Vo in this simple circuit with forward biased diode

Question

So what's $$ V_{o}?$$ if V_s is 4V

$$R_1 = 3k$$

$$R_o = 1k$$

It seems like such a simple question, do I assume diode is short and then combine the resistors in series?

Do I find current across R1 and then multiply it by Ro?

Or is it just KVL??

Like 0 = V0 - Vs + (voltage across R_1) + 0.7V (diode) ????

Or do I just intuitively say, oh I start with 4V then I go clockwise, experience first drop of 4V-0.7V across R1, then again 0.7V across diode, to finally 2.6V across R_o ???

Also, I know that diode is on if cathode is not at a higher potential, since the cathode side is 0V, diode will conduct (be on).

Answer 1

First, if your diode is on (you haven't analyzed the circuit yet, so you don't know, but you can assume and then check), you have a total voltage in your circuit of 3.3V (=4V-0.7V). Then, you can treat the circuit like this:

^{simulate this circuit – Schematic created using CircuitLab}

Now, you can find the current in your circuit. Your total voltage is 3.3V (remember, the diode 'voltage source' is in the opposite direction as the main supply, so they subtract), and your total resistance is 4kΩ. This gives you 3.3V/4kΩ = 0.825mA. Since this is positive, you can say your assumption before was correct and the diode is on. If you want to be more precise, though, this amount of current is low enough that the diode voltage will probably vary a fair bit from 700mV--however, unless you're trying to make things hard for yourself, 700mV is probably a good enough assumption.

Now you have the current, so Vo should be easy to find with Ohm's law: V = IR. Vo, then, is 825μA·1kΩ, or 825mV.