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This question originates from a diode question in which we modelled diodes using the simple piece-wise linear model. According to the lecture notes, the diode acts as a battery after attaining Vf(forward bias voltage). This creates a seemingly paradoxical relation.

enter image description here

becomes

enter image description here

In the battery-equivalent circuit above, quick circuit analysis shows that if unknown= unknown1 = 6V, there should be 4A flowing upwards from the 5V source. However this is impossible in the diode circuit since no current can flow through the diode from - to +(In the model that we are using at least).

So this creates a very confusing situation, is there actually current flowing through the diode or not? If so, how can this be explained?

EDIT: So I understand that I made a wrong assumption which led to the inconsistency. However, a new question arises, if I had unknown = 10V, that reaches the minimum requirement to activate the diode, but KCL shows that no current flows through the diode. I'd like to know why this is? More specifically in terms of intuition, what makes the resistor path more prefered for the current to travel through?

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  • \$\begingroup\$ what are the values of resistors? \$\endgroup\$ – Raj Feb 8 '17 at 9:16
  • \$\begingroup\$ sorry, I did it in a rush, the resistor are all 1ohm for simplicity \$\endgroup\$ – Frank Feb 8 '17 at 9:17
  • \$\begingroup\$ A diode conduct when it is forward biased from your circuit 1 the diode conducts only when the anode voltage is more than the voltage applied to cathode + forward voltage i.e (4.4V + 0.6 = 5V) \$\endgroup\$ – Raj Feb 8 '17 at 9:20
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"According to the lecture notes, the diode acts as a battery after attaining Vf(forward bias voltage)." The model should have gone on to say if it does not get to forward bias it acts as an open circuit (reverse biased) so your 4.4V source is not connected. If the diode is reverse biased.

You can determine if it is in reverse bias by looking at the current through the diode if current could flow from cathode to anode it is reverse biased.

In your case this occurs if the voltage created by the R1,R2 divider is less than 5V in this case you have an open circuit instead of a diode. If the undefined voltage source is high enough to drive the diode into forward bias you have the case in your second diagram.

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Look at Kirchoff's Amps law in parallel circuits. It looks like you have a parallel circuit but you don't because 100% of the circuit current can flow through the R2.

Diodes also need a flow of current to stay on. Zero current means they are off.

What you really have here is a series circuit with 2 ohms of total resistance. That gives you 5 amps of total current.

Answer by a First year, non-engineer, electronics technology Dunwoody student.

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The constant voltage drop (0.7V in many textbooks) across the diode are just estimation to make life simpler for engineer. The estimation is fine if you series a resistor with the diode. In your case, you may need consider to series a internal resistant within the power source.

Look at the diode equation. On many diode when voltage across is around 0.7V, current become very big (almost like open circuit). One will need to apply Kirchoff voltage law using the diode equation and there will not be any inconsistency.

enter image description here

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  • \$\begingroup\$ I did a calculation with unknown = 10V, this gives the minimum requirement to activate the diode, but no current flows through it at this condition. Does this mean that the simple piecewise linear model is a bad model that doesn't model what really happens? \$\endgroup\$ – Frank Feb 8 '17 at 10:24
  • \$\begingroup\$ Your best bet is to pull a data sheet for a real diode. In this you will find an I-V curve. The piecewise linear is good enough for most practical situations. You have to remember that in practical circuit that there are tolerances on all of the circuit elements. Usually these errors swamp the errors in your diode model. Part of the art of engineering is recognizing when you need to move from simple models to the more accurate ones. \$\endgroup\$ – RoyC Feb 8 '17 at 12:03

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