Use of KVL to solve a diode circuit 30 October 2013 problem 2

Use KVL principle to find the current I and voltage V_o in the diode circuit below.

Solution:

Use of KVL to solve a diode circuit 30 October 2013 problem 1

Use KVL principle to find the current I_D and voltage V_o in the diode circuit below.

Solution:

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.