Operating points of diodes 25 February 2014

Find the operting (Q) points of the upper diode (D1) and lower diode (D2) showing the current passing through and voltage across each diode.

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:

Average negative DC output of the full-wave rectified circuit using silicon diode.

With input voltage of 10 Vrms, find the average DC output V_DC of the circuit below.

Average positive DC output of the full-wave rectified circuit using silicon diode.

With input voltage of 10 Vrms, find the average DC output V_DC of the circuit below.

Average negative DC output of the half-wave rectified circuit using silicon diode.

With input voltage of 10 Vrms, find the average DC output V_DC of the circuit below.

Average positive DC output of the half-wave rectified circuit using silicon diode.

With input voltage of 10 Vrms, find the average DC output V_DC of the circuit below.

Node analysis of a DC circuit with silicon diode and dual supply voltages (2)

Use node analysis to find the voltage at point A in the circuit.

Node analysis of a DC circuit with silicon diode and dual supply voltages

Use node analysis to find the voltage at point B in the circuit.

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.