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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?

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  • \$\begingroup\$ ID2 is 6 volts across 43 kohms = 139 uA \$\endgroup\$
    – Andy aka
    Oct 18 '14 at 10:53
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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 \$

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  • \$\begingroup\$ @Asger Bjørn Jørgensen Thank you :) but I have another similar question where the circuit looks like this link. When I assume that D1 is off and D2 is on, I can show that D1 should be on so the assumption is wrong. The problem is when I assume that both diodes are on. When I calculate ID1 as you have done, the result is negative!. So all assumptions (D1 and D2 on, D1 on and D2 off, D1 off and D2 on, D1 and D2 on) are wrong in this circuit?? What is my wrong? ... thanks \$\endgroup\$
    – ammar
    Oct 18 '14 at 11:34
  • \$\begingroup\$ @David Fixed text and MathJax equations. Should be searchable now. \$\endgroup\$
    – asgerbj
    Oct 18 '14 at 11:44
  • \$\begingroup\$ .I understand you and thanks for information but this is my work -following your instructions- which suggest that this assumption is wrong. I don't know. Click here \$\endgroup\$
    – ammar
    Oct 18 '14 at 12:06
  • \$\begingroup\$ @ammarx I am sorry for misguiding you, i was wrong. I found the correct to be both diodes on. My calculations are here: imgur.com/7pfBDfW \$\endgroup\$
    – asgerbj
    Oct 18 '14 at 12:18
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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).

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  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

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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.

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