Ideal Diode: Exploring Negative Voltage Drop and Its Feasibility

In summary, the ideal diode cannot have a positive voltage drop because the Shockley diode equation, which describes the behavior of a PN junction, becomes highly non-linear for even small voltage drops. This non-linearity makes it difficult to find a solution to a circuit using the ideal diode. To make the model more accurate, a voltage source is added when the diode is on, with different values depending on the semiconductor material. Another step is to add a resistor in series with this voltage source to account for the non-infinite slope of the curve.
  • #1
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I am a little unsure about why the ideal diode can't have positive voltage drop... Looking back at my notes, it just simply says that "It's not a feasible solution" and that only when voltage drop is zero or negative can there be a solution.
 
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  • #2
This is because you're using the completely ideal model for a diode.

Shockley's diode equation for the static behavior of a PN junction is given by

[tex]I=I_\mathrm{S} \left( e^{V_\mathrm{D}/(n V_\mathrm{T})}-1 \right),\,[/tex]

If you graph this for normal values of the parameters, you'll see that it's practically 0 for all values of the voltage drop less than 0, and it starts to have a very high slope for a very small voltage drop and the curve is practically vertical.

One of the main reasons this is done is because this equation is highly non-linear, and even a simple, one mesh circuit will yield an algebraic equation which cannot be solved in terms of elementary functions, so an approximate solution to a circuit is obtained by idealizing this non-linear equation.

Another step to make the model more accurate is to incorporate a voltage source when the diode is on. The value of this voltage source depends highly on what semiconductor the diode is made of. I will list some standard values:

Si: 0.7 V
GaAs: 1.2 V
Ge: 0.25 V

Yet another step taken to get closer to the Shockley equation is to add a resistor in series with this voltage source to account for the non-infinite slope of the curve.

Hope this helps. :)
 

1. What is an ideal diode voltage drop?

An ideal diode voltage drop refers to the amount of voltage that is lost when current flows through a diode in the forward direction. It is typically measured in volts and can vary depending on the specific diode used.

2. How does an ideal diode voltage drop affect circuit performance?

The ideal diode voltage drop can affect circuit performance by causing a decrease in the overall voltage and power available to the circuit. This can lead to reduced efficiency and performance, especially in low voltage circuits.

3. What factors influence the ideal diode voltage drop?

The ideal diode voltage drop can be influenced by a variety of factors including the material used to make the diode, the temperature of the diode, and the amount of current flowing through the diode. Additionally, the diode's design and construction can also play a role in the voltage drop.

4. How can the ideal diode voltage drop be minimized?

The ideal diode voltage drop can be minimized by using diodes made from materials with lower voltage drop, such as Schottky diodes. Lowering the temperature of the diode and reducing the amount of current flowing through it can also help to decrease the voltage drop.

5. Are there any disadvantages to having a low ideal diode voltage drop?

While a low ideal diode voltage drop can be beneficial in terms of circuit performance, it can also lead to higher power dissipation and increased heat generation in the diode. This can potentially damage the diode and other components in the circuit, so it is important to consider the trade-off between voltage drop and power dissipation.

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