# Vo in this simple circuit with forward biased diode

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).

• Consider using a formula for the diode instead of a fixed 0.7 drop. For a resistor you use $I=V/R$, for a diode you'd use this: en.wikipedia.org/wiki/Shockley_diode_equation Also consider trying out the CircuitLab editor.
– uhoh
May 2, 2017 at 23:24
• Would there be any difference in $V_O$ if you switched the places of $R_1$ and the diode? No, there would not. So, suppose you did that swap and asked yourself the same question? Given the logic you provide in your 2nd to last paragraph, you might go: "Start with $4\:\textrm{V}$ and go clockwise, experience diode drop, so then there is $3.3\:\textrm{V}$ remaining." Would you be able to work out $V_O$ given the remaining two resistors, now?
– jonk
May 2, 2017 at 23:24
• yes KVL applies. and if any shunt parts use Norton equiv. Req and Veq. May 2, 2017 at 23:24
• @TonyStewart.EEsince'75 well, according to KVL, $0 = V_0 - 4V + (4-0.7) + 0.7V. Vo = 0V$. So is $V_o$ = 0 V?
– Jack
May 3, 2017 at 2:23
• Jack , it's a diode, so treat as a constant voltage sink (same as a source) except has ESR or Rs bulk resistance which can be neglected here May 4, 2017 at 18:57

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.

• Why doesn't it work intuitively?? Like you start out with 4V, then go clockwise experiencing 2 voltage drops?
– Jack
May 3, 2017 at 2:26
• You can't determine what the voltage drops would be without first finding the current. May 3, 2017 at 2:38
• you know what's bizarre? That current is the same across both resistors (3k and 1k), you'd think that if first resistor has low resistance, current will flow easily and more of it will flow through it, but then, as it encounters another resistor which would have a high resistance, current would struggle flowing through it, so at the end of the resistor current value would be less... yet according to electrical engineering, current is supposed to flow in same amount in both resistors... like wtf. *where resistors are in series
– Jack
May 5, 2017 at 20:19
• What do you have trouble understanding here? Think of it like water flowing through a pipe--if there's a blockage in the pipe, the water upstream from it can't flow any more than the water right next to the blockage. You might want to look up Kirchoff's Current Law. May 5, 2017 at 20:36
• also why did you say total resistance is 3k+1k if there's a voltage (from diode) source in between them?
– Jack
May 24, 2017 at 15:16