How will I know how many ohms I have to resist with 11 leds in Phat PSP?

[SEA DAWG]

Should I use a 30k ohm resistor?

For four leds, the llamma.com PSP led tut says to use an 10k ohm resistor. How can I find this out on my own?

Thx!

[Ben.]

If You want To Find Out What resistor is needed, try here.. http://ledcalculator.net/

[DarkShot]

It depends on the forward voltage, and how you wire them.

If you wire them in parallel to a 5v point, than a 100ohm resistor will generally suffice unless they are different, or a colour like red or yellow.

30Kohm Is a little overkill.

[SEA DAWG]

If You want To Find Out What resistor is needed, try here.. http://ledcalculator.net/

Thanks, I tried this, but it says I would need to hook up 91 ohm resistors to each led (I did eight). If I add all of them up I get 728 ohms, so could I just put in a resistor that resists more than 728 ohms? Like possibly a 10k ohm resistor?

It depends on the forward voltage, and how you wire them.

If you wire them in parallel to a 5v point, than a 100ohm resistor will generally suffice unless they are different, or a colour like red or yellow.

30Kohm Is a little overkill.

What if I am wiring them directly to the power switch? Is that a 5v point?

[DarkShot]

All you would need is the one resistor, and have the leds branch off that one.

As for the power switch, what exactly do you mean?

[bustinthejustin]

he means is the power switch a 5v point. nope its a ground.

[SEA DAWG]

All you would need is the one resistor, and have the leds branch off that one.

As for the power switch, what exactly do you mean?

ok, but what resistor should I use? 10k ohm? 100 ohm? Like how do would I know what one to use? Does it have to me a resistor the resists above 728 ohms?

he means is the power switch a 5v point. nope its a ground.

hmm ok well I have two leds hooked up. The negative wire from the led is hooked up to the negative point on the power switch, while the positive wire of the led is connected, with a 10k ohm resistor, to the positive point on the power switch. Both of the leds work so if the switch is ground, how am I getting power to them? And is the amount of power 5v?

[Blizzrad]

The relationship between voltage, resistance, and current is described in one of the fundamental laws of electronics, Ohm's law, as :

V=IR (voltage = current x resistance) *when expressed in Volts, Amps, and Ohms.

With this, along with it's equivalents...

I=V/R (current = voltage / resistance)
R=V/I (resistance = voltage / current)

we can calculate what resistance should be applied to our circuit to achieve a desired current flow (this is also how the calculator linked above comes up with it's answer). First we have to know our voltage source (we will say 5 volts for this example) and the amount of current we want running through the LED. More current will increase the brightness of the LED, but too much will cause it to burn out. The maximum current rating for a given LED is always specified in its datasheet, but we will just assume 20mA which is a common rating for ordinary LEDs. Since 20mA is the max rated current, we will go with a safe 15ma for our desired current through the LED (though for this type of thing you should be fine with 20mA if you want the extra brightness). There is also a voltage drop across the LED which must be factored in, we will just say it is 2v here (also specified in the datasheet). So now we plug these values into our equation:

R=V/I
Resistance = (5v - 2v) / 0.015A
Resistance = 3v / 0.015A
Resistance = 200 ohms

[SEA DAWG]

The relationship between voltage, resistance, and current is described in one of the fundamental laws of electronics, Ohm's law, as :

V=IR (voltage = current x resistance) *when expressed in Volts, Amps, and Ohms.

With this, along with it's equivalents...

I=V/R (current = voltage / resistance)
R=V/I (resistance = voltage / current)

we can calculate what resistance should be applied to our circuit to achieve a desired current flow (this is also how the calculator linked above comes up with it's answer). First we have to know our voltage source (we will say 5 volts for this example) and the amount of current we want running through the LED. More current will increase the brightness of the LED, but too much will cause it to burn out. The maximum current rating for a given LED is always specified in its datasheet, but we will just assume 20mA which is a common rating for ordinary LEDs. Since 20mA is the max rated current, we will go with a safe 15ma for our desired current through the LED (though for this type of thing you should be fine with 20mA if you want the extra brightness). There is also a voltage drop across the LED which must be factored in, we will just say it is 2v here (also specified in the datasheet). So now we plug these values into our equation:

R=V/I
Resistance = (5v - 2v) / 0.015A
Resistance = 3v / 0.015A
Resistance = 200 ohms

ohhh k, but what does that mean for the number of resistors I should use? I know that means I need a 200 ohm resistor for each led (or whichever resistor they make that is closest to being 200 ohms). So, could I just add it all up and get one resistor, meaning 200 X 8 = 1600, so hooking up all the leds to a resistor that can resist 1600 ohms?

[rceckspurt13]

No. Just use a single resistor. Attach to resistor to the 5 volt point and then attach all of the LEDs to the single resistor.

[SEA DAWG]

No. Just use a single resistor. Attach to resistor to the 5 volt point and then attach all of the LEDs to the single resistor.

A resistor that can resist 1600 ohms---maybe a 20k ohm resistor?

[pspupgrade]

Try using the SickMods calculator: http://diy.sickmods.net/Tutorials/Case_Modding/Resistor_Calculator/
 You can just specify what colour your 0603 SMD LED is with the amount of them and, assuming you're using a 5v power source, it should provide you witht the correct type.

[SEA DAWG]

Try using the SickMods calculator: http://diy.sickmods.net/Tutorials/Case_Modding/Resistor_Calculator/
 You can just specify what colour your 0603 SMD LED is with the amount of them and, assuming you're using a 5v power source, it should provide you witht the correct type.

Ok wait, I got 12 ohm 1/2 watts, but for the when I used ohm's law, I got a much different answer! I got a total of 200 ohms, and 200 (ohms X 8 (leds) = 1600 ohms. But with the calculator, I got 12 ohm 1/2 watts. How could a 12 ohm 1/2 watts resistor resist 1600 ohms?

But anyways, I got 12 ohm 1/2 watt. So does that mean I use a 12 ohm 1/2 watt resistor for each led (8 of them) meaning I need 8, 12 ohm 1/2 watt resistors?

Or do I use just one? And hook all of the leds up to it?

[Blizzrad]

Ok wait, I got 12 ohm 1/2 watts, but for the when I used ohm's law, I got a much different answer! I got a total of 200 ohms, and 200 (ohms X 8 (leds) = 1600 ohms. But with the calculator, I got 12 ohm 1/2 watts. How could a 12 ohm 1/2 watts resistor resist 1600 ohms?

The resistors would only add up to 1600 ohms if they were all in series, when placed in parallel their resistance decreases. So, using the example of eight 200 ohm resistors...

resistance in series = R1+R2+... R8
resistance in series = 1600 ohms

resistance in parallel = 1/(1/R1+1/R2+... 1/R8)
resistance in parallel = 25 ohms

The calculators all use Ohm's law as well, the difference you are seeing is that one resistor is calculated for a parallel circuit of 8 LEDs (sharing one resistor) and the other is for a series circuit of 1 LED. Current is divided equally between the LEDs in the parallel circuit (assuming they are all identical), so the current limiting resistor would be of a much smaller resistance than in a series circuit of only one LED. The wattage rating is also higher because all current in the circuit is flowing through this single resistor. The resistors used depends heavily on the configuration of the circuit.

If the forward voltage of your LEDs is low enough to accommodate placing two in series from your 5v source, I recommend wiring them in four parallel sets of two LEDs in series with a resistor. Placing several LEDs in parallel off a single resistor is generally not a good idea (though I have done it plenty of times when I am too lazy to rummage around for more resistors.)

[SEA DAWG]

The resistors would only add up to 1600 ohms if they were all in series, when placed in parallel their resistance decreases. So, using the example of eight 200 ohm resistors...

resistance in series = R1+R2+... R8
resistance in series = 1600 ohms

resistance in parallel = 1/(1/R1+1/R2+... 1/R8)
resistance in parallel = 25 ohms

The calculators all use Ohm's law as well, the difference you are seeing is that one resistor is calculated for a parallel circuit of 8 LEDs (sharing one resistor) and the other is for a series circuit of 1 LED. Current is divided equally between the LEDs in the parallel circuit (assuming they are all identical), so the current limiting resistor would be of a much smaller resistance than in a series circuit of only one LED. The wattage rating is also higher because all current in the circuit is flowing through this single resistor. The resistors used depends heavily on the configuration of the circuit.

If the forward voltage of your LEDs is low enough to accommodate placing two in series from your 5v source, I recommend wiring them in four parallel sets of two LEDs in series with a resistor. Placing several LEDs in parallel off a single resistor is generally not a good idea (though I have done it plenty of times when I am too lazy to rummage around for more resistors.)

Ok so if I had four sets of two leds wired up in a parallel circut, I would use a total of four resistors?

And also, a parallel circut is having all of the + wires of leds hooked up two one wire and all of the - hooked to another wire. While a series circuit is having the - and + attached to each other. Am I correct?

[Blizzrad]

Yes, yes, and yes. So now to find the resistance needed in each branch of your parallel circuit using Ohm's law...

resistance = (supply voltage - LED1 forward voltage - LED2 forward voltage) / desired current

[SEA DAWG]

Yes, yes, and yes. So now to find the resistance needed in each branch of your parallel circuit using Ohm's law...

resistance = (supply voltage - LED1 forward voltage - LED2 forward voltage) / desired current

Ok what do you mean by desired current though?

[Blizzrad]

Desired current is the amount of current you want to have flowing through your LEDs, this will determine their brightness, and the rate at which your battery drains. Less resistance will allow more current flow, while more resistance will yield the opposite. Current through an LED should be kept within the specifications provided in it's datasheet.

[SEA DAWG]

Desired current is the amount of current you want to have flowing through your LEDs, this will determine their brightness, and the rate at which your battery drains. Less resistance will allow more current flow, while more resistance will yield the opposite. Current through an LED should be kept within the specifications provided in it's datasheet.

Ok this is what I got (here is the site I bought my exact leds at: http://www.unique-leds.com/index.php?target=products&product_id=1734 )

resistance = (5v-3.4v-3.4v)(divided by)(20 mA) = -0.05 mA (V)

I don't know (and don't know how to) if I did the math correct, could you explain if I was correct or not, cause I don't think -0.05 mA (V) is a legit answer.

Also, could you explain how to do it, I think I am having trouble with the conversions.

[Blizzrad]

There are a few problems here. Firstly since you are using blue LEDs with a 3v forward voltage, the combined forward voltage of two in series will exceed the 5v source voltage. These LEDs can not be placed in series (unless the source voltage is raised), this is why the equation returns a negative result. Secondly, the equation above is meant to solve for resistance with voltage and current already known, so the result should yield the required resistance in ohms for each parallel branch of the circuit. Finally, for the equation to work, all values must be expressed in volts, ohms, and amps, meaning that for 20mA you would enter it as 0.02A.

[SEA DAWG]

There are a few problems here. Firstly since you are using blue LEDs with a 3v forward voltage, the combined forward voltage of two in series will exceed the 5v source voltage. These LEDs can not be placed in series (unless the source voltage is raised), this is why the equation returns a negative result. Secondly, the equation above is meant to solve for resistance with voltage and current already known, so the result should yield the required resistance in ohms for each parallel branch of the circuit. Finally, for the equation to work, all values must be expressed in volts, ohms, and amps, meaning that for 20mA you would enter it as 0.02A.

Wait, I thought that equation was for a parallel circuit. Well, since it isn't, how would I do it for a parallel circuit?

Also, how is it that we can divide by amps, wouldn't it have to cross out? Cause I got -168.4, so what would that be? Amps? Volts?

[Blizzrad]

The equation will tell you the resistance required in each identical branch of your parallel circuit, the LEDs you are planning to use cannot be placed in series with a 5v source because any more than one of them will amount to a combined forward voltage that is greater than 5. So to find the resistance needed in a parallel branch with a 5v source using one of your LEDs we calculate...

resistance = (source voltage - LED forward voltage) / desired current
resistance = (5v - 3v) / 0.02A
resistance = 2v / 0.02A
resistance = 100ohms

I'm not sure how you arrived at -168.4, or what you mean by "cross out", but since we are solving for resistance the result will always be in ohms. Also, just a side note, it's not necessary to quote entire replies in every single post, as this tends to clutter/bloat the thread.

[SEA DAWG]

Oh ok, so this equation is 5v - 3v (which is one led) so I only use one led with a 100ohm resistor?

Oh ok sry, thx.

[Blazinkaos]

Man i remember back in the day when i didnt know how to do this.

Oh ok, so this equation is 5v - 3v (which is one led) so I only use one led with a 100ohm resistor?

Oh ok sry, thx.

Yes your correct. Learning fast nice job.

[SEA DAWG]

oh ok thx! yes!

now one last question. Since I need 100ohms for each led, could I just use 200ohms for 2 leds?

Or would the LED forward voltage be to high (exceed 5v) if I connected them all together.

[Blazinkaos]

Nono lol ok every time you add one more the the resistance change here this should help.
Ok this is an example 5-3=2 divided by .04(each led is .02 amps) once that is divided you get 50 ohm so you need a resistor thats above 50 and closest to it.

[SEA DAWG]

Oh! so if I was to have 8, I would go 5-3=2
Now 2/.16= 12.5ohms

So a resistor that is above 12.5ohm and closest to it.

But is 8 leds to one resistor recommended?

[Blizzrad]

Exactly, the current is divided in a parallel circuit, so a smaller resistance is needed to provide the desired current through each LED.

It's generally not recommended to wire LEDs in parallel with a single resistor, though given the limited space for cramming resistors into a PSP it might be difficult to avoid. One reason this can be problematic is that if the forward voltages of the LEDs in parallel are too far apart, the one with the lowest forward voltage will pass all of the current while the others will not conduct. You can see this clearly if you put a red, green, and blue led together in parallel with one resistor, only the red led will light up, and will pass all of the current because of it's lower forward voltage. Engineers would of course avoid this in a commercial product, but there are probably others here who have wired their LEDs this way and not had problems. Be sure to use a resistor that is rated for at least 1/2 watt.

[Blazinkaos]

Oh! so if I was to have 8, I would go 5-3=2
Now 2/.16= 12.5ohms

So a resistor that is above 12.5ohm and closest to it.

But is 8 leds to one resistor recommended?

There go man now you got nice job.

[SEA DAWG]

Blizzrad, what do you mean by, "forward voltages of the LEDs in parallel are too far apart, the one with the lowest forward voltage will saturate and pass all of the current while the others will not conduct."?

Could you please explain that more? Is it because the red takes more voltage than the blue and green, thus the blue and green won't light up?