# How do I solve this network of diodes?

Here is the assignment:

The assignment is to calculate various values across the circuit.

Here are the important values:

$R=3.3k\Omega$

$U_1 = U_2 = 4 V$

The diodes are ideal with the following current voltage relation:

Here is my thought process what I tried and where I failed: I am assuming all diodes are on because there is nothing directly blocking them, like an air valve.

Lets say I want to calculate $I_5$. I know $I_5$ is half of $I_2$ and is equal to $I_6$ and $I_7$.

The resistance should be $2R$ and the Voltage should be because the diodes are in parallel $4V- 1.4V = 2.6 V$

But then it all stops making sense after I try to calculate it, something must be wrong with my approach.

Here are my questions:

How do I go about this circuit,how do I split up this circuit in to the correct Kirchhoff meshes ?

Can you give me a hint or a direction how I would go about solving this ?

Update: here is the marked solution:

• I guess you're forced to examine the 8 possible cases where you consider each diode is conducting or blocking, and the case(s) where there is no contradiction anywhere are the solutions to the exercise. Also, I5 is not necessarly half of I2. Dec 18, 2016 at 18:04
• You started off right. You made an assumption and the math failed. Good. See electronics.stackexchange.com/a/158040/9409 Dec 18, 2016 at 18:05
• @Bregalad what could i5 else be, I dont see any other possibility Dec 18, 2016 at 18:35
• i5 can literally be anything. Why do you think it is? And I do not think all diodes are on... Dec 18, 2016 at 18:59
• This is similar to yesterday's diode problem: electronics.stackexchange.com/questions/275417/… Dec 18, 2016 at 19:15

In principle all combinations have to be checked, however often a few simplification can be made by inspecting the circuit.

The diode with the voltage U6 across it will always be on. There is no possibility to bring the anode below 0.7V.

The cathode of the diode with U7 is off because there is a resistor between the cathode and ground. If it was on there would be a voltage drop across the resistor reducing the voltage across the diode. This voltage is 0.7V minus the voltage drop across the resistor. However, below 0.7V it would turn off.

The diode with U3 across it is on. It has 8V at the anode, the remaining 7.3V split equally across the resistors. We can verify that the diode with U7 is still off.

The currents are a result of the applied voltages, they are not needed for anything.

• thanks for your reply, the problem is now more clear to me, i appreciate it, I will accept your answer in a bit in case no one else does Dec 18, 2016 at 19:51

By simple inspection: U6 = 0.7 V
U8 > 0 V so will turn off the diode therefore I7 = 0

That leaves two strings to calculate:

I5 = I6 = (4 - 0.7) / 3300 = 1 mA

I3 = I4 = I8 = (8 - 0.7) / 6600 = 1.1 mA

U8 = 0.0011 * 3300 = 3.63 V