Imposing voltage

I have two questions from the circuit below, with an ideal diode and $$\v_1(t) = 20\sin(\omega t)\$$:

1. If $$\V_2 = +5\ V\$$ I guess the diode would be ON (because $$\V_2 = u_D\geq 0\$$). In that case, would $$\V_3\$$ be always $$\5\text{ V}\$$? I mean, would $$\V_2\$$ impose the +5 V voltage over node A and would $$\V_1\$$ always have to drop $$\V_1 - IR\$$ such that the voltage at A is the +5 V imposed by $$\V_2\$$?
2. If $$\V_2=-10\ V\$$, because we have $$\V_2 = u_D \leq 0\$$ would the diode be in OFF mode? If that's the case, why don't we have $$\V_3(t) = V_1(t)\$$?

I know that's not the case for question 2 because the multiple choice option

c)$$\V_2=-10\ V\ \Rightarrow \text{diode is OFF and } v_3(t)=v_1(t)\$$

is not the correct one (besides being the one I thought was correct, and don't understand why it isn't).

simulate this circuit – Schematic created using CircuitLab

First, let's define what we mean by an ideal diode:

• off current: 0A
• on incremental resistance: 0ohm
• on forward voltage: 0V

if V2=+5V I guess the Diode would be ON (because V2=uD≥0).

$$\u_D\$$ is the voltage across the diode. It's not absolute. Thus, whether the diode would be on or off depends on the voltage put out by both $$\V_1\$$ and $$\V_2\$$, since $$u_{DC}(t)=V_2(t)-V_1(t).$$

Now, this is a simplification, since it assumes no current flow across R. If current flows into/out of $$\V_3\$$, then we would have $$u_{DC}(t)=V_2(t)-\left(V_1(t)-I_3\cdot R\right),$$ with $$\I_3\$$ being positive flowing out of $$\V_3\$$.

The diode control voltage is $$\u_{DC}\$$. That's the voltage that would be across the diode if the diode was an open-circuit. But we know that sometimes the diode conducts.

The diode voltage, then, is $$u_D(t) = \begin{cases} u_{DC}(t)\le 0: u_{DC}(t),\\ u_{DC}(t)>0: 0{\,\rm V}\\ \end{cases}$$

Obviously, when the diode is on, the voltage across it drops to zero, since it's ideal.

In that case, would V3 be always 5V?

If the diode was ON, and that depends on $$\u_{DC}\$$, and that depends on both $$\V_1\$$ and $$\V_2\$$, then yes. When the diode is on, it acts like a piece of wire (it is very ideal!).

if V2=−10V, because we have V2=uD≤0 would the Diode be in OFF mode

Again, that's only assuming $$\u_{DC}\le0\$$, so it depends on $$\V_1\$$ and $$\V_2\$$.

If that's the case, why don't we have V3(t)=V1(t)?

Who said that? If that's the case, then we do have $$\V_3=V_1\$$ indeed - but only if the output current from the $$\V_3\$$ terminals is 0A. Otherwise, there would be drop across R.

When the diode is OFF, it acts as if it wasn't there! Open circuit.

So, your analysis is mostly correct, you just wrongly assumed that $$\u_D(t)=V_2(t)\$$. That's not the case, as shown above.

1. For an ideal 0V diode, it conducts and "clips"any input voltage below V2 and passes any input above V2. But V2 is +5V not +10V.

2. The second case with V2 at -10V clips any input below -10V and otherwise passes thru input to output.

This blocking action thus sets the lower limit for the output voltage set by V2 regardless of its value. (Within reason)