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

2011-06-18 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.