Diode modelling question

[Karamel]

Hi everyone,

Nowadays, I am working on diodes and their mathematical assumptions. I have, almost, understood all assemptions but, I am really wondering one case which is whole the new ball game.

We know small voltage changing case. Diode resistance calculates with this formula: Rd = nVt/I

Okay. We change the case like that, if I use 3sin(2000piT) instead of DC or DC + ripple etc. from input signal, What will the Rd of diode be? Can we use this formula? of Can we use piecewise model?

Best Regards,
Karamel 

[Kleinstein]

The formula for Rd is for small changes in voltage changes superimposed on a DC current. There is no good equivalent resistance for a pure AC voltage (except for very small values in the 10 mV range).

[Karamel]

I thought like that, too. Okay, there is no any model for alternating voltage but, which model is more convenient for pure ac signal? Piecewise model could be?

[T3sl4co1l]

The derivative (called incremental resistance) is also the small-signal AC resistance.

There is no single model that works for large AC amplitudes as well.  In that case, the diode will be switching on and off, and you can only analyze it one step at a time.

Tim

[G0HZU]

In the past I've used a simple IV lookup table that models the diode as a non linear resistor. I created the model myself using a couple of DMMs and a PSU to plot an IV curve for the diode. This model works surprisingly well (for DC or AC or both) for a jellybean diode like the 1N4148 up to several MHz on a simulator. So the model is just a line of text containing my IV measurements as a crude lookup table or 'map' of the characteristic.

[Ratch]

Karamal,

You had better check your formula for Rd again. Using Schockley's diode equation, when I calculate the dynamic resistance by finding the derivative of the applied voltage Vd with respect to the saturation current Is, I see an exponential term in the denominator of the formula. 

Let me understand what you are trying to do.  You are applying a "small" voltage of 3 volts P-P across a diode whose DC voltage is probably less than 1 volt?  If that AC voltage is small, what do you consider large?

Keep in mind that the dynamic resistance of a diode depends on characteristics of the diode and the voltage applied across its terminals.  Therefore, if you change its operating point, you will change its dynamic resistance.

Ratch

[G0HZU]

Karamal,

You had better check your formula for Rd again. Using Schockley's diode equation, when I calculate the dynamic resistance by finding the derivative of the applied voltage Vd with respect to the saturation current Is, I see an exponential term in the denominator of the formula. 

Let me understand what you are trying to do.  You are applying a "small" voltage of 3 volts P-P across a diode whose DC voltage is probably less than 1 volt?  If that AC voltage is small, what do you consider large?

Keep in mind that the dynamic resistance of a diode depends on characteristics of the diode and the voltage applied across its terminals.  Therefore, if you change its operating point, you will change its dynamic resistance.

Ratch

I think you are misinterpreting the Shockley equation. The current through the diode (I) changes exponentially with applied V but for small signals where the DC bias current I is large wrt the ac signal, the equation Rd = nVt/I  is a classic equation for the Rd of a diode and I've used it countless times throughout my career. Stick a jellybean diode like the 1N4148 on a VNA and measure the Rd of the diode in a suitable test circuit using a small test signal level (eg -30dBm) from the VNA and you will see that n is about 2 for a diode like this. The equation will work into the VHF region with this diode for small signals.

If you need to work with large AC test signals that can cause significant waveform distortion then the lookup/piecewise/SPLINE model I described is surprisingly good although not as good as a detailed SPICE model. Things get harder to model accurately (with big AC test waveforms) at higher frequencies or if you use a PIN diode.

[T3sl4co1l]

Karamal,

You had better check your formula for Rd again. Using Schockley's diode equation, when I calculate the dynamic resistance by finding the derivative of the applied voltage Vd with respect to the saturation current Is, I see an exponential term in the denominator of the formula.

Substitute the exponential with the I_f you started with.

Tim

[Ratch]

Karamal,

You had better check your formula for Rd again. Using Schockley's diode equation, when I calculate the dynamic resistance by finding the derivative of the applied voltage Vd with respect to the saturation current Is, I see an exponential term in the denominator of the formula. 

Let me understand what you are trying to do.  You are applying a "small" voltage of 3 volts P-P across a diode whose DC voltage is probably less than 1 volt?  If that AC voltage is small, what do you consider large?

Keep in mind that the dynamic resistance of a diode depends on characteristics of the diode and the voltage applied across its terminals.  Therefore, if you change its operating point, you will change its dynamic resistance.

Ratch

I think you are misinterpreting the Shockley equation. The current through the diode (I) changes exponentially with applied V but for small signals where the DC bias current I is large wrt the ac signal, the equation Rd = nVt/I  is a classic equation for the Rd of a diode and I've used it countless times throughout my career. Stick a jellybean diode like the 1N4148 on a VNA and measure the Rd of the diode in a suitable test circuit using a small test signal level (eg -30dBm) from the VNA and you will see that n is about 2 for a diode like this. The equation will work into the VHF region with this diode for small signals.

If you need to work with large AC test signals that can cause significant waveform distortion then the lookup/piecewise/SPLINE model I described is surprisingly good although not as good as a detailed SPICE model. Things get harder to model accurately (with big AC test waveforms) at higher frequencies or if you use a PIN diode.

Yes, Rd = nVt/I  is a good approximation when the bias current is large in comparison to the AC signal.  There is no misinterpretation or miscalcuation of Schockley's equation about that.  But, what small diode will withstand to have 3 volts p-p across it?  That is hardly a small signal.

Ratch

[Ratch]

Karamal,

You had better check your formula for Rd again. Using Schockley's diode equation, when I calculate the dynamic resistance by finding the derivative of the applied voltage Vd with respect to the saturation current Is, I see an exponential term in the denominator of the formula.

Substitute the exponential with the I_f you started with.

Tim

Agreed, that will give you a good approximation of the dynamic resistance if the total current is much greater than the saturation current, which it usually is.

Ratch