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How can a circuit where a diode is put in parallel to a resistor be described mathematically? Yes a solution would be to linearize the diode function in a specific point but in my case I need a general model of this situation. It would be interesting to take the whole diode curve into account.

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    \$\begingroup\$ Can you describe the diode on its own mathematically in the way you want? \$\endgroup\$
    – Andy aka
    Jan 10 at 15:35
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    \$\begingroup\$ Yes. It can be done. Read up on the Shockley diode equation and the product-log or branch 0 of the W-Lambert function. \$\endgroup\$ Jan 10 at 15:38
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    \$\begingroup\$ Do you need to include reverse leakage? Reverse breakdown? If you have a resistor in parallel, you can probably assume that the leakage is negligible. \$\endgroup\$
    – Mattman944
    Jan 10 at 15:39

5 Answers 5

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$$I = \frac{V}{R} + I_s\left(\exp(\frac{V}{nV_T})-1\right)$$

where \$V\$ is the voltage across the combination, \$R\$ is the resistance of the resistor, and \$I_s\$, \$n\$, and \$V_T\$ are parameters describing the diode.

This is just combining Ohm's Law (\$I=V/R\$) for the resistor and the Schockley equation for the diode (\$I=I_s\left(\exp(\frac{V}{nV_T})-1\right)\$) by summing the currents, as is appropriate for a parallel connection.

If you want to also include reverse breakdown behavior of the diode, then you will need to add additional terms to the diode model.

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If you have an V/I mathematical model for your diode (ideal or otherwise), then as the voltage across both devices is the same, to get the total V/I model you simply need to add the linear V/I model for the resistor.

If you are actually asking about V/I models for diodes, then there are a wide range of paper describing how to do that modelling, including this one. Almost all describe the diode in a sectional manner, according to reverse bias / forward bias < nominal voltage drop / forward voltage > nominal voltage drop.

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  • \$\begingroup\$ Thanks, but why do the function of the resistor and the diode have to be added to retain the behavior of the whole circuit? \$\endgroup\$ Jan 10 at 15:40
  • \$\begingroup\$ If you are looking at the V/I curve, then you are controlling the voltage over the pair of devices in parallel, so both see the same voltage, The total current flow will be the sum of the current through the two devices. \$\endgroup\$
    – colintd
    Jan 10 at 16:39
  • \$\begingroup\$ In reality, when reverse biased the resistor will tend to dominate the response. When forward biased, below the forward voltage the diode will play an increasingly large role, and once the forward voltage is reached, will dominate. \$\endgroup\$
    – colintd
    Jan 10 at 16:43
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Resistor is put in parallel with a diode in a "special" case.

It is the ONLY ("serious") case of using a "special" diode (like "tunnel") which has a "negative" part in its behavior.

When a resistor is put in parallel with that kind of diode, we can use the "whole" thing as an "passive" amplifier.

Here is an "example" (simulated) of how it works ...

enter image description here

Curve in black is the "curve" of a component with "negative" resistance (simulated).
Straight lines are the current through a resistor (proportional to voltage).
The upper curves are the sum of the preceedings (3 cases).

The down curves are the numerical "differential" of the upper curves (3 cases), which are the equivalent "gain".

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    \$\begingroup\$ It is also used to provide asymmetric RC delays, with the diode providing fast discharge in one direction, and the resistor providing delay in the other. However, have an upvote! \$\endgroup\$
    – colintd
    Jan 10 at 17:04
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    \$\begingroup\$ Putting a diode in parallel with resistor is nothing special really, it's done all the time. For example many microcontrollers have reset pin which requires a pull-up, a capacitor, and a diode to discharge the cap at power-off, so the cap delays the power-on reset properly. The situation you describe is indeed very special. \$\endgroup\$
    – Justme
    Jan 10 at 17:39
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    \$\begingroup\$ Interesting... A kind of "dynamic current divider" is formed. I think the more common one is a "dynamic voltage divider" where the two resistors - positive and negative - are connected in series. \$\endgroup\$ Jan 10 at 18:10
  • \$\begingroup\$ @Justme I spoke about the applications in RF where the characteristic of the diode (Tunnel diode or similar with a "negative" resistance) is really "modified" with a resistor ... and the whole system is used to make RF "passive" amplifiers with a 20 -> 60 dB gain ... \$\endgroup\$
    – Antonio51
    Jan 10 at 19:26
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Kirchhoff's current law says the current is the sum of those of the diode and the resistor: $$I(V) = I_D(V) + I_R(V).$$ This is required by charge conservation. Note that the voltage \$V\$ is the same for both devices.

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IV curve

As other answers also say, the resultant IV curve of a diode and resistor in parallel is a "vertical" (current) sum of the individual diode and resistor IV curves. Let's see this experimentally with the help of the CircuitLab DC Sweep Simulation.

Diode

Since the diode is a voltage stabilizer, it would be best to explore it with a variable current source. But then CircuitLab will put the current on the abscissa and the voltage on the ordinate, which is contrary to what is generally accepted. Therefore, we will examine the diode by voltage but changing it carefully (within small limits) so as not to exceed the maximum allowable current.

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see from the IV curve, the diode behaves like a variable (dynamic) resistor. At low voltage, its resistance is high, but from a certain point on, it starts to decrease and the curve becomes steeper.

STEP 1

Resistor

A resistor does not have this problem, and we can measure its IV curve with both voltage (like Ohm two centuries ago) and current. We choose the first so that we can compare the IV curves.

schematic

simulate this circuit

STEP 2

Diode and resistor

When we connect the two elements in parallel, each of them consumes current according to its IV curve.

schematic

simulate this circuit

The total curve is flatter than that of the diode alone.

R = 10 Ω

STEP 3a

R = var

STEP 3b

Applications

A typical application of this trick (resistor in parallel to a diode) is to limit the voltage across diode elements below the threshold voltage.

CE transistor stage

For example, in order to reliably turn off a transistor, a resistor is connected in parallel to its base-emitter junction.

schematic

simulate this circuit

STEP 4

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