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### Network # Series resistor for LED?

B

#### Bill Bowden

Jan 1, 1970
0
Anybody know a formula to work out the resistor needed for several
series LEDs operating from a rectified AC source with no filter
capacitor?

Say the AC rectified source is 24 volts RMS, 34 volts peak,and we have
5 white LEDs at 3 volts each in series drawing 20mA RMS. What series
resistor and power rating is needed?

Since the LEDs will not light until the voltage rises above 15 volts,
the conduction time will be some fraction of the half cycle. The peak
current will be 34 minus 15 divided by whatever resistor is used, which
will yield some RMS value. But since the duty cycle will be less than
100%, the RMS value needs to be higher by some amount as a function of
duty cycle.

Is there some formula to work out the needed resistor?

-Bill

J

#### John Fields

Jan 1, 1970
0
Anybody know a formula to work out the resistor needed for several
series LEDs operating from a rectified AC source with no filter
capacitor?

Say the AC rectified source is 24 volts RMS, 34 volts peak,and we have
5 white LEDs at 3 volts each in series drawing 20mA RMS. What series
resistor and power rating is needed?

Since the LEDs will not light until the voltage rises above 15 volts,
the conduction time will be some fraction of the half cycle. The peak
current will be 34 minus 15 divided by whatever resistor is used, which
will yield some RMS value. But since the duty cycle will be less than
100%, the RMS value needs to be higher by some amount as a function of
duty cycle.

Is there some formula to work out the needed resistor?

---
Since the diodes are going to be conducting even if they haven't lit
up, I think if you use the RMS value of the AC source it'll come out
right. That is:

Vrms - Vled 24V - 15V
Rs = ------------- = ----------- = 450 ohms
Iled 0.02A

Closest safe 5% is 470, so the RMS current will go down to:

Vrms - Vled 24V - 15V
Iled = ------------- = ----------- ~ 0.019A
Rs 470R

And the power dissipation in the resistor will be:

P = Iled * Vrms - Vled = 0.019A * 9V ~ 0.172 watts

B

#### Bill Bowden

Jan 1, 1970
0
Since the diodes are going to be conducting even if they haven't lit
up, I think if you use the RMS value of the AC source it'll come out
right.

Yes, that's probably close, but the diodes only conduct about 600
nanoamps at 2 volts each, and probably nothing below that, so there is
about 17 degrees out of 90 with no current.

So, I'm figuring the RMS value might be 20% lower, but I don't have a
true RMS meter to measure it.

-Bill

F

#### Fritz Schlunder

Jan 1, 1970
0
John Fields said:
---
Since the diodes are going to be conducting even if they haven't lit
up, I think if you use the RMS value of the AC source it'll come out
right. That is:

Vrms - Vled 24V - 15V
Rs = ------------- = ----------- = 450 ohms
Iled 0.02A

Hmm... This is an interesting question... I hope people were paying
attention in math class.

So uhh... What is the justification for your above formula? When I use my
method I calculate a resistance of 440 Ohms, which is amazingly close to
yours, but I am having trouble seeing how your method could be accurate.
Suppose you hooked up a set of LEDs with V(f) of say 24V. In that case Vrms
is still 24V, but then your formula would result in a resistance of 0 Ohms.
That won't do. But even from 24Vrms AC, an LED string with V(f) of 24V
should still be drivable.

The way I calculated 440 ohms as appropriate took more than one step.

Step 1: find out when (in seconds) the instantaneous AC voltage exceeds 15V,
and also when in the cycle it also drops below 15V.

Suppose we look at one half wave of the sinusoid which at 60Hz would last
8.33ms. The instantaneous voltage is V(inst.)=34sin(377t). Solve for
V(inst.)=15V. I get 1.212ms and 7.121ms.

Step 2: apply formula:
I(dc average in LED) = {integral[(34sin(377t) - 15)/R]dt}/0.00833 seconds
with upper limit at 0.007121 seconds, and lower limit at 0.001212 seconds.
When I evaluate that only mildly ugly integral and solve for R I get:

R= 0.07323/[I(dc average in LED)*0.00833]

For 20mA this is 440 Ohms. For 470 ohms, DC average current in LED is
18.7mA.

If we integrate the instantaneous voltage formula with respect to time
(during that interval when V(inst.) is above 15V) divided by the resistance
R, when we should get the total coulombs of charge that flows through the
LEDs during that half cycle. To get the equivalent DC average current
through the LEDs, we simply divide that result by the duration of that total
half cycle (8.33ms). After all, the definition of current is charge per
unit time, or in other words, one ampere = one coulomb per second.

When I use the above expression and plug in the desired I(dc average in LED)
equal to 20mA and solve for R, I get 440 ohms. Hopefully I pressed the
right buttons on my calculator, but that number seems somewhat believable.

Note that the DC average current through the LEDs will be 20mA, so assuming
the efficiency of the LEDs doesn't decrease at higher peak currents, then
the LEDs should be equally bright as if powered from pure 20mA DC. In
practice the forward voltage of the LEDs does increase somewhat at higher
current densities, so the result won't be perfectly accurate, and the
dissipation in the LEDs will be slightly higher than what it would be from
20mA pure DC. One should probably derate the forward current slightly for
this kind of use.

To find the power dissipation in the resistor, one would have to integrate
the instantaneous power in the resistor (during the interval any current
actually flows through the resistor) divided by the total time of the cycle.

Or in other words:

Average power dissipation in resistor =
{integral([(34sin(377t)-15)^2]/R)dt}/0.00833 where the limits of integration
are also 1.212ms to 7.121ms. Evaluating I get:

Average power dissipation in resistor R = (1.1/R)/0.00833

For 440 ohms, this would be approximately 300mW. For 470 ohms this would be
approximately 280mW.

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