----------------------------

#937 Sep 5, 2012

Hi, I have a simulation going for an LED driver; I am going to be using a few zillion Kingbright LED's: 660 nm peak, 20 mA forward current, 1.85 forward voltage drop. I made up an LTspice model using a curve-fitting program from an LTspice contributor, but only for the forward voltage vs current:

.model Red_Led_660nm3mm d(

+ IS=4.58E-13 N=2.973 RS=1.197388

+ EG=1.43 XTI=3)

The datasheet only gives me the following data: 1.85 forward voltage drop at 20 mA...is it possible for me to calculate...or at least approximate...the forward resistance?

Do I simply do R = E/I, as in: 1.85/.020 = 92.5 ohms?

Now, the datasheet DOES say max 30 mA If and max 75 mW power dissipation, so....if I assume that it is dissipating 75 mA at 30 mA If...and I can see from the graph that at 30 mA the forward voltage drop is circa 1.89 volts...

So, E/I = 1.89/.03 = 60 ohms at 30 mA....if that were true, then

75 mW should = I2xR = .03 x .03 x 60 = 54 mW, so it doesn't work out super well...75 mW is in the same ballpark as 43 mW but they're not equal....

could someone please point out the errors of my ways, and the true path here to enlightenment?

VB, Maturin

----------------------------

#938 Sep 5, 2012

How you normally make an LED is by subtracting the VFWD from the supply voltage. Say you are going to hit it with 5v and the forward drop is 1.8...that gives you 3.2v left. Now, to get the current through the LED you will want to divide that 3.2v by the amount of current you want. For 20 mA this works out to about 160 ohms. I would use a 180 ohm resistor just in case. Now, multiply 3.2v by 0.02 amps to get the power in the resistor...64 mW, so a 1/4w resistor could easily take on this burden. The 180 ohm resistor will restrict current a little more so you'll get less power loss in the circuit but just a little less brightness in the LED. The LED is going to dissipate 1.8 * 0.02 = 36 mW. That isn't anything you can fix but it's so low that it's not worth worrying about.

LEDs have a forward drop that is fairly stable with amounts of current. You will get a little drift based on the current but it won't be so much that you can't plan the driver with a simple circuit. With the LED, the forward resistance is basically the voltage drop divided by the current. It's more helpful to multiply the current by the forward voltage drop for power dissipated. I hope I helped.

Ed

On Tue, Sep 4, 2012 at 11:11 PM, Gandolf gandolfreefer@...> wrote:

.Hi, I have a simulation going for an LED driver; I am going to be using a few zillion Kingbright LED's: 660 nm peak, 20 mA forward current, 1.85 forward voltage drop. I made up an LTspice model using a curve-fitting program from an LTspice contributor, but only for the forward voltage vs current:

.model Red_Led_660nm3mm d(

+ IS=4.58E-13 N=2.973 RS=1.197388

+ EG=1.43 XTI=3)

The datasheet only gives me the following data: 1.85 forward voltage drop at 20 mA...is it possible for me to calculate...or at least approximate...the forward resistance?

Do I simply do R = E/I, as in: 1.85/.020 = 92.5 ohms?

Now, the datasheet DOES say max 30 mA If and max 75 mW power dissipation, so....if I assume that it is dissipating 75 mA at 30 mA If...and I can see from the graph that at 30 mA the forward voltage drop is circa 1.89 volts...

So, E/I = 1.89/.03 = 60 ohms at 30 mA....if that were true, then

75 mW should = I2xR = .03 x .03 x 60 = 54 mW, so it doesn't work out super well...75 mW is in the same ballpark as 43 mW but they're not equal....

could someone please point out the errors of my ways, and the true path here to enlightenment?

VB, Maturin

----------------------------

#939 Sep 5, 2012

1. Go to the manufacturer���s website�� -�� www.kingbrightusa.com/

----------------------------

#940 Sep 5, 2012

The datasheet only gives me the following data: 1.85 forward voltage drop at 20 mA...is it possible for me to calculate...or at least approximate...the forward resistance?

Do I simply do R = E/I, as in: 1.85/.020 = 92.5 ohms?

Now, the datasheet DOES say max 30 mA If and max 75 mW power dissipation, so....if I assume that it is dissipating 75 mA at 30 mA If...and I can see from the graph that at 30 mA the forward voltage drop is circa 1.89 volts...

So, E/I = 1.89/.03 = 60 ohms at 30 mA....if that were true, then 75 mW should = I2xR = .03 x .03 x 60 = 54 mW, so it doesn't work out super well...75 mW is in the same ballpark as 43 mW but they're not equal....

could someone please point out the errors of my ways, and the true path here to enlightenment?

----------------------------

#941 Sep 5, 2012

--- On Wed, 9/5/12, Chris Albertson albertson.chris@...> wrote:

> From: Chris Albertson albertson.chris@...>

> Subject: Re: [funwithtransistors] how to calculate forward resistance LED?

> To: funwithtransistors@yah

> Date: Wednesday, September 5, 2012, 4:08 PM

>

> Could it be that diodes don't act like resisters. .

They don't. They have a VERY strong exponential relationship

between the current and the voltage.

en.wikipedia.org/wiki/

Or, for those more comfortable with a visual presentation:

pw1.netcom.com/~wa4qal

> I think they have a near constant drop because of the PN

> junction and then once they conduct might act like resistors. .

There is some slope to the curve, since there is a bit of

internal resistance. But, at low currents, the exponential

portion of the curve factors in.

> I don't really know but some how you'd need to account for the

> about 0.7V forward voltage drop

The voltage drop depends upon the material and doping. It's about

1.5 Volts (or thereabouts) for red LEDs and goes up to about 2.5

Volts for blue LEDs (or, thereabouts).

Grab some and do an I-V plot.

> Chris Albertson

> Redondo Beach, California

Dave

----------------------------

#942 Sep 5, 2012

Thanks, Dave and all....

....the answers portray the same head-scratcher I have been puzzling over:

a diode has a forward voltage drop. In plain-old-diode land, that's about 0.7 volts.

But an LED is not a plain-old-diode; it is converting some portion of that forward voltage drop to light. It is emitting photons. It is also, like a plain-old-diode, emitting plain-old-heat; presumably the power dissipation.

So far so good - (BTW Jerry the datasheet I'm quoting does come from Kingbright and it does not include the forward resistance parameter, which is why I posted this in the first place) - but, when I look at spice models for diodes, there is an equation for the forward voltage drop that is related to current in an exponential fashion, and this graph, If vs Vf (forward current vs forward voltage drop), is on any self-respecting LED datasheet.

But there is also the parameter "forward resistance" in spice models that go further in modeling LED's... and herein lies the mystery.

If current vs voltage defines an LED sufficiently, why even have forward resistance? Or is forward resistance just another way of stating current vs voltage (R = E/I)? I think forward resistance is something additional...but when it's not in the datasheet, and I don't find a graph that gives me forward resistance vs current (or voltage drop), then I don't know how to add it to the spice model.

So, anybody really know what the heck "forward resistance" is in an LED? (or in a diode, for that matter)

Very Best, Maturin

--- In funwithtransistors@yah

>

> --- On Wed, 9/5/12, Chris Albertson albertson.chris@...> wrote:

>

> > From: Chris Albertson albertson.chris@...>

> > Subject: Re: [funwithtransistors] how to calculate forward resistance LED?

> > To: funwithtransistors@yah

> > Date: Wednesday, September 5, 2012, 4:08 PM

> >

> > Could it be that diodes don't act like resisters. .

>

> They don't. They have a VERY strong exponential relationship

> between the current and the voltage.

>

> en.wikipedia.org/wiki/

>

> Or, for those more comfortable with a visual presentation:

>

> pw1.netcom.com/~wa4qal

>

> > I think they have a near constant drop because of the PN

> > junction and then once they conduct might act like resistors. .

>

> There is some slope to the curve, since there is a bit of

> internal resistance. But, at low currents, the exponential

> portion of the curve factors in.

>

> > I don't really know but some how you'd need to account for the

> > about 0.7V forward voltage drop

>

> The voltage drop depends upon the material and doping. It's about

> 1.5 Volts (or thereabouts) for red LEDs and goes up to about 2.5

> Volts for blue LEDs (or, thereabouts).

>

> Grab some and do an I-V plot.

>

> > Chris Albertson

> > Redondo Beach, California

>

> Dave

>

----------------------------

#943 Sep 5, 2012

Forward resistance can only be defined at a specific point. If you set the

current to a specific value and read the resulting voltage you can calculate

a resistance value that will hold for that particular value of current and

voltage. If you change the current or the temperature, which will change

the voltage at the same current, the resistance will change. You can't

calculate the resistance at one current and apply it to another current. In

other words PN junctions don't obey ohms law.

Regards.

Max. K 4 O DS.

Email: max@...

Transistor site www.funwithtransistors

Vacuum tube site: www.funwithtubes.net

Woodworking site

www.angelfire.com/elec

Music site: www.maxsmusicplace.com

To subscribe to the fun with transistors group send an email to.

funwithtransistors-sub

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funwithtubes-subscribe

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funwithwood-subscribe@

----- Original Message -----

From: "Gandolf" gandolfreefer@...>

To: funwithtransistors@yah

Sent: Wednesday, September 05, 2012 1:11 AM

Subject: [funwithtransistors] how to calculate forward resistance LED?

> Hi, I have a simulation going for an LED driver; I am going to be using a

> few zillion Kingbright LED's: 660 nm peak, 20 mA forward current, 1.85

> forward voltage drop. I made up an LTspice model using a curve-fitting

> program from an LTspice contributor, but only for the forward voltage vs

> current:

>

> .model Red_Led_660nm3mm d(

> + IS=4.58E-13 N=2.973 RS=1.197388

> + EG=1.43 XTI=3)

>

> The datasheet only gives me the following data: 1.85 forward voltage drop

> at 20 mA...is it possible for me to calculate...or at least

> approximate...the forward resistance?

>

> Do I simply do R = E/I, as in: 1.85/.020 = 92.5 ohms?

>

> Now, the datasheet DOES say max 30 mA If and max 75 mW power dissipation,

> so....if I assume that it is dissipating 75 mA at 30 mA If...and I can see

> from the graph that at 30 mA the forward voltage drop is circa 1.89

> volts...

>

> So, E/I = 1.89/.03 = 60 ohms at 30 mA....if that were true, then

> 75 mW should = I2xR = .03 x .03 x 60 = 54 mW, so it doesn't work out super

> well...75 mW is in the same ballpark as 43 mW but they're not equal....

>

> could someone please point out the errors of my ways, and the true path

> here to enlightenment?

>

> VB, Maturin

>

>

>

---------------

>

> Always keep electrons and holes flowing in opposite directions.Yahoo!

> Groups Links

>

>

>

>

----------------------------

#944 Sep 5, 2012

Thanks, Max. I emailed Kingbright, and their reply was that the forward resistance was simply E/I for a particular current, as read off the I/V graph.

What got me confused about this - my bad - was a previous model for the TI LM4041 shunt voltage reference, which had a forward resistance of about 0.7 ohms....and a voltage drop of 1.225 volts....but I could not find a model for it, and the device is actually quite complex, so I ended up - at the suggestion of modeling it as a plain-old-diode with a voltage drop of 1.225 volts and forward resistance of 0.7 ohms, which worked fine...

....but, LOL, I then FORGOT I was approximating a shunt by pretending it was a plain-old-diode and - then got all puzzled by the forward resistance thingie.

I THINK I can simply model each LED with the forward voltage drop for that LED - at that particular current, that is, it wouldn't be valid at any other current, - but since I AM current controlling at 20 mA anyway in the circuit, the result should be correct, without adding an additional resistance. I think.

VB, Maturin

--- In funwithtransistors@yah

>

> Forward resistance can only be defined at a specific point. If you set the

> current to a specific value and read the resulting voltage you can calculate

> a resistance value that will hold for that particular value of current and

> voltage. If you change the current or the temperature, which will change

> the voltage at the same current, the resistance will change. You can't

> calculate the resistance at one current and apply it to another current. In

> other words PN junctions don't obey ohms law.

>

> Regards.

>

> Max. K 4 O DS.

>

> Email: max@...

>

> Transistor site www.funwithtransistors

> Vacuum tube site: www.funwithtubes.net

> Woodworking site

> www.angelfire.com/elec

> Music site: www.maxsmusicplace.com

>

> To subscribe to the fun with transistors group send an email to.

> funwithtransistors-sub

>

> To subscribe to the fun with tubes group send an email to,

> funwithtubes-subscribe

>

> To subscribe to the fun with wood group send a blank email to

> funwithwood-subscribe@

>

> ----- Original Message -----

> From: "Gandolf" gandolfreefer@...>

> To: funwithtransistors@yah

> Sent: Wednesday, September 05, 2012 1:11 AM

> Subject: [funwithtransistors] how to calculate forward resistance LED?

>

>

> > Hi, I have a simulation going for an LED driver; I am going to be using a

> > few zillion Kingbright LED's: 660 nm peak, 20 mA forward current, 1.85

> > forward voltage drop. I made up an LTspice model using a curve-fitting

> > program from an LTspice contributor, but only for the forward voltage vs

> > current:

> >

> > .model Red_Led_660nm3mm d(

> > + IS=4.58E-13 N=2.973 RS=1.197388

> > + EG=1.43 XTI=3)

> >

> > The datasheet only gives me the following data: 1.85 forward voltage drop

> > at 20 mA...is it possible for me to calculate...or at least

> > approximate...the forward resistance?

> >

> > Do I simply do R = E/I, as in: 1.85/.020 = 92.5 ohms?

> >

> > Now, the datasheet DOES say max 30 mA If and max 75 mW power dissipation,

> > so....if I assume that it is dissipating 75 mA at 30 mA If...and I can see

> > from the graph that at 30 mA the forward voltage drop is circa 1.89

> > volts...

> >

> > So, E/I = 1.89/.03 = 60 ohms at 30 mA....if that were true, then

> > 75 mW should = I2xR = .03 x .03 x 60 = 54 mW, so it doesn't work out super

> > well...75 mW is in the same ballpark as 43 mW but they're not equal....

> >

> > could someone please point out the errors of my ways, and the true path

> > here to enlightenment?

> >

> > VB, Maturin

> >

> >

> >

---------------

> >

> > Always keep electrons and holes flowing in opposite directions.Yahoo!

> > Groups Links

> >

> >

> >

> >

>

----------------------------

#945 Sep 6, 2012

That's right.

Regards.

Max. K 4 O DS.

Email: max@...

Transistor site www.funwithtransistors

Vacuum tube site: www.funwithtubes.net

Woodworking site

www.angelfire.com/elec

Music site: www.maxsmusicplace.com

To subscribe to the fun with transistors group send an email to.

funwithtransistors-sub

To subscribe to the fun with tubes group send an email to,

funwithtubes-subscribe

To subscribe to the fun with wood group send a blank email to

funwithwood-subscribe@

----- Original Message -----

From: "Gandolf" gandolfreefer@...>

To: funwithtransistors@yah

Sent: Wednesday, September 05, 2012 5:56 PM

Subject: [funwithtransistors] Re: how to calculate forward resistance LED?

> Thanks, Max. I emailed Kingbright, and their reply was that the forward

> resistance was simply E/I for a particular current, as read off the I/V

> graph.

>

> What got me confused about this - my bad - was a previous model for the TI

> LM4041 shunt voltage reference, which had a forward resistance of about

> 0.7 ohms....and a voltage drop of 1.225 volts....but I could not find a

> model for it, and the device is actually quite complex, so I ended up - at

> the suggestion of modeling it as a plain-old-diode with a voltage drop

> of 1.225 volts and forward resistance of 0.7 ohms, which worked fine...

>

> ....but, LOL, I then FORGOT I was approximating a shunt by pretending it

> was a plain-old-diode and - then got all puzzled by the forward

> resistance thingie.

>

> I THINK I can simply model each LED with the forward voltage drop for that

> LED - at that particular current, that is, it wouldn't be valid at any

> other current, - but since I AM current controlling at 20 mA anyway in the

> circuit, the result should be correct, without adding an additional

> resistance. I think.

>

> VB, Maturin

>

> --- In funwithtransistors@yah

>>

>> Forward resistance can only be defined at a specific point. If you set

>> the

>> current to a specific value and read the resulting voltage you can

>> calculate

>> a resistance value that will hold for that particular value of current

>> and

>> voltage. If you change the current or the temperature, which will change

>> the voltage at the same current, the resistance will change. You can't

>> calculate the resistance at one current and apply it to another current.

>> In

>> other words PN junctions don't obey ohms law.

>>

>> Regards.

>>

>> Max. K 4 O DS.

>>

>> Email: max@...

>>

>> Transistor site www.funwithtransistors

>> Vacuum tube site: www.funwithtubes.net

>> Woodworking site

>> www.angelfire.com/elec

>> Music site: www.maxsmusicplace.com

>>

>> To subscribe to the fun with transistors group send an email to.

>> funwithtransistors-sub

>>

>> To subscribe to the fun with tubes group send an email to,

>> funwithtubes-subscribe

>>

>> To subscribe to the fun with wood group send a blank email to

>> funwithwood-subscribe@

>>

>> ----- Original Message -----

>> From: "Gandolf" gandolfreefer@...>

>> To: funwithtransistors@yah

>> Sent: Wednesday, September 05, 2012 1:11 AM

>> Subject: [funwithtransistors] how to calculate forward resistance LED?

>>

>>

>> > Hi, I have a simulation going for an LED driver; I am going to be using

>> > a

>> > few zillion Kingbright LED's: 660 nm peak, 20 mA forward current, 1.85

>> > forward voltage drop. I made up an LTspice model using a curve-fitting

>> > program from an LTspice contributor, but only for the forward voltage

>> > vs

>> > current:

>> >

>> > .model Red_Led_660nm3mm d(

>> > + IS=4.58E-13 N=2.973 RS=1.197388

>> > + EG=1.43 XTI=3)

>> >

>> > The datasheet only gives me the following data: 1.85 forward voltage

>> > drop

>> > at 20 mA...is it possible for me to calculate...or at least

>> > approximate...the forward resistance?

>> >

>> > Do I simply do R = E/I, as in: 1.85/.020 = 92.5 ohms?

>> >

>> > Now, the datasheet DOES say max 30 mA If and max 75 mW power

>> > dissipation,

>> > so....if I assume that it is dissipating 75 mA at 30 mA If...and I can

>> > see

>> > from the graph that at 30 mA the forward voltage drop is circa 1.89

>> > volts...

>> >

>> > So, E/I = 1.89/.03 = 60 ohms at 30 mA....if that were true, then

>> > 75 mW should = I2xR = .03 x .03 x 60 = 54 mW, so it doesn't work out

>> > super

>> > well...75 mW is in the same ballpark as 43 mW but they're not equal....

>> >

>> > could someone please point out the errors of my ways, and the true path

>> > here to enlightenment?

>> >

>> > VB, Maturin

>> >

>> >

>> >

---------------

>> >

>> > Always keep electrons and holes flowing in opposite directions.Yahoo!

>> > Groups Links

>> >

>> >

>> >

>> >

>>

>

>

>

>

---------------

>

> Always keep electrons and holes flowing in opposite directions.Yahoo!

> Groups Links

>

>

>

>