I think SteveMann nailed it. That's a dead ringer for an E10 bulb and as somebody mentioned, it looks like
they put an aesthetic clear bulb around it.
Similar, measurements are slightly different, and it obviously has the aesthetic 'bulb'
Agreed, the external dimensions are different but the E10 bulb dimensions are the same.
PerryBebbington:
I suspect from that they have a resistor built in. I guess the right value to operate from 3V.
So so probably just over a 300ohm resistor if there's 9.75mA at 3v?
flyagaricus:
So so probably just over a 300ohm resistor if there's 9.75mA at 3v?
Err, no.
Think again.
Err, no.
Think again.
HINT: a resistor is linear, a led is not.
RCurrent Limiting =(V-Vf)/Iled@Vf
ie:
Let V=3V
Let Vf = 1.2V
Let Iled = 0.00975V (@Vf)
.'. 3-1.2V=1.8V
IRcurrent limiting = Iled
.'. R=VR/IR = 1.8V/0.00975V = 184.6 ohms
That
raschemmel:
HINT: a resistor is linear, a led is not.RCurrent Limiting =(V-Vf)/Iled@Vf
ie:
Let V=3V
Let Vf = 1.2V
Let Iled = 0.00975V (@Vf)
.'. 3-1.2V=1.8V
IRcurrent limiting = Iled
.'. R=VR/IR = 1.8V/0.00975V = 184.6 ohms
[/quote]
That sucks. Here I was thinking I was wrapping my head around ohm's law. I really like electronics, but math is a major weakness for me
I don't get how you know the forward voltage is 1.2v....
I don't get how you know the forward voltage is 1.2v....
I didn't and I don't know the forward voltage because I can't find the datasheet for that led and even if
I did, it is quite possible the forward voltage would NOT be given because the manufacturer has done
all the work for you. That information is literally "X-DON'T CARE" because you do NOT have to calculate
the resistor value because it has already been done.
The example given:
FYI, 'ie:" means ("for EXAMPLE")
says "Let If =1.2V"
which means for my example , I have CHOSEN 1.2 V in order to show you HOW to calculate the resistor
value IF YOU NEEDED TO DO SO, (ie, you want to use a led that does NOT have an INTERNAL RESISTOR !)
hence, I have created a HYPOTHETICAL case of a led that does NOT have an INTERNAL resistor and therefore DOES have a DATASHEET that DOES give the forward voltage.
In retrospect , I realize that my arbitrary choice of 1.2 V was unrealistic because typical led forward
voltage starts at 1.8V (MY BAD !)
While we are already on the subject of led forward voltage I would be remiss in my duties if I failed to
mention that led forward voltage is a function of color !
That being said, if you scroll down to page 4 of the white led datasheet you will see the Vf is 3V, whereas for the red led it is 2V.
So, in conclusion, for the purposes of this discourse, pretend the led you have does NOT have an
INTERNAL resistor and read the example given, substituting real world values from the two datasheets
provided.
ergo,
DO THE MATH...
FYI, 'ie:" means ("for EXAMPLE")
Errrr, no, I thionk you will fine 'eg' means for example.
'ie' is when you want to specify something exactly, pretty much the opposite of an example.
raschemmel:
I didn't and I don't know the forward voltage because I can't find the datasheet for that led and even if
I did, it is quite possible the forward voltage would NOT be given because the manufacturer has done
all the work for you. That information is literally "X-DON'T CARE" because you do NOT have to calculate
the resistor value because it has already been done.
The example given:
FYI, 'ie:" means ("for EXAMPLE")says "Let If =1.2V"
which means for my example , I have CHOSEN 1.2 V in order to show you HOW to calculate the resistor
value IF YOU NEEDED TO DO SO, (ie, you want to use a led that does NOT have an INTERNAL RESISTOR !)hence, I have created a HYPOTHETICAL case of a led that does NOT have an INTERNAL resistor and therefore DOES have a DATASHEET that DOES give the forward voltage.
In retrospect , I realize that my arbitrary choice of 1.2 V was unrealistic because typical led forward
voltage starts at 1.8V (MY BAD !)[led datasheet color red/url
led datasheet color whiteWhile we are already on the subject of led forward voltage I would be remiss in my duties if I failed to
mention that led forward voltage is a function of led color !That being said, if you scroll down to page 4 of the white led datasheet you will see the Vf is 3V, whereas for the red led it is 2V.
So, in conclusion, for the purposes of this discourse, pretend the led you have does NOT have an
INTERNAL resistor and read the example given, substituting real world values from the two datasheets
provided.
ergo,
DO THE MATH...
I appreciate the response, but maybe you missed the part where I said 'I suck at math'. That's not to say I don't want to do it, or try it - but it's a MAJOR obstacle for me. So, when I see a number and I don't know where it came from, my brain shuts down.
But, I get what your saying now (I think).
raschemmel:
FYI, 'ie:" means ("for EXAMPLE")
.
I said I was bad at MATH, not English. No need to be condescending.
flyagaricus:
I said I was bad at MATH, not English. No need to be condescending.
If you look at the attribution, you will see that the comment was not directed at you. For what it's worth.
flyagaricus:
I said I was bad at MATH, not English. No need to be condescending.
I wasn't commenting to you, I was commenting to raschemmel! I think he knows me well enough that I can get away with that, or if not he'll tell me 
He corrects me sometimes and I do not mind at all.
We're all grownups here. (that is to say children in adult bodies).
thanks for the heads up on "ie: vs eg". (how embarassing !)
You can find the approximate forward voltage, by applying a much smaller current to the bulb, and measuring the voltage. For example, put it in series with a 100k ohm resistor. The actual loaded forward voltage will be approximately a hundred or two hundred mV greater than that. In this way, you can make an estimation.
Interesting ! I never heard that before.
aarg:
You can find the approximate forward voltage, by applying a much smaller current to the bulb, and measuring the voltage. For example, put it in series with a 100k ohm resistor. The actual loaded forward voltage will be approximately a hundred or two hundred mV greater than that. In this way, you can make an estimation.
++Karma; // Never thought of that, now it seems obvious!
To be reasonably close in estimation, it would be best to compare with the most similar resistor-less LED you can find... then you can get a better curve match. Some playing around with different LEDs could give you some ballpark bench formulas.
Notwithstanding all the discussion, I have explained twice that these are LED assemblies with no built-in resistor.
If you run a white LED up with a variable supply, you will find it starts to light somewhere around 2½ Volts and is reasonably bright at 3 V. Most multiple-LED torches and the string lights I illustrated in #13 actually use three cells or a LiPo so are running between 3.7 and 4.5 V - with no resistor. Note the curve below, reproduced in various places:
So can we forget about the resistor thing? It is using the internal resistance of the LED. 
Unless of course, you wish to operate them from 5 V on an Arduino, in which case I explained it way back in #4 and did the maths. 
Oh yes, I meant to mention.
"i.e." - abbreviation for Latin "id est" meaning "that is" or "that is to say".
This may help.
And yes, it is an E10 bulb as explained in #4.
PerryBebbington:
I wasn't commenting to you, I was commenting to raschemmel! I think he knows me well enough that I can get away with that, or if not he'll tell me
I know it wasn't you that made that comment.
Last night I was trying to work out how to solve 2 equations with 2 unknowns (Vf and R) to get both from measuring the current at 2 different voltages. I was good at this at school....
There isn't any convincing reason to believe that it has, or doesn't have a resistor built in. Of course you might find one without a resistor, but the main reason for the "candelabra" base is for drop in replacement of incandescent bulbs. I have several of those in my collection, they have the correct resistor to be powered directly from 12V. The only way to know for sure, is for the OP to perform some controlled tests, but (no slight intended) I don't believe he has the equipment or specific knowledge to perform the tests.
