Power that is consumed by diode?

Question

I’m wondering if I’m making the right approach when I’m trying to calculate the power consumed by the diode in the circuit.

Here is the problem:

So my approach is:

Power loss by diode $$V_f \times I_f$$ where

\$V_f\$ = Diode forward voltage drop.

\$I_f\$ = The current flowing through the diode.

So the first thing I did was to calculate the current

$$I = V/R = 5/1000 = 5 mA$$

And then I use $$ V_f \times I_f= 2 \times 0.005 = 10 mW $$

Is this correct? I feel like there is more to the problem with the fact that it states ”red color light”. Should I check on a sheet or similar?

Answer 1

I think that you have neglected the voltage drop across the diode which limits the current.

My solution is : $$ \because V_{R_1}=V_0-V_f=5-2=3\,\text{V} $$ $$ \therefore I_f=I_{R_1}=V_{R_1}/R_1=3/1000=3\,\text{mA} $$ $$Power Loss =V_f\times I_f=2 \times0.003=6\,\text{mW} $$

Answer 2

You calculated the current wrong. Apply KVL. You will see that the voltage accross the resistor is actually 3 volts. Other than that, your approach seems correct.

Answer 3

You just forgot the voltage drop across the resistor so...

Power loss by diode $$V_f×I_f$$ where

\$V_f\$ = Diode forward voltage drop.

\$I_f\$ = The current flowing through the diode = current through the resistor.

So the first thing I did was to calculate the current through the resistor.

\$V_R=V_0-V_f = 5\,\text{V} - 2\,\text{V} = 3\,\text{V}\$.

\$I=V_R/R=3/1000=3\,\text{mA}\$

And then I use $$V_f×I_f=2×0.003=6\,\text{mW}$$

Answer 4

\$V_R=V_0-V_f = 5 - 2 = 3\,\text{V}\$.

\$I=V_R/R=3/1000=3\,\text{mA}\$.

\$V_f×I_f=2×0.003=6\,\text{mW}\$.

Answer 5

\$P_{diode} = 2\,\text{V} \times 0.003\,\text{A} = 6\,\text{mW}\$.

\$P_{resistor} = 3\,\text{V} \times 0.003\,\text{A} = 9\,\text{mW}\$.

\$P_{total} = 6\,\text{mW} + 9\,\text{mW} = 15\,\text{mW}\$ consumed for supplying the diode.

The efficiency of the system is $$ \frac{P_{diode}}{(P_{diode} + P_{resistor})} = \frac{6\,\text{mW}}{(6+9)\,\text{mW}} = 0.4 = 40\%. $$ Efficiency can be improved by using a supply with voltage closer to the diode voltage drop.
This will reduce the \$9\,\text{mW}\$ loss from heating the resistor.

Answer 6

\$I_F = (V_O - V_F) / R_1 = 3 / 1k = 3 mA\$.

\$P_F = V_F \times I_F = 2 × 3 = 6 mW\$.

Answer 7

\$V = 5 - 0.7 - 2= 2.3\,\text{V}\$.

\$I = ?\$.

\$R = 1 \times 10^3\Omega\$.

\$P = V^2 ÷ R\$.

Hence:

\$ P = 2.3^2 ÷ 1000 = 5.29 × 10^-3\,\text{W}\approx 5\,\text{mW}\$.

Answer 8

First of all, led has voltage drop about 1.3V for red, 1.7 for green, 3.3 for White led etc. So on your dwg it is red so the current I=5-1,3=3.7V/1000ohms=3.7mA and so P=Vd*I=1.3*0.0037=2.9mW

Usually i set signali g led current to 10 mA, so in this case R=3.7/.01=370ohm. In this case I use nearest lower standard resistance i. e. 330ohm, so I=11mA and power 15mW.