I have to find the output voltage Uo related to the input voltage Ui knowing that the input values range from -10V to 30V. I started by assuming that D1 is reversed biased. But then how do i know if D2 is forward or reversed biased? I think i have to take that D2 is forward biased so i can find the Voltage at the node with resistors 10kΩ-10kΩ-10kΩ and then be able to define the range where D1 conducts. However, i am not sure. All diodes are ideal. Any help is appreciated! Thanks in advance!
Well, I’m assuming those are ideal diodes, right? Start by simplifying your thoughts. D2 in series with 10V is the same as a diode conducting only above 10V, let’s call it D2’. So, before D2’ reaches 10V this is a non-conductive path. This means you can ignore the third 10K resistor since it doesn’t have current. The first two resistors act as a voltage divider, so D2’ will conduct only when the input voltage is 20V, which gives 10V over D2’, are you following? After 20V, D2’ starts conducting and the third resistor cannot be ignored anymore. But this part is easy too. Since D2’ is now conducting, any further raise in the input voltage is now clamped on D2’ and it stays at 10V.
To broadly answer your question regarding the bias of D2. It will be forward biased if the calculated voltage at Vo exceed the rated forward voltage of D2 (let's say 0.7 V for the sake of argument) and the 10v power supply.
D1 will be forward biased if the input voltage is above the rated forward voltage of D1. For example: if Vi i 3.7 V, you will have 3 V available after D1. You know that 3V will not be enough to overcome the 10.7V of D2. Therefore, since we don't know what load is connected to Vo, we can assume that the ''in-line'' 10k resistor is floating and that you will have a voltage divider between the two other 10k. End result is that Vo will be 1.5 V.
When D2 will become forward biased, the best analytical approach will be to either use a Thevenin equivalent on D1/10V or use Millman theorem to solve the circuit. Before doing Thevenin, you can get rid of D1 in your circuit analysis since you know that it will be forward biased and that it will reduce the input voltage by 0.7 V. For the actual circuit output, it will be the combo of the forward voltage of the diode and the 10 V (10.7 V). Thevenin and Millman will only be required if you want to compute the node in between all the resistors.