newbie series-parallel circuit question

T

the_doug

Guest
I'm a software developer that has recently decided that electronics
would be a *fun* hobby. i'm not too far into it and I'm already
stumped... In this little project that I've invented for myself I'm
trying to light an array of twelve flashlight bulbs simultaneously...
and can't seem to manage it. :| Please excuse my idiocy in
advance...

the bulbs say they are rated for 2.33V and 270mA. I got a 6V, 300mA
dc wall adapter that I'm trying to use to power the deal. If i throw
22 Ohms worth of 1/2 watt resistors in front of a single bulb - i have
no problem lighting it... but those resistors start getting Hot.
Eventually - I'll want to be able to selectively switch each bulb on
and off, so i'm aware that i'll need some sort of parallel
configuration... but I'm just not sure if i'm doing the right thing
by generating so much damn heat.

feel free to give me hints, suggestions, schematics - or anything else
you have a moment to impart upon an idiot. :)
 
"the_doug" <st_november@yahoo.com> wrote in message
news:6219d7e3.0405241438.7890602f@posting.google.com...
I'm a software developer that has recently decided that electronics
would be a *fun* hobby. i'm not too far into it and I'm already
stumped... In this little project that I've invented for myself I'm
trying to light an array of twelve flashlight bulbs simultaneously...
and can't seem to manage it. :| Please excuse my idiocy in
advance...

the bulbs say they are rated for 2.33V and 270mA. I got a 6V, 300mA
Click to expand...
You can't do it with this transformer because you will not have
enough wattage to light all bulbs. What sort of bulbs are these, 2.33V
is a strange rating, usually one finds 1.5/3/4.5/6 V Have you considered
working with LED's, you can do it with LEDS. If you want to do it this
way simply connect all the leds in parallel and put a resistor of 500ohms
after each LED. You can also have switches as required.

dc wall adapter that I'm trying to use to power the deal. If i throw
22 Ohms worth of 1/2 watt resistors in front of a single bulb - i have
no problem lighting it... but those resistors start getting Hot.
Eventually - I'll want to be able to selectively switch each bulb on
and off, so i'm aware that i'll need some sort of parallel
configuration... but I'm just not sure if i'm doing the right thing
by generating so much damn heat.

feel free to give me hints, suggestions, schematics - or anything else
you have a moment to impart upon an idiot. :)
Click to expand...
 
Hi,

st_november@yahoo.com (the_doug) wrote in
news:6219d7e3.0405241438.7890602f@posting.google.com:

In this little project that I've invented for myself I'm
trying to light an array of twelve flashlight bulbs simultaneously...

the bulbs say they are rated for 2.33V and 270mA.
Click to expand...
That would be 2.33 * .27 = 0.63W (or 630 mW)
12 of these would amount to a little over 7.5W

I got a 6V, 300mA dc wall adapter that I'm trying to use to power
Click to expand...
It is only able to deliver 6 * 0.3 = 1.8W, not nearly enough !


If i throw 22 Ohms worth of 1/2 watt resistors in front of a single
bulb - i have no problem lighting it... but those resistors start
getting Hot.
Click to expand...
Of course, it gets more than 0.8W, you would need a higher wattage
resistor to keep ot from "melting" :)


Eventually - I'll want to be able to selectively switch each bulb on
and off, so i'm aware that i'll need some sort of parallel
configuration... but I'm just not sure if i'm doing the right thing
by generating so much damn heat.
Click to expand...
First of all, you would need a beefier supply to get all 12 lamps to
light up at once.

Resistors are not the best way to keep your lamps alive, a voltage
regulator (delivering 2.33V @ 3.3A) would be a better solution.


If you want to use resistors, you should choose the right values:
A lamp of 2.33V/270mA has a resistance of 2.33 / 0.27 = 8.63 Ohms.

If your supply is 6V, the needed resistance would be:
(6V - 2.33V) / 0.27 = 13.6 Ohms.
You then select the next _higher_ standard value of 15 Ohms.
The current possible is then: 6 / (15 + 8.63) = 0.254A.

The power dissipated in the resistor is then:
15 * 0.254^2 = 0.97W so you need at least a 1W


The current is slightly reduced compared to the nominal rating, but this
will give the lamp a longer life :)


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *
 
"Soeren" <Look@iNO-SPAMt.dk.invalid> wrote in message
news:Xns94F4EA5BF87B4o8oLOOKatHOMEo8o@212.242.40.196...
Hi,

st_november@yahoo.com (the_doug) wrote in
news:6219d7e3.0405241438.7890602f@posting.google.com:

[...] In this little project that I've invented for myself I'm
trying to light an array of twelve flashlight bulbs simultaneously...

the bulbs say they are rated for 2.33V and 270mA.

That would be 2.33 * .27 = 0.63W (or 630 mW)
12 of these would amount to a little over 7.5W

I got a 6V, 300mA dc wall adapter that I'm trying to use to power

It is only able to deliver 6 * 0.3 = 1.8W, not nearly enough !


If i throw 22 Ohms worth of 1/2 watt resistors in front of a single
bulb - i have no problem lighting it... but those resistors start
getting Hot.

Of course, it gets more than 0.8W, you would need a higher wattage
resistor to keep ot from "melting" :)


Eventually - I'll want to be able to selectively switch each bulb on
and off, so i'm aware that i'll need some sort of parallel
configuration... but I'm just not sure if i'm doing the right thing
by generating so much damn heat.

First of all, you would need a beefier supply to get all 12 lamps to
light up at once.

Resistors are not the best way to keep your lamps alive, a voltage
regulator (delivering 2.33V @ 3.3A) would be a better solution.
Click to expand...
A voltage regulator *is* a resistor, albeit a semiconductor one.
What is required is to match the voltage of the supply to the
voltage of the bulbs. Therefore, it would be better to choose
standard 3V bulbs and a 3V transformer that can supply enough
current, any other way will be a waste of energy.

If you want to use resistors, you should choose the right values:
A lamp of 2.33V/270mA has a resistance of 2.33 / 0.27 = 8.63 Ohms.

If your supply is 6V, the needed resistance would be:
(6V - 2.33V) / 0.27 = 13.6 Ohms.
You then select the next _higher_ standard value of 15 Ohms.
The current possible is then: 6 / (15 + 8.63) = 0.254A.

The power dissipated in the resistor is then:
15 * 0.254^2 = 0.97W so you need at least a 1W


The current is slightly reduced compared to the nominal rating, but this
will give the lamp a longer life :)


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *
Click to expand...
 
I greatly appreciate the help everyone. I found this interesting...
This calculation should probably have been somewhat obvious to me -
but I hadn't made the connection to apply Ohm's law to this info.

If you want to use resistors, you should choose the right values:
A lamp of 2.33V/270mA has a resistance of 2.33 / 0.27 = 8.63 Ohms.
Click to expand...
It also explains why my calculations were causing me so much grief...
I had done a manual measurement of the bulb resistance and found it to
be ~1.6 Ohms. Is this difference between spec and measurement due to
lamp resistance changing under heat?

This is probably as good a time as any to post a follow-up question,
and bear with me if i do a poor job of explaining myself... If I were
to decide to run this circuit in a series - And Still try to
incorporate switching the bulbs on and off individually... i'm
imagining something like this... (don't laugh):

|- NPN - Bulb -
--| |-----
|- PNP - R -

....with the base V coming in from a 5V pin on a pic... I'm thinking
that a high or low signal would alternately push the current through
the bulb -OR- the resistor... Giving me switching in a "pseudo"
series circuit. Of course, I've been doing electronics for about two
weeks - and made this up myself... SO - would this actually work?!

Thanks in advance,
Doug
 
It also explains why my calculations were causing me so much grief...
I had done a manual measurement of the bulb resistance and found it to
be ~1.6 Ohms. Is this difference between spec and measurement due to
lamp resistance changing under heat?
Click to expand...
Exactly!

As the filiment heats up, the resistance increases.

This is probably as good a time as any to post a follow-up question,
and bear with me if i do a poor job of explaining myself... If I were
to decide to run this circuit in a series - And Still try to
incorporate switching the bulbs on and off individually... i'm
imagining something like this... (don't laugh):

|- NPN - Bulb -
--| |-----
|- PNP - R -

...with the base V coming in from a 5V pin on a pic... I'm thinking
that a high or low signal would alternately push the current through
the bulb -OR- the resistor... Giving me switching in a "pseudo"
series circuit. Of course, I've been doing electronics for about two
weeks - and made this up myself... SO - would this actually work?!
Click to expand...
I have no idea what you're trying to do here, sorry :)

Al

Thanks in advance,
Doug
Click to expand...
 
"bench" <bench@neti.com> wrote in news:c90p20$mhs$1@news.epidc.co.kr:

A voltage regulator *is* a resistor, albeit a semiconductor one.
Click to expand...
No !
It might fill out the same function *if* the driving voltage is free of
any fluctuations, but in any real life applications, it is a grossly
simplified view, espscially when the mains supply deviates maybe +/-
10%.


What is required is to match the voltage of the supply to the
voltage of the bulbs. Therefore, it would be better to choose
standard 3V bulbs and a 3V transformer that can supply enough
current, any other way will be a waste of energy.
Click to expand...
Well, if that means "discard what you allready have and go out and spend
money on new stuff", your solution will be a waste of both energy (the
energy he has to spend shopping) *and* money ;)


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *
New forum: <URL:http://www.elektronikteknolog.dk/cgi-bin/SPEED/>
 
Hi,

st_november@yahoo.com (the_doug) wrote in
news:6219d7e3.0405252123.548c2b41@posting.google.com:

It also explains why my calculations were causing me so much grief...
I had done a manual measurement of the bulb resistance and found it to
be ~1.6 Ohms. Is this difference between spec and measurement due to
lamp resistance changing under heat?
Click to expand...
Yes, you could try to measure the resistance of a thin piece of steel
wire (mounted in free air) and while measuring, heat (some of) it to a
cherry red with a blow torch - quite an eye opener :)


This is probably as good a time as any to post a follow-up question,
and bear with me if i do a poor job of explaining myself... If I were
to decide to run this circuit in a series - And Still try to
incorporate switching the bulbs on and off individually... i'm
imagining something like this... (don't laugh):

|- NPN - Bulb -
--| |-----
|- PNP - R -

...with the base V coming in from a 5V pin on a pic... I'm thinking
that a high or low signal would alternately push the current through
the bulb -OR- the resistor... Giving me switching in a "pseudo"
series circuit. Of course, I've been doing electronics for about two
weeks - and made this up myself... SO - would this actually work?!
Click to expand...
Do you mean something like this ?

+V O----+
|
+--+--+ First stage
| |
b |/e c\| b
--| PNP NPN |--
|\c e/|
| |
| |
[Resistor] [Bulb]
| |
+--+--+
|
Etc.
|
+--+--+ Last stage
| |
b |/e c\| b
--| PNP NPN |--
|\c e/|
| |
| |
[Resistor] [Bulb]
| |
+--+--+
|
-V O----+

If so, I'm sorry, but it won't work.

What is the actual purpose, perhaps there are better ways to do it.

If you are using a PIC, you could use PWM to reduce the current through
the lamps.


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *
New forum: <URL:http://www.elektronikteknolog.dk/cgi-bin/SPEED/>
 
I have no idea what you're trying to do here, sorry :)
Click to expand...
No problem - somehow - the lines got messed up upon saving. what the
diagram attempted to show was two branches of a "pseudo" parallel
circuit attached to a larger series circuit. I say pseudo because my
intention would be to use switching to have the current run through
only one "branch" of this parallel hookup at a time - thus, either
lighting the bulb or running through the parallel resistor.

My question is simply: is this sort of configuration simply the
imaginings of a deranged amateur? =)

Thanks again,
Doug
 
Do you mean something like this ?

+V O----+
|
+--+--+ First stage
| |
b |/e c\| b
--| PNP NPN |--
|\c e/|
| |
| |
[Resistor] [Bulb]
| |
+--+--+
|
Etc.
|
+--+--+ Last stage
| |
b |/e c\| b
--| PNP NPN |--
|\c e/|
| |
| |
[Resistor] [Bulb]
| |
+--+--+
|
-V O----+

If so, I'm sorry, but it won't work.
Click to expand...
Yes - this seems like a pretty accurate description. Won't work, eh?
I was afraid of that...

A concise description of the project would be this:

3x4 array of lights which will end up mounted pixel-like behind 1-inch
square pieces of frosted glass - it will need to generate enough light
to function as an ambient end table lamp.

I've got a PIC driving 12 leds directly now - and doing all the
patterns i want - etc (the programming aspect is the only part i've
got experience with - so it was no problem). The problem is - I need
it to be brighter. So I started playing with the idea of a separate
power loop being controlled by the PIC via transistors... little did
i realize how much Amperage it was going to take to drive 12 bulbs
independently in parallel. The reason i was trying to make a series
solution work is because higher voltage seems to come cheaper and in
smaller packaging than higher amperage (and this could be BAD newbie
reasoning).

I don't really want to swallow a lot of your time on this... but if
you're enjoying the discourse - I'll take all the help I can get.
 
"the_doug" <st_november@yahoo.com> wrote in message
news:6219d7e3.0405260747.1fbc8e08@posting.google.com...
I have no idea what you're trying to do here, sorry :)


No problem - somehow - the lines got messed up upon saving. what the
diagram attempted to show was two branches of a "pseudo" parallel
circuit attached to a larger series circuit. I say pseudo because my
intention would be to use switching to have the current run through
only one "branch" of this parallel hookup at a time - thus, either
lighting the bulb or running through the parallel resistor.

My question is simply: is this sort of configuration simply the
imaginings of a deranged amateur? =)
Click to expand...
Probably. :) What you've described will probably work - apparently
you want to put resistor/lamp pairs in series, and swap out the
bulb with a resistor to get the same series resistance, ergo the
same current, right? Well, it would waste power and be a clooge
to drive.

Since you have a pic, just use 12 NPN transistors with a suitable
current rating, connect the bulbs directly to the supply, and
PWM them to cut down the _average_ voltage/current. Voila! Total
new parts count: 12 transistors. And that's only if you need new
ones. :)

And with a suitable input, you could ramp up the pulse width
slowly, and watch the bulbs get brighter until they're at about
95%, then define that as the upper limit.

And you _might_ be able to get by with your existing supply, as
the total load will be a percentage (the duty cycle) of the bulbs'
rating.

Cheers!
Rich
 
"the_doug" <st_november@yahoo.com> wrote in message
news:6219d7e3.0405261048.5c1c3209@posting.google.com...
I've got a PIC driving 12 leds directly now - and doing all the
patterns i want - etc (the programming aspect is the only part i've
got experience with - so it was no problem). The problem is - I
need
it to be brighter. So I started playing with the idea of a separate
power loop being controlled by the PIC via transistors... little
did
i realize how much Amperage it was going to take to drive 12 bulbs
independently in parallel. The reason i was trying to make a series
solution work is because higher voltage seems to come cheaper and in
smaller packaging than higher amperage (and this could be BAD newbie
reasoning).

I don't really want to swallow a lot of your time on this... but if
you're enjoying the discourse - I'll take all the help I can get.
Click to expand...
If you are using LEDs now, you should know that you can get really
bright ones (that will, sadly, take more current to light.) Take a
look at www.whitelightled.com for some ideas. Also, these are sold on
E-Bay.

Using incandescent bulbs should work, but its going to take 'more
power, Scotty!'.

You can get cheap wall warts that put out 12V with higher current
ratings at various places, including here:

http://sales.goldmine-elec.com/prodlist.asp?catid=2077

I see a nice 12V 800mA adapter for $3 USD. (plus shipping, sigh)

Regards,
Bob Monsen
 
Hi Rich,


And you _might_ be able to get by with your existing supply, as
the total load will be a percentage (the duty cycle) of the bulbs'
rating.
Click to expand...
Ahemmm...
12 lamps of 2.33V/270mA = ~7.5W
1 power adapter of 6V/300mA = 1.8W

Now I am very anxious to get the specs of your power-from-thin-air
component ;)


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *
New forum: <URL:http://www.ElektronikTeknolog.dk/cgi-bin/SPEED/>
 
"Soeren" <Look@iNO-SPAMt.dk.invalid> wrote in message
news:Xns94F69BF5F90A0o8oLOOKatHOMEo8o@212.242.40.196...
Hi Rich,


And you _might_ be able to get by with your existing supply, as
the total load will be a percentage (the duty cycle) of the bulbs'
rating.

Ahemmm...
12 lamps of 2.33V/270mA = ~7.5W
1 power adapter of 6V/300mA = 1.8W

Now I am very anxious to get the specs of your power-from-thin-air
component ;)
Click to expand...
Well, I said "might," since I didn't remember the spec from the
1st post. Just sloppiness on my part, sorry. :)

Thanks!
Rich
 
Hi Rich,


Well, I said "might," since I didn't remember the spec from the
1st post. Just sloppiness on my part, sorry. :)
Click to expand...
Just wondered :)


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *
New forum: <URL:http://www.ElektronikTeknolog.dk/cgi-bin/SPEED/>
 
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