keep the relay closed between pulses. To make the calculation of the RC time constant, you need to know R for the relay. And you need to know what time you're shooting for, the time between pulses. We usually assume 60Hz but if this thing is pulsed DC, it could be generating an even higher frequency. That'd be good, as it reduces C. Note that you may need a diode too, to prevent reverse current during the "off" phase.
     
  6. wayneh

    Expert

    Sep 9, 2010
    12,100
    3,034
    +1
    You really should explain your overall goal
     
  7. circa27

    Thread Starter Member

    Feb 22, 2011
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    0
    Okay I'll try to make it more clear, I've been told of PWM's before tho I'll look that up next.

    So basically, the ORIGINAL apparatus was just a halogen bulb with OFF HIGH and LOW

    off was off

    low was pulsed

    high was continuous

    NOW I have the original power supply triggering a relay to give the bulb power off of another 12v source (a battery in this case) and all works well. I just dont like the idea of having LOW trigger the relay with pulsing current, as it seems it may cause failure from the mechanical wear.

    Regardless, I thought LOW was 6v instead of 12v, but instead its just 12v pulsing. so the relay does not do what it was intended to (give a constant state of HIGH from the 12v battery even when triggered by the "6v" I thought it was getting) It just clicks rapidly.

    Correct me if I'm wrong, the PWM takes a continuous DC current and chops it up into pulses with varying times to simulate varying wattage in the case of a filament bulb? I want the opposite! I have pulsing current telling the relay to switch on and off rapidly. I WANT it to actually convert the pulsing to be a continuous smooth current so the relay will not rapidly engage/disengage, it will then just stay on as if it was getting the HIGH signal

    Would a capacitor not simply store the voltage and let it out in a *more* continuous flow? In which case could I not have multiple capacitors to even the flow until the relay is no longer pulsing?

    EDIT: I feel like I'm over explaining a very simple issue. The relay is getting the ON signal from a pulsing source. It needs to be continuous instead. I thought a capacitor would achieve this, any help now?
     
    Last edited: Jun 20, 2011
  8. WellGrounded

    Member

    Jun 19, 2011
    32
    2
    circa27,

    I can simplify this for you. Get an NO/OFF/NO toggle switch. Substitute your 12 Volt battery for two 6 Volt batteries in series. Attach the common of the switch to your light. Attach one of the NO contacts to the 12 volt output of the two batteries and attach the other NO contact to the middle point 6 volt output. You now can have either the 6 or 12 voltage with the toggle switch.

    Let the lower battery be of a higher ampere hour rating since it will be used more often since it will be used for both 6 and 12 volts.

    If the battery is in a charging circuit then special precautions have to be taken when charging batteries in series with different ampere hour ratings. I can go deeper into that later with you if that is the situation. It's a simple workaround.

    Well Grounded
     
  9. circa27

    Thread Starter Member

    Feb 22, 2011
    38
    0
    For the application I have here its on a boat actually, so the power is the boat battery. The switch is built into the boat therefore I cannot substitute it. The socket that powered the original light on the boat is now the 'trigger' power supply to tell the relay to take power directly from the battery to the lights.

    So please can someone just help explain or set me in the direction of how to do this just by adding a component before the relay? That is my one and only option at this point
     
  10. wayneh

    Expert

    Sep 9, 2010
    12,100
    3,034
    Well, you still need to know the current (R, resistance) needed by the relay. That's what determines how big a cap you need. If you've got electrolytic capacitors in your parts box and can test to see if one works, another approach is to simply start out with the biggest one you have. Big, because relays in general need a lot of current to remain energized, and because more is better in this particular app.

    Since you're on a boat, I'd look for a voltage rating on the capacitor of at least 20v, and 50v might be better. That's because the system is noisy and you need to protect against spikes.

    Put your cap across the power supply to the relay, so that it's in parallel with the relay. Pay attention to the polarity. In the "+" power to the relay, insert a diode. This also has a polarity. If it doesn't work, turn it around. The size of this diode depends on the current needed by the relay, so again we need to know something about that relay. The diode is for good measure to prevent reverse discharge of the cap when the power pulse is in the off state, forcing the cap to discharge ONLY thru the relay. It's possible that the diode is not needed, but I'd assume you need it until you can prove otherwise.

    I want to emphasize this a bit more: If your pulse cycle changes the positive lead from positive to ground instead of from positive to open, then the cap will see a short across its poles on every cycle. This will drive a lot of current thru the cap - possibly destroying it - and may also overload your pulse generator. That's because, if the cap discharges, it looks like a dead short when the next pulse comes along. And the cap will not be performing any useful purpose, just causing these problems. The diode solves all concerns over this.
     
  11. WellGrounded

    Member

    Jun 19, 2011
    32
    2
    circa27,

    This is starting to sound a little confusing when you talk of taking the power from "another battery" to feed your light. You will have to describe in detail the series power flow through the devices in 12 Volt mode and in 6 Volt mode. It sounds like you have programmable relay when it is electronically pulsing (toggling) fast with a 50% Duty Cycle. Explain better how the light socket ties in to the power arrangement and is the light bulb still part of the original boat lighting circuit with the bulb in the socket? Does the relay have sets of SPDT contacts (NO and NC). Give the make and model and refer to the pin out connections of your circuit if you can, that will be so helpful. The relay may be the key to the answer.

    The smoothing capacitor will work if the + pulsed output first goes to the + of a large diode and then the diode - is wired in to the + of the capacitor, which is wired which is in parallel with your light bulb. The resulting voltage pattern across the light bulb will look like shark fins back to back.

    There is still another simple solution if you can give an accurate description of your power circuit and an accurate description of your relay.

    WellGrounded
     
  12. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    Have you??

    Wow, Your project is unique! It has the ability to morph. This same lighting system was first in your automobile, then your house and now it's in your boat. I'm getting dizzy! Which is it??

    http://forum.allaboutcircuits.com/showthread.php?t=53813 <<< It's head lights here
    http://forum.allaboutcircuits.com/showthread.php?t=53921 <<< It's home lighting here

    Bottom line.. If you're having trouble getting accurate answers, it's because you haven't been giving accurate information.

    FYI: Boat electronics modifications fall under the same AAC TOS.
     
    circa27 likes this.
  13. circa27

    Thread Starter Member

    Feb 22, 2011
    38
    0
    Okay that sounds good, So the diode would go AFTER the cap, and before the relay right? That way its keeping the current smooth and in the same direction before it triggers the relay?
    Or would I just put a diode pre cap as well just to protect the capacitor?

    And its okay CDDRIVE, I tried for automotive lighting with my DRL before and you guys said it was against the site to discuss (lol) so I made it morally correct for you with a house and boat analogy to achieve the same help. I'm just here to learn about the circuitry that I need to do, and it will get done whether you guys approve of my vehicular lighting or not so helping is not a crime. Not to mention I just completed a full OEM retrofit for HID projectors, so I'd say my vehicles lighting is probably more appropriate than yours anyways. No need to lecture me about my ride.

     
  14. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    Initially, I questioned AAC's TOS regarding automotive electrical mods but not any longer. One of the Mods once opined that if a TS doesn't possess the knowledge to do the work on his own he has no business messing with it. Reading back over your three same topics with different titles, I now totally agree.
     
  15. wayneh

    Expert

    Sep 9, 2010
    12,100
    3,034
    Nope, other way. Sorry I didn't make it clearer before. Murphy's law rules again. Look up "peak detector" if you want to understand it better.

    But, this is really a hack. Just because it may work doesn't make it an elegant to solution to whatever it is you're doing. And just because I've helped you accomplish what you asked for doesn't mean you asked for the right thing. It appears you duped me into violating the TOS of the forum, so I'm done for now.
     
  16. WellGrounded

    Member

    Jun 19, 2011
    32
    2
    circa27,

    The more info you add the more it becomes a little more confusing. What is needed is a power flow chart of the light in 12 Volt mode and in 6 Volt mode. Also, identify the make and model of the relay. It sounds like it may be a programmable type. Does it have sets of NC and NO contacts? Show how the second battery source ties in and identify it differently from the boat's 12 Volt battery supply. Also, in your flow chart identify the specific pin out connections in the series or parallel connections. A sample flow chart is shown.

    High mode - Boat 12 Volt Source -> Light switch -> 12 Volt Light
    Low mode - Boat 12 Volt Source -> ????? -> ?????

    WellGrounded
     
  17. circa27

    Thread Starter Member

    Feb 22, 2011
    38
    0
    Haha okay well this may be my last post here since I'm working on it tomorrow based on what I have researched

    But just so everyone can sleep at night, here is exactly what I have and what I'm doing., no analogies to avoid breaking the TOS as some of the members here are too sensitive to read what I have to ask

    Initially, being a North American built vehicle I have 2 light settings. DRL and ON. one for daytime obviously, one for full power at night. Now I have retrofit OEM HID projectors from an Infiniti SUV to improve my light output and to avoid glaring people with a cheap PnP HID kit in my factory reflector housings.

    The one and only problem is that when I turn my headlights "off" its actually DRL, so its a pulsing 12v source that basically gives a PWM effect for the halogen filament, making it appear dimmer for daytime driving. If anyone here is familiar with gas filled lighting, it usually requires a ballast to spike the voltage and ignite the gaseous bulb when there is no filament. This is the problem. Since the ballasts for my headlights pull too much current initially to ignite the HID bulbs, it would fry my thin factory wiring in the car. So what most of us do is use a relay with a separate (lower gauge) harness to safely pull power from the battery to ignite the lights and not harming the factory wiring. The factory wiring is still used to trigger the relay which as you all know will allow current to run from the battery through the new harness to the ballasts that will then ignite and power the HID lights.

    The point of that being that youre factory headlight switch still functions as normal, just without risking your factory wiring.

    Since the DRL current is pulsing through the factory wiring to the relay, the relay is rapidly engaging and disengaging which still gives the ballasts pulsed current through the new wiring harness. This is veryy bad for an HID ballast to receive pulsing current. So what I was trying to ask was how could I smooth out the current so that the relay does not flicker when it receives the pulsed current from the factory DRL mode signal. I have a capacitor for my stereo to protect the amplifier from power spikes, so I assumed a similar principal could apply for this application.

    If a mod chooses to lock this thread or ban me that is irrational, but thats fine. I spent a very long time with this lighting setup to be as safe for myself as well as others on the road, I tried to be good on the forum and keep in the TOS while getting proper advice, but it seems I've hurt a few peoples feelings while doing so. Now that you know why I was asking about the capacitor, maybe you'll understand that what I'm doing is completely safe and I just came to learn. Or be ignorant and assume anything automotive is dangerous and lock the thread, up to the mods I guess.

    If the thread stays open I'd still appreciate the info you guys have. If not, I'll figure this out another way. Thanks

    EDIT: to clear things up more, there never was a boat or a house involved. Just my car
     
  18. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    It doesn't really matter how big the capacitor is just so long that it keeps the relay energized. Go to Radio Shack and look for a couple of 2200uF/35 volt capacitors. If they don't have 2200, get whatever is close. The more the merrier. Get a 1N4001 diode too. They sell them 10 for a buck or so.
    Use some clip leads and put a cap across the relay coil. Energize it with a 12v battery (watch the +'s and -'s). Remove the power and see how long the relay stays energized. If it's longer than the chatter was, you're set. If it's too quick, add another cap in parallel with the first one and try again. It can't be too long (within reason).
    Solder the cap(s) and the diode to the relay. The banded side of the diode (cathode) goes to (+) of the cap and one relay pin. The other side of the diode (anode) goes to (+) of your PWM/pwr supply/whatever it is. The other side of the cap (-) and the other relay pin go to pwr supply (-)
    If the cap keeps the relay for a long period, it won't matter. Once the power is removed from the lamp when it is turned off, the relay is a moot point.
     
    circa27 likes this.
  19. circa27

    Thread Starter Member

    Feb 22, 2011
    38
    0
    Thank you! Sorry for the confusion everyone, I was just trying to please the mods with the analogies.

    I'll try that, seems to do exactly what I was looking for. Only problem now is that radioshack was taken over in Canada by "the source" which sells $12 toy cars and no electronics parts. Gotta drive across town to pick all this up.

    Thanks a lot man, thats all I was looking for

    EDIT: is the diode just to protect the relay from spikes and the cap from reverse current? And is it in series between the power supply and + pin from the relay?
     
    Last edited: Jun 21, 2011
  20. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    The diode prevents the cap from discharging back through the power supply instead of through the relay's coil.
     
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