TECH : Series circuit advice needed for homebrew pin
Cyborg
Archived from groups: rec.games.pinball (More info?)
Howdy!
Quick question about series connections, I've been using nothing but
parallel so far so I'm a total series noob. I wanted to create a low current
draw Aux. GI system for decorative accent lighting on my table, but I'm not
sure if my understanding of one fundamental of series connections is right :
Let's say I want to put two #555 bulbs in series on a 12 volt power
supply (I know they're only rated for 6.3 volts).
If the 555 has specs of : 6.3 volts, 0.25 amp, ~25 ohm impedance, is
this correct :
At 12 volts, two 555 bulbs in series will consume : 12v / 50ohm = 0.24
amps. Ok.
Now... what I'm wondering is does this mean : this is the total current
which will flow through both bulbs; or, does each one get that amount of
current (ie total current draw off that series 2 bulb connection = 0.48
amps?)
I sure hope it's the first one, I'd love to save about 4 amps! ;)
Thanks for any answers;
cyBORG
Howdy!
Quick question about series connections, I've been using nothing but
parallel so far so I'm a total series noob. I wanted to create a low current
draw Aux. GI system for decorative accent lighting on my table, but I'm not
sure if my understanding of one fundamental of series connections is right :
Let's say I want to put two #555 bulbs in series on a 12 volt power
supply (I know they're only rated for 6.3 volts).
If the 555 has specs of : 6.3 volts, 0.25 amp, ~25 ohm impedance, is
this correct :
At 12 volts, two 555 bulbs in series will consume : 12v / 50ohm = 0.24
amps. Ok.
Now... what I'm wondering is does this mean : this is the total current
which will flow through both bulbs; or, does each one get that amount of
current (ie total current draw off that series 2 bulb connection = 0.48
amps?)
I sure hope it's the first one, I'd love to save about 4 amps! ;)
Thanks for any answers;
cyBORG
5
answers
Last reply
More about tech series circuit advice needed homebrew

Archived from groups: rec.games.pinball (More info?)
It is indeed the first option, each bulb will carry 0.24A. The only
problem is that one bulb failing will also make its partner go dark,
but that is easy to live with. 
Archived from groups: rec.games.pinball (More info?)
The resistance/("impedance") part is incorrect but something you do
not need to calculate answers. Total current for two bulbs in series
at twice rated voltage will be same as one at rated voltage.
Ken S
On Fri, 08 Jul 2005 10:35:00 GMT, "cyBORG" <cyborg@hotmail.com> wrote:
>Howdy!
>
> Quick question about series connections, I've been using nothing but
>parallel so far so I'm a total series noob. I wanted to create a low current
>draw Aux. GI system for decorative accent lighting on my table, but I'm not
>sure if my understanding of one fundamental of series connections is right :
>
> Let's say I want to put two #555 bulbs in series on a 12 volt power
>supply (I know they're only rated for 6.3 volts).
>
> If the 555 has specs of : 6.3 volts, 0.25 amp, ~25 ohm impedance, is
>this correct :
>
> At 12 volts, two 555 bulbs in series will consume : 12v / 50ohm = 0.24
>amps. Ok.
>
> Now... what I'm wondering is does this mean : this is the total current
>which will flow through both bulbs; or, does each one get that amount of
>current (ie total current draw off that series 2 bulb connection = 0.48
>amps?)
>
>I sure hope it's the first one, I'd love to save about 4 amps! ;)
>
>Thanks for any answers;
>cyBORG
> 
Archived from groups: rec.games.pinball (More info?)
Ohm's Law is your friend. V=IR. Since V=12 and R=50, then I=12/50 or
..24 amps
K2 
Archived from groups: rec.games.pinball (More info?)
> If the 555 has specs of : 6.3 volts, 0.25 amp, ~25
>ohm impedance, is this correct :
only when the bulb is up to full temp. Resistance will be lower when
the bulb is cooler.
> At 12 volts, two 555 bulbs in series will consume :
>12v / 50ohm = 0.24 amps.
Ok.
> Now... what I'm wondering is does this mean : this >is the total current which will flow through both bulbs; >or, does each one get that amount of current (ie total >current draw off that series 2 bulb connection = 0.48
>amps?)
THe total going through both bulbs. You are seeing 12V/50ohm AT THE
POWER SUPPLY. That is .24 amps going out of the power supply and
through both bulbs in one loop.
Problem with series circuits one bulb blows, all the bulbs in that
string are out.
Kirb 
Archived from groups: rec.games.pinball (More info?)
Ok, thanks for the infos guys.
cyBORG
"cyBORG" <cyborg@hotmail.com> wrote in message
news:oFsze.58432$Ph4.1710937@ursanb00s0.nbnet.nb.ca...
> Howdy!
>
> Quick question about series connections, I've been using nothing but
> parallel so far so I'm a total series noob. I wanted to create a low
current
> draw Aux. GI system for decorative accent lighting on my table, but I'm
not
> sure if my understanding of one fundamental of series connections is right
:
>
> Let's say I want to put two #555 bulbs in series on a 12 volt power
> supply (I know they're only rated for 6.3 volts).
>
> If the 555 has specs of : 6.3 volts, 0.25 amp, ~25 ohm impedance, is
> this correct :
>
> At 12 volts, two 555 bulbs in series will consume : 12v / 50ohm = 0.24
> amps. Ok.
>
> Now... what I'm wondering is does this mean : this is the total
current
> which will flow through both bulbs; or, does each one get that amount of
> current (ie total current draw off that series 2 bulb connection = 0.48
> amps?)
>
> I sure hope it's the first one, I'd love to save about 4 amps! ;)
>
> Thanks for any answers;
> cyBORG
>
>
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