2011-07-01 Biasing a red light emitting diode.

When a red light-emitting diode (LED) is forward biased, it gives out red light. In this state, the voltage across its terminals will always be 1.4 V. Bob wants to drive this red LED with a 5 V power supply so that the driving current is approximately 13 mA. What is the value of the current-limiting resistor he would need? If a resistor of such value is not available at electronics stores. What is the closest commercial value he can use instead? Also find the power rating of the resistor and LED he would need to specify when buying them at the stores.

Solution

Since the potential +5 V is higher than 0V, the red LED tends to be forward biased.

KVL;

0 V+ 1.4 V + (13 mA)(R) = +5 V

R = (5 V - 1.4 V) / 13 mA = 277 Ω

The resistor of this value is not available at stores. Bob would have to use the closest one which is 270 Ω.

KVL;

0 V+ 1.4 V + (I)(270 Ω) = +5 V

I = (5 V - 1.4 V) / 270 Ω = 13.33 mA

The actual current flowing through the 270 Ω resistor and the red LED is 13.33 mA.

P_270Ω = (270 Ω)(13.33 mA)² = 48 mW ==> A 1/4 W resistor would be enough.

P_{red LED} = (1.4 V)(13.33 mA) = 19 mW ==> A 1/4 W LED would be enough.

Bob needs a 1/4 W 270 Ω resistor and a 1/4 W red LED. Ans

2011-07-01 Finding the potential and current in a circuit with one Si diode

Use KVL and KCL to find the potential V_C and the current flowing through Si diode.

Solution

Since the potential +5 V is higher than 0V, the Si diode tends to be forward biased.

KVL;

0 V+ 0.7 V = V_C

V_C = 0.7 V Ans

KVL;

0 V + (4 kΩ)(I_4kΩ) = V_C = 0.7 V

I_4kΩ = 0.7 V / 4 kΩ = 0.18 mA

KVL;

V_C + (2 kΩ)(I_2kΩ) = 5 V

I_2kΩ = (5 V - 0.7 V) / 2 kΩ = 2.15 mA

KCL;

I_Si + I_4kΩ = I_2kΩ

I_Si = 2.15 mA - 0.18 mA = 1.97 mA Ans

2011-07-01 Finding the potential and current in a diode circuit.

Find V_X and the current flowing in the Si diode.

Solution

Since the potential +20 V is higher than 0V, the Ge and Si diodes tend to be forward biased.

KVL;

+20 V - 0.3 V - 0.7 V = V_x

V_x = 19 V Ans

KVL;

0 V + (1 kΩ)(I_1kΩ) = V_x = 19 V

I_1kΩ = 19 mA

KVL;

V_x + 0.7 V = V_x + (2 kΩ)(I_2kΩ)

I_2kΩ = 0.7 V / 2 kΩ = 0.35 mA

KCL;

I_Si + I_2kΩ = I_1kΩ

(6 kΩ)(I_4kΩ) = 9.7 V

I_Si = 19 mA - 0.35 mA = 18.65 mA Ans

2011-06-20 Finding the potential and current in a two-diode circuit.

Find V_H and the current flowing in the 4 kΩ resistor.

Solution

Since the potential +10 V is higher than +5 V and 0V, the Ge diode and Si diode tend to be forward and reverse biased, respectively. Assuming that the potential V_H is higher than +5 V but lower than +10 V so that the Ge diode is forward biased and Si diode is reverse biased. Therefore, the same current flows through Ge diode and 4 kΩ resistor.

KCL;  

I_Ge + I_Si = I_4kΩ ;  I_Si = 0 A

I_Ge = I_4kΩ

KVL;

0 V + (4 kΩ)(I_4kΩ) + 0.3 V + (2 kΩ)(I_4kΩ) = +10 V 

(6 kΩ)(I_4kΩ) = 9.7 V 

I_4kΩ = 1.62 mA          Ans

KVL;

0 V + (4 kΩ)(I_4kΩ) = V_H 

V_H = (4 kΩ)(1.62 mA) =  6.48 V      Ans

Check assumptions;

+5 V < V_H < +10 V ; OK.

V_H > +5 V ; Si diode is reverse biased; OK.

The assumptions are true.

2011-06-18 Generating a constant current with diodes.

Find the potential V_C and current I_X flowing in 2.7 kΩ resistor when the potential V_X is -5 V and -10 V, respectively. Observe if there is a change in I_X.

Solution

We develop equations that relate V_X with V_C and I_X. 

KVL;  

V_X + 0.3 V + 0.3 V = V_C

V_C = V_X + 0.6 V  

KVL;

V_X + 0.3 V + 0.3 V + 0.7 V + 0.7 V = V_X + 0.3 V + (2.7 kΩ)(I_X)

I_X = (1.7 V) / (2.7 kΩ) = 0.63 mA

For V_X = -5 V;

V_X + 0.6 V = V_C = -4.4 V       Ans

I_X = 0.63 mA                         Ans

For V_X = -10 V;

V_X + 0.6 V = V_C = -9.4 V       Ans

I_X = 0.63 mA                         Ans

We can see that the potential V_C depends on V_X but the current I_X is constant.

2011-06-17 Finding the potential at point B in a three-diode circuit using KVL and KCL.

Find V_B and the current flowing in the Ge diode.

Solution

Assuming that the potential V_B is higher than 0 V but lower than +5 V so that the current I_Si flows from +5 V to point B and I_Ge flows from point B to 0 V.
From this assumption, V_B is higher than -5 V. Therefore, the ideal diode is reverse biased and the current -I_ideal = I_ideal = 0 A.

KCL;  

I_Ge = I_ideal + I_Si = I_Si

KVL;

0 V + 0.3 V + (1 kΩ)(I_Ge) + 0.7 V + (1 kΩ)(I_Si)  = +5 V 

(2 kΩ)(I_Ge) = 4 V 

I_Ge = 2 mA          Ans

KVL;

0 V + 0.3 V + (1 kΩ)(2 mA) =  V_B 

V_B = 2.3 V       Ans

Check assumptions;

0 V < V_B < +5 V ; OK.

V_B > -5 V ; ideal diode is reverse biased; OK.

The assumptions are true.

2011-06-17 Finding the potential at point A in a resistive circuit using KVL.

Find V_A and the current flowing in the 2 kΩ resistor.

Solution

Collapse the resistance network between point A and the potential -10V to get

2 kΩ || (1 kΩ + 1 kΩ)

= 2 kΩ || 2 kΩ

= [ (2*2)/(2+2) ] kΩ

= 1 kΩ

as shown in the picture below.

Let the current I flows from +10 V to -10 V.

KVL;  

-10 V + (1 kΩ)( I ) + (1 kΩ)( I ) = +10 V

( I )(2 kΩ) = 20 V

I = 10 mA

Then we find V_A from one of these approaches.

KVL;

-10 V + (1 kΩ)( 10 mA ) = V_A 

V_A = 0 V   Ans

or

KVL;

+10 V - (1 kΩ)( 10 mA ) =  V_A 

V_A = 0 V   Ans

Finally, we find I flowing through the 2 kΩ resistor in the first picture.

Ohm's law;

I_2 kΩ = [ V_A - (-10 V) ]  / 2 kΩ = [ 0 V + 10 V ] / 2 kΩ = 5 mA  Ans