How to calculate Vx in this circuit?

Question

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?

2. 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?

Answer 1

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

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:

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 \$

Answer 2

The currents are just Ohm's law, keeping in mind that the ID1 = 9V/22K - ID2 = 9V/22K- 6V/43K (KCL). No need to redraw it, it's easy to solve by inspection.

As to the diode assumptions- if you calculate ID1 to be < 0 (when you assume it to be 'on') then it cannot be 'on'. Similarly, if the voltage you calculate to be across it is positive (when you assume it to be 'off') then it cannot be 'off'.

To calculate Vx, it's just a voltage divider (D1 off D2 on) so the voltage Vx will be:

Vx = (9/22K - 6/43K) * (22K || 43K) = 3.92V. Since it is > 0, you know D1 is on (and the assumption is therefore invalid, D2 is on so Vx = 0 for an ideal diode).

Answer 3

1. Prove by contradiction. You assumed that D1 is reverse biased but upon calculating you got Vx=3.92, which means the diode must be conducting. So your assumption was false and the diode is forward biased.

2. Now calculating current is pretty trivial. We know Vx=0 (reference).

So Id1 + Id2 = (9-0)/22k =.409mA

Id2 = (0-(-6))/43k = .139mA

So, Id1 = .409-.139 = .27mA

Answer 4

To make the numbers easier, let's assume that the 43K Ohm resistor is 44K Ohm.

Then, without D1, the voltage divides 1:2 across the two resistors, hence Vx would be 4V. So D1 is on. Both diodes disappear, so Vx is, in fact, zero.