Electronic – How to calculate Vx in this circuit

circuit analysisdiodes

In this circuit:

Assuming ideal diodes

We are required to calculate VD1, VD2, ID1, and ID2.

  1. If I assume that D1 is off and D2 is on which is a wrong assumption (as I think). How I prove that it is wrong? how to calculate Vx to show that D1 should be on here?

Edit: I calculated Vx as following:

I = (9 + 6) / (22k + 43k) = 2.31 * 10^-4

V(of 23k resistor) = 43k * 2.31 * 10^-4 = 9.93 v

=> Vx = 9.93 – 6 = 3.93 v

Is that correct?

  1. If I assume that D1 and D2 are both on, then the circuit will look like:

How can I calculate ID1 and ID2?
Can this resulting circuit redrawn to be easier to solve?

Best Answer

Question 1: Assuming that D1 is off and D2 is on, we have the following circuit: Circuit for question 1

We can first calculate the current throught diode 2

\$ i_{D2} = \frac{9V-(-6V)}{22k \Omega + 43k \Omega} = 0.2308 mA \$

and then the voltage across diode 1

\$ V_{D1} = 9V-0.2308mA \cdot 22 k \Omega = 3.92 V \$

As a diode may only block negative voltages, the assumption of diode 1 being off and D2 being on is invalid.

Question 2: Assuming that both diodes are turned on yields the following circuit: Schematic used for question 2

Calculating the current through diode 2

\$ i_{D2} = \frac{0V-(-6V)}{43 k \Omega} = 0.1395 mA \$

and then the current denoted as ix

\$ i_x = \frac{9V - 0V}{22 k \Omega} = 0.4091 mA \$

Using Kirchoffs Current Law (KCL) stating that all currents entering a node should be equal to the current leaving the node we get

\$ i_x = i_{D1} + i_{D2} \$

Which may be rearranged to

\$ i_{D1} = i_x - i_{D2} = 0.4091 mA - 0.1395 mA = 0.2696 mA \$