# Resistor problem with LEDs

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#### ibrahim03

##### Full Member level 1
I am facing a rather simple problem and want to confirm my doubts. I wanted to use LEDs in a particular circuit of mine,the voltage level was 32V DC and I wanted about 10mA of current to pass through the LED (green Color) to get the appropriate brightness so I used the good old Ohm's Law to calculate the resistance as :

R = 32V/10mA = 3.2 KOhm

So I took the ordinary 3.3KOhm .25Watt resistors and attached them with the LEDs in my circuit. BUT when I turned on the circuit , the resistors started to heat up !!!. Ok so now I Calculated the power by :

P = VI

P = 32V * 10mA = 0.32W

the problem was found, I needed a higher wattage resistors of 3.2KOhms and 0.5 Watts.All seemed to be fine uptil this point. But the problem is that in another device(an Extension Socket), I found an Led attached across 220V AC with a 100K series resistance the resistor was 0.25 W !!!! . When we perform the same calculations as above in this scenario:

I = 220V/100KOhm = 2.2mA

P = VI = 220V*2.2mA = 0.484 W

But the resistance attached was 0.25W ! Can anyone tell me why was a 0.5W resistor not attached

in this case? What if I want to pass 10mA at 220V?

I might be missing something very basic over here, please point out the problem.

Also I found a similar topic at:

in it a circuit diagram is given for attaching LED with mains what will happen if I dont attach the diode bridge ? The frequency of the mains voltage is 50Hz so in my view it shouldnt have any visible effect right? Also the voltage of the capacitor in the circuit is required to be about 500V isnt that too high? wont such a capacitor be a bit expensive(I might be wrong) ?

#### Sal

##### Full Member level 4
Hi

I see your point, but remember you are talking about different kinds of voltage, your case (32V) which I assume is DC and the other case (220V) which must be AC. In the AC circuit the LED diode is acting like a half wave rectifier, the power dissipated by the resistor is half of your calculations because it is been affected by a half wave rectified voltage.

cheers

Sal

#### jayc

##### Member level 3
It is true that the LED will act as half wave rectifier, halving your signal power, but even more importantly, the power of an AC signal is not calculated the same way as DC!

In fact, the power of an AC signal is
$P = \frac{1}{2}IV = \frac{1}{2}\frac{V^2}{R}$

This can be seen through some simple calculations. Let's take for example and AC voltage source connected directly to a resistor. Say the source has a voltage of
$v(t) = v_o \sin\omega t$ and the resistor has a value R.

Let's calculate the power dissipated through the resistor.

The voltage across the resitor is always the supply voltage, which is
$v_r(t) = v_o \sin\omega t$.

The current through the resistor is then
$i(t) = \frac{v_r(t)}{R} = \frac{v_o \sin\omega t}{R}$.

The power can be calculated as the following:
$P_{avg} = \lim\limits_{T\rightarrow\infty} \frac{1}{T}\int\limits_{-T/2}^{T/2} v_r(t) i(t) dt = \lim\limits_{T\rightarrow\infty} \frac{1}{T}\int\limits_{-T/2}^{T/2} \frac{v_o^2 \sin^2\omega t}{R}$

This signal is periodic, so the integral is equivalent to
$P_{avg} = \frac{1}{T}\int\limits_{0}^{T}\frac{v_o^2 \sin^2\omega t}{R}$, where T is the period of the signal. In this case, $T = \frac{2\pi}{\omega}$.

$P_{avg} = \left(\frac{v_o^2}{R}\right)\frac{\omega}{2\pi}\int\limits_{0}^{\frac{2\pi}{\omega}} \sin^2\omega t dt$

I'll leave you to evaluate the integral for yourself, but you will find that the result is
$P_{avg}=\frac{v_o^2}{2R}$

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#### Sal

##### Full Member level 4
Hi again

I thought for a while and finally found what is "strange" with the definition given by jayc (sorry bro, it is a friendly disagrement). In AC and DC the definition of power is the same P = V*I, the point is that in AC we are talking about the rms value of the voltage, which is the case of V = 220 volts. Looking at the last equation on the above explanation, it refers to Vo which is the peak voltage, if you replace Vo = sqrt(2)*V, you will find that P= V*I,

cheers

Sal

#### IanP

in it a circuit diagram is given for attaching LED with mains what will happen if I dont attach the diode bridge ? The frequency of the mains voltage is 50Hz so in my view it shouldnt have any visible effect right? Also the voltage of the capacitor in the circuit is required to be about 500V isnt that too high? wont such a capacitor be a bit expensive(I might be wrong) ?

One of the bridge's function is to protect LEDs against reverse voltage (Vr)..
Most of LEDs are rated for Vr≈5Vdc and exposing them to higher reverse voltages will just kill them ..

Capacitors for single-phase supply should be rated for 250V AC, which, in some data sheets, is an equivalent to 630Vdc ..
(some other ratings are: 63Vdc-->40Vac, 250Vdc-->160Vac, 400Vdc-->200Vac ..)
Are they expensive? I don't know, 0.1µF/250Vac may cost you <1\$ (single quantity) ..

#### MikeOhio

##### Member level 3
Ibrahim03;
there is just one small error in your initial calculation in that you didn't include the forward voltage drop required by the LED. To calculate the value of the dropping resistor the correct formula would be:
R=(Vs-Vf)/If, where Vs=source voltage, Vf=forward voltage drop of the LED, and If the forward current. The forward voltage drop and max current for the particular LED is obtained from the manufacturer's datasheet. Different LED types (standard, high current, low current, standard blue, etc) all have different parameters.

In your case, let's assume a 2 volt forward voltage for your LED and 10mA of current. So R=(32-2)/.01 = 30/.01 = 3000 ohms. Likewise, the power rating of the dropping resistor would be P=I*E or P=.01 * 30 = .3W

I realize there isn't much difference in the end results between your calculations and mine, but you may as well use the correct math in the future.

Mike

#### jayc

##### Member level 3
Sal said:
Hi again

I thought for a while and finally found what is "strange" with the definition given by jayc (sorry bro, it is a friendly disagrement). In AC and DC the definition of power is the same P = V*I, the point is that in AC we are talking about the rms value of the voltage, which is the case of V = 220 volts. Looking at the last equation on the above explanation, it refers to Vo which is the peak voltage, if you replace Vo = sqrt(2)*V, you will find that P= V*I,

cheers

Sal

I suppose we are both right. In my calculation, I defined the average power in terms of the sin wave amplitude. It was not specified that the 220V AC was the RMS value. In that case, you are right that $P_{avg} = IV_{rms}$.

#### jeffttan

##### Member level 2
i think you should include the voltage drop across the diode in your calculation of the current using ohms law...

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