diodes
I have to solve the task shown on the image below. (The diodes should be ideal with a forward voltage of 0.7V)
I tried solving it but always got the wrong current for D1 and V. First I drew it new like the schematics below.
simulate this circuit – Schematic created using CircuitLab
How do I need to approach such a task? I don't fully understand the -5V source.
From the sounds of it, the diode model you are using is the simple "ideal diode" with a fixed forward voltage. This model is an open circuit when \$V_{\textrm{Anode}} - V_{\textrm{Cathode}} < V_D\$ (reverse biased), and a fixed \$V_D\$ voltage supply otherwise (forward biased).
Start by making assumptions about the state of D1 and D2 (for example, D1 is forward biased, and D2 is reverse biased).
Your circuit would then look like this:
simulate this circuit – Schematic created using CircuitLab
What is \$V_o\$ here? The last step is to check your assumptions on each diode. If the assumptions are correct, the model is applicable. If not, permute your assumptions (ex.: what if D1 is reverse biased, and D2 is forward biased? or D1 and D2 are reverse biased, etc.) Only one of the 4 possible permutations will have a consistent answer.
As a side note, the answer you get will not match what the circuit simulator gives you (though it will be close). This is because the circuit simulator uses a more advanced diode model.
Since it is clear that the diode is forward biased, it acts as a short circuit but with a constant voltage drop. So you can replace the diode with a 0.7V and the circuit can be redrawn as shown.
simulate this circuit – Schematic created using CircuitLab
Now it is just a matter of writing and solving two linear equations.
Best Answer
I'd start with the assumption that both diodes are forward biased and see what happens.
D1 connected to ground means the anode is 0.7V. D2 will have a forward drop of 0.7 meaning that it's actually creating a virtual ground at V(out).
Then it's just voltages over resistors. (5-0.7)/10k = current through 10k. (0-(-5))/5k = current through 5k.
Then use KCL to determine what I is. Current through 10k flows into node, current through 5K flows out of node, and current I flows out of node. So you should have an equation like this:
$$I_{10k}-I_{5k}-I=0$$
When you do this, you'll see that I is actually negative which can't be because the diode would then be reversed biased and would block all current. That means D1 becomes an open in this circuit.
Now you just have one series line of voltage across resistors and a diode. Ohms law states this:
$$\frac{5-(-5)-0.7}{10k+5k}=I_{series}$$
With Iseries and 5k, you can find Vout. You know that I is 0 because D1 is reverse biased.