# Confusion regarding piece wise linear diode model

I understand that when you have a diode that is in forward bias, it can be modeled by a resistor and a voltage source, as seen in the top right of this diagram.

However, I'm confused about the negative bias region (modeled by the bottom left corner resistor/source combination). If the anode of a diode has positive voltage with respect to the cathode, why isn't the resistor/source combination modeled like this?

I feel like I'm missing something really fundamental here.

EDIT to include further questions requiring clarification:

(1) Using the forward bias diode model. If I use a positive voltage (1.6V) source greater than the "turn-on-voltage(0.6V)", I get a positive (clock-wise) current which is the same as the value given in the piece-wise linear diode model graph.

However, when I use a Voltage less than the 0.6V "turn-on-voltage", I get a negative current. What is this negative current telling me?

(2)

Also, assigning a voltage of -7.2V and using a clockwise current loop. Why do I get a value of -110mA when I should be getting -100mA according to the graph?

• In your 2nd added question, you have the wrong polarity on the external voltage source. And $(7.2 - 6)/12 = 0.1$, giving the expected answer. Oct 27 '17 at 1:35
• With your arrangement you should have gotten 1.1 A, not 0.11 A. Oct 27 '17 at 1:36
• So the Voltage should be saying +7.2V? Oct 27 '17 at 2:26
• Yes, It should be +7.2 Oct 27 '17 at 3:00
• But doesn't that then give me a positive clockwise current via KVL whereas I should be getting a negative current/anticlockwise current because the the diode is in reverse bias? Oct 27 '17 at 3:25

If the anode of a diode has positive voltage with respect to the cathode, why isn't the resistor/source combination modeled like this?

When the device is reverse biased, the anode does not have a more positive voltage than the cathode, it has a more negative voltage.

While the term anode technically means the electrode into which current flows, in practice it is used to mean the electrode into which current most typically flows. When a reverse bias is applied to the device, we don't rename the cathode to be the anode and the anode to be the cathode.

Even in zener diodes, which are most often used in "reverse" mode, we still call the p-doped side of the junction the anode and the n-doped side of the junction the cathode.

However, when I use a Voltage less than the 0.6V "turn-on-voltage", I get a negative current. What is this negative current telling me?

Notice that in the piecewise linear model, the model with a 0.6 V source and small resistor only applies when the applied voltage is greater than 0.6 V.

So when you get a negative current, it's telling you that you're using the wrong piece of the piecewise linear model and you should be using the piece where the current is 0 for applied voltages between the reverse breakdown voltage (-6 V in your example) and the forward turn-on voltage (+0.6 V in your example).

• Hi "The Photon", thank you for your answer. However, I'm still a little confused with something related to this, but I think I'll require a couple of images to show what I'm confused about. Do you know how I can can attach images to a comment, or should I edit my question so I can add them? Oct 26 '17 at 23:51
• You'll need to edit your question. If you are restricted from adding enough images, put links in your question and someone with higher rep can change them to inline images. Oct 26 '17 at 23:54
• @tapeside The terms anode and cathode come from chemistry and if you study their use there (batteries, electroplating, electrolysis, etc), then you find even more to consider (and perhaps be confused by.) Eventually, with enough knowledge acquired, it all makes sense again. But then you have a much more nuanced understanding that helps you navigate. I think the terms originally came out of electrolysis work done in and around the 1830's. Long before semiconductors and dopants.
– jonk
Oct 27 '17 at 0:19
• @ThePhoton I've now edited my question Oct 27 '17 at 0:28
• @ThePhoton, Thank you so much for your help. I just have one more question (I hope). Oct 27 '17 at 0:53