# What is the value of the node voltage, V1?

The value of the Vz (voltage across the Zener-diode) is 4V. There is also a current flowing against the diode. We know that when a diode is in the zener region, Vd = -Vz. I tried a simple KVL and got the equation 5V + V1 -(-4V) = 0. Is this correct?

simulate this circuit – Schematic created using CircuitLab

• As shown the value of the node voltage V1 is zero. One end of the 5V source is floating and no current is flowing. Oct 6, 2019 at 3:36
• @JohnD there is a current flowing against the diode. Oct 6, 2019 at 3:38
• Not in the circuit you showed. Oct 6, 2019 at 3:41
• @Hector What John is saying is that both nodes of a voltage source needed connection. It is a technical issue with the way you made the schematic that more likely suggests you are unfamiliar with drawing them well. What you probably mean is that there is a +5 V source, referenced to ground, at the top end of R1. John probably knows that but is making a point to you about learning to draw.
– jonk
Oct 6, 2019 at 5:22
• @jonk is exactly right, the way you have the circuit drawn the voltage source can't provide current. I THINK you mean that the top of R1 is at 5V, but your drawing doesn't show that. If you were to ground the top end of your voltage source it would be at -5V. So consider redrawing your circuit with the plus side of the 5V source connected to the top of R1 and the minus side connected to ground. Or just put a +5V label on the top of R1, which by convention is referenced to ground. (Thanks, Jonk, I didn't have time earlier for a more complete explanation.) Oct 6, 2019 at 5:44

It's been a while and no one has decided to close your question, so I'l write something short.

$$0\:\text{V} + 5\:\text{V}-I\cdot R_1-V_{Z_1}=0\:\text{V}$$
The top node of the zener diode, the node at $$\V_1\$$, will be more positive than the opposite zener node (bottom, at ground) and so you substract $$\V_{Z_1}=4\:\text{V}\$$ (or else add the negative voltage you worked out.)
You could merely observe that $$\V_1=4\:\text{V}\$$ and easily work out the current in $$\R_1\$$ as well as its voltage drop. Either way works okay.