Confused about resistance

 
  gombul Station Master

Location: Canberra, Australian Capital Territory
Hi,

I was planning on simply adding a single resistor to the common cathode of my blue/yellow bipolar LED. Works fine except of course, the blue is much brighter than the yellow. Not ideal.

Then I noticed a big resistor in my scraps box. I thought I'd do a simple test of adding separate resistors to each of the anode legs - to see if I could control the brightness of the yellow/blue individually.

I "thought" the large resistor would offer more resistance (I now see it says 56R on the packet, which may explain a few things). It didn't dull the blue brightness at all. What's more, after a few moments the blue burnt out (the resistor was very hot). The yellow - despite being quite dull - continued to tick along nicely with the 470R resistor.

Is this a simple case of the huge resistor offering less resistance (56R vs 470R)??? I just assumed the chunky one would be more limiting.

regards,
Rowan (electronics fizzkid)

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  Poath Junction Chief Commissioner

Location: In front of a computer most of the time.
Assuming you were using a 12 volt DC power supply, the 56 ohm resister would have given a current of ~.2 amps and power of ~2.5 watts. The resister got hot as it's rated for just 1 amp, and the diode blew as 200 milliamps is way above its rating. A 5 volt power supply would have dropped those figures by a bit over half, but that's still above the resistor and diode ratings.

The 470 ohm resister with 12V DC gives ~0.02 amps (20 milliamps) and ~0.3 watts which are within the rated specs of the diode & resister hence no problem. On 5V you're down to 10 milliamps and 0.15 watts which of course is also within spec.

LEDs don't use much energy, hence you want a larger rating (in ohm's) for the resister. The higher the resistance the lower the power and current.
  Poath Junction Chief Commissioner

Location: In front of a computer most of the time.
PS. The physical size of a resister is determined by various factors which don't necessarily directly relate to the rated resistance. In your examples the resistors use different technology (carbon v metal film) and also have a different power rating - the physically larger one is rated for 1 watt, while the smaller is rated at 1/2 watt.

edit: fixed my amp/watt faux pas
  Pressman Spirit of the Vine

Location: Wherever the Tin Chook or Qantas takes me

I "thought" the large resistor would offer more resistance (I now see it says 56R on the packet, which may explain a few things). It didn't dull the blue brightness at all. What's more, after a few moments the blue burnt out (the resistor was very hot). The yellow - despite being quite dull - continued to tick along nicely with the 470R resistor.

Is this a simple case of the huge resistor offering less resistance (56R vs 470R)??? I just assumed the chunky one would be more limiting.

regards,
Rowan (electronics fizzkid)
gombul
The physical size has nothing to do with the resistance value.
the one marked 56R is 56 Ohms, the one marked 470R is 470 ohm

Different LED's have different voltages and current draws
LED Voltage Sometimes "Forward Voltage" but usually just abbreviated "V".
LED Current Sometimes "Forward Current". This is listed in milliamps or "mA".

Typically
Red LED: 2V 15mA
Green LED: 2.1V 20mA
Blue LED: 3.2V 25mA
While LED: 3.2V 25mA

Then use an LED Resistance calculator like this one http://ledcalc.com/#calc
  TheBlacksmith Chief Commissioner

Location: Ankh Morpork
The other issue is that not all LEDs put out the same luminance, so a single resistor value may not get them to the same level of brightness.
  apw5910 Deputy Commissioner

Location: Location: Location.
PS. The physical size of a resister is determined by various factors which don't necessarily directly relate to the rated resistance. In your examples the resistors use different technology (carbon v metal film) and also have a different power rating - the physically larger one is rated for 1 amp, while the smaller is rated at 1/2 amp.
Poath Junction
Watts, not amps.
  gombul Station Master

Location: Canberra, Australian Capital Territory
The goal is to achieve consistency of brightness across the 2 colours. There are 6 blocks of LEDs and the switches are centre off. This means, there could be 0, 1, 2, 3, 4, 5 or 6 lights powered at any given time depending on switch positions.

To my way of thinking I'd need to offer resistance for the "worst case scenario" which would be ensuring "only 1 bulb on" doesn't receive too much juice. But this is where I get confused with series vs parallel... Below is my wiring setup. Those LED specs were all I got with the LEDs.

  Aaron The Ghost of George Stephenson

Location: University of Adelaide SA
PS. The physical size of a resister is determined by various factors which don't necessarily directly relate to the rated resistance. In your examples the resistors use different technology (carbon v metal film) and also have a different power rating - the physically larger one is rated for 1 amp, while the smaller is rated at 1/2 amp.
Watts, not amps.
apw5910
Correct.

Assuming 12v supply and blue LED Vf of 3.5v the 56R resistor is passing about 151mA, (which tells us something about rating resistors in terms of current - and provides clear causation as to the LED's death#) the power it's dissipating is the square of the current multiplied by R, sufficiently close to 1.28W, the resistor is likely* rated at 1W and hence it gets somewhat warm, to hot.

Assuming 12v supply and a yellow LED Vf of 2.1v, the 470R resistor is passing about 21mA, (that's sufficiently close to pefect for a common variety LED) the power it's dissipating is the square of the current multiplied by R, sufficiently close to 207mW, the resistor is likely* rated at 300mW and hence has some margin available to remain cool.

#The power the LED was dissipating is the current multiplied by the Vf, likely 529mW, it can't push that into photons (which in itself is a REALLY interesting study in physics but well beyond the scope of this forum) so instead the LED generates heat at the junction and burns up.

*Further, whilst Poath makes a reasonable assumption aside from the amps instead of watts unit, one cannot simply determine the power dissipation by a cursory look at physical size. I have some 300mW resistors that are physically larger than my 600mW packages.
  Poath Junction Chief Commissioner

Location: In front of a computer most of the time.
PS. The physical size of a resister is determined by various factors which don't necessarily directly relate to the rated resistance. In your examples the resistors use different technology (carbon v metal film) and also have a different power rating - the physically larger one is rated for 1 amp, while the smaller is rated at 1/2 amp.
Watts, not amps.
apw5910
Boy is my face red. Redder than a 1W resistor being try to dissipate 3W Embarassed
  Pressman Spirit of the Vine

Location: Wherever the Tin Chook or Qantas takes me
The goal is to achieve consistency of brightness across the 2 colours. There are 6 blocks of LEDs and the switches are centre off. This means, there could be 0, 1, 2, 3, 4, 5 or 6 lights powered at any given time depending on switch positions.

To my way of thinking I'd need to offer resistance for the "worst case scenario" which would be ensuring "only 1 bulb on" doesn't receive too much juice. But this is where I get confused with series vs parallel... Below is my wiring setup. Those LED specs were all I got with the LEDs.

gombul
Ok, first Series v Parallel connections .... This applies when you connect a number of LED's using one resistor.
In your wiring diagram you are using single LED's with Single resistors so forget about series/parallel as it does not apply
When you use the resistor calculator I posted in my previous post you select the Single LED page.

Next, as the two colours have a different LED voltage, you need to calculate the individual colours separately.
The Yellow Vf is 3.2Volt, assume 20mA current and the calculator comes up with a resistance required of 440 ohm. Use a 470 Ohm 1/4 Watt resistor.
The Blue Vf is 3.6Volt again assume 20mA and the calculator comes up with 420 ohm required again the next resistor size is 470 Ohm 1/4 Watt

Now to play with the difference in brightness. Try a larger ohm resistor on the brighter colour side, a larger resistance will lower the voltage across the LED and hence reduce it's luminosity.
The standard sizes in that area are 430, 470, 510, 560, 620.
Remember the resistors need to be rated at least 1/4Watt.



Now the rating of the Bi-colour LED

....... mcd means millicandela which is the units for the luminous intensity of the LED
so from the figures on your diagram (Yellow = 7000mcd, Blue = 8000mcd) the blue will appear "brighter" than the yellow.

As each of the six indicator circuits will only draw approximately 20mA each the total current draw on your power supply will be around 120mA with all circuits turned on. As the power supply is rated at 1.5Amp (1,500mA) the power supply will hardly 'know' there is any load and the individual LED's brilliance won't change with respect to how many are turned on
  Aaron The Ghost of George Stephenson

Location: University of Adelaide SA
The goal is to achieve consistency of brightness across the 2 colours. There are 6 blocks of LEDs and the switches are centre off. This means, there could be 0, 1, 2, 3, 4, 5 or 6 lights powered at any given time depending on switch positions.

To my way of thinking I'd need to offer resistance for the "worst case scenario" which would be ensuring "only 1 bulb on" doesn't receive too much juice. But this is where I get confused with series vs parallel... Below is my wiring setup. Those LED specs were all I got with the LEDs.

Ok, first Series v Parallel connections .... This applies when you connect a number of LED's using one resistor.
In your wiring diagram you are using single LED's with Single resistors so forget about series/parallel as it does not apply
When you use the resistor calculator I posted in my previous post you select the Single LED page.

Next, as the two colours have a different LED voltage, you need to calculate the individual colours separately.
The Yellow Vf is 3.2Volt, assume 20mA current and the calculator comes up with a resistance required of 440 ohm. Use a 470 Ohm 1/4 Watt resistor.
The Blue Vf is 3.6Volt again assume 20mA and the calculator comes up with 420 ohm required again the next resistor size is 470 Ohm 1/4 Watt

Now to play with the difference in brightness. Try a larger ohm resistor on the brighter colour side, a larger resistance will lower the voltage across the LED and hence reduce it's luminosity.
The standard sizes in that area are 430, 470, 510, 560, 620.
Remember the resistors need to be rated at least 1/4Watt.



Now the rating of the Bi-colour LED

....... mcd means millicandela which is the units for the luminous intensity of the LED
so from the figures on your diagram (Yellow = 7000mcd, Blue = 8000mcd) the blue will appear "brighter" than the yellow.

As each of the six indicator circuits will only draw approximately 20mA each the total current draw on your power supply will be around 120mA with all circuits turned on. As the power supply is rated at 1.5Amp (1,500mA) the power supply will hardly 'know' there is any load and the individual LED's brilliance won't change with respect to how many are turned on
Pressman
Thanks Pressman, I didn't pay as much attention to this as I should have the first time around.

Gombul, do you have a model number or data sheet for the LEDs you're using?

Vf 3.2v for a yellow LED seems unusually high for me, and 7000mcd is going to be pretty (actually bloody) bright.

8000mcd from a blue LED probably shouldn't be viewed straight on without eye protection.

I'd run each of them down to around 10mA and the up the current slightly if required, running these puppies at full intensity of a control panel designed to be directly looked at is probably not going to be a good idea!
  ParkesHub Chief Commissioner

LEDs are current devices so the voltage source applied has little relevance. The resistor value is determined by the voltage drop across the entire circuit divided by the current the LED is rated at.

The wattage of the resistor is determined by I2R and if your calculations have anything bigger than 1/4W, then I reckon your calculations are up the creek.

You can tweak the resistor value to alter the brightness (without exceeding the LED's maximum current flow).

Edit: Additionally, the brain interprets "brightness" differently according to the colour of the light source. Blue will appear different to yellow or red even though the light being emitted from the LED is the same value.
  Aaron The Ghost of George Stephenson

Location: University of Adelaide SA
You can't calculate If without knowing Vf...
  Aaron The Ghost of George Stephenson

Location: University of Adelaide SA
If = (Vdd - Vf)/R
  ParkesHub Chief Commissioner

You can't calculate If without knowing Vf...
Aaron
The point I'm making is a subtle one.
  Parkeston Station Staff

Just in case Gombul is confused I'll offer a summary:

Gombul, you had the correct approach and were nearly there.  You only made one small error.

The wiring you have is correct and, as has been mentioned before, there are no series/parallel concerns.

You do indeed need to start with the 470 ohm value and increase it gradually until the brightness matches for you.

The size of a resistor does not necessarily have any correlation to its resistance value.  There are two ways to determine the resistance value:

1.  Read the colour code or written value.
2.  Use a multimeter set to "resistance" and measure the value using that.

Hope this helps.
  Gremlin Assistant Commissioner

OK, silly question...perhaps...

Could one use a variable pot for the resistors and fine-tune to preferred visible outcome (heuristic approach), rather than the strictly mathematically rigorous and electronically correct process Smile

Are there any problems using a variable pot such as induced noise, resistance creep, something else...?
  ParkesHub Chief Commissioner

OK, silly question...perhaps...

Could one use a variable pot for the resistors and fine-tune to preferred visible outcome (heuristic approach), rather than the strictly mathematically rigorous and electronically correct process Smile

Are there any problems using a variable pot such as induced noise, resistance creep, something else...?
Gremlin
Yeah, that'd work.
  apw5910 Deputy Commissioner

Location: Location: Location.
OK, silly question...perhaps...

Could one use a variable pot for the resistors and fine-tune to preferred visible outcome (heuristic approach), rather than the strictly mathematically rigorous and electronically correct process Smile

Are there any problems using a variable pot such as induced noise, resistance creep, something else...?
Gremlin
No problems, it's just that pre-set pots are much more expensive than single resistors and this sort of circuit is a "set and forget, do it once and do it right" arrangement. You could use a pot to fine-tune the desired brightness, measure its set resistance, then use the nearest preferred value resistor for the installation.
  Pressman Spirit of the Vine

Location: Wherever the Tin Chook or Qantas takes me
You could use a pot to fine-tune the desired brightness, measure its set resistance, then use the nearest preferred value resistor for the installation.
apw5910
That's the way I'd go about it Wink
  ParkesHub Chief Commissioner

You could use a pot to fine-tune the desired brightness, measure its set resistance, then use the nearest preferred value resistor for the installation.
That's the way I'd go about it Wink
Pressman
Indeed, a good idea.
  apw5910 Deputy Commissioner

Location: Location: Location.
If you do, make sure there's a minimum value fixed resistor in series (150 ohms?) with the variable one. Otherwise you could set the pot to zero ohms and burn out the LED with the resultant high current flow. Like so:

                         v------------------
---v^v^v^-------v^v^v^
   Rmin           pot

I hate ASCII graphics...
  Aaron The Ghost of George Stephenson

Location: University of Adelaide SA
Back in 'the old days' we used to have 'cheap' decade resistance boxes, these days I think they're fairly specialised, I'd say it's fairly standard to have them in ppb accuracy and price tags into four figures (mine was).

Those aside, the next best thing you will get is probably one of these:
http://www.jaycar.com.au/PRODUCTS/Passive-Components/Resistors/Other-Resistors/Resistance-Wheel/p/RR0700
  gombul Station Master

Location: Canberra, Australian Capital Territory
Aaron, unfortunately I don't have data sheets. They are quite bright if looked at from the end but with ambient lights I think the yellow is pretty good.

Here is a test using R470 for yellow, R620 blue. Interestingly, the photos make the blue look brighter than it does with naked eyes. I might add yet more resistance to the blue - another run down to Jaycar but I'm very happy they've opened another store closer to my house.

  gombul Station Master

Location: Canberra, Australian Capital Territory
Back in 'the old days' we used to have 'cheap' decade resistance boxes, these days I think they're fairly specialised, I'd say it's fairly standard to have them in ppb accuracy and price tags into four figures (mine was).

Those aside, the next best thing you will get is probably one of these:
http://www.jaycar.com.au/PRODUCTS/Passive-Components/Resistors/Other-Resistors/Resistance-Wheel/p/RR0700
Aaron
Nice gadget. At $30 though, a packet of 8 resistors is 50cents... I can buy a packet of each. Downside is just more stuffing around. What I've learned today - 470 to 620 makes bugger all difference for blue.

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