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The 1N4001 diode's junction capacitance versus reverse voltage plot is given as follows:

As an example https://www.vishay.com/docs/81857/1n4148.pdf

I guess to measure this capacitance they applied 1MHz 50mV pk-pk sine wave signal as reverse biased and varied the DC reverse bias from 100mV to 100V(?).

I tried the following:

enter image description here

But I obtain the following:

enter image description here

For the capacitance I used 1/(6.28*1000000*(V(n002)/I(R1))) which comes from

Xd = |Vd|/|Id| diode capacitive reactance

C = 1/(2×pi×f×(|Vd|/|Id|))

So C = 1/(2× pi × f × Xd)

Obviously something is terribly wrong.

How can we obtain a similar plot by using LTspice? Is it even possible?

user16307
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3 Answers3

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Why don't you try an X-Y plot, like you might on an oscilloscope. (I'll assume the \$50\:\Omega\$ signal generator output, too.) So I used the exact same schematic you used in posting your question.

One caution your plot suggested to me was that the diode has a breakdown specification. Probably the BV specification. So I decided to stay well away from that. The model I used is:

.MODEL DI1N4001G D( IS=65.4p RS=42.2m BV=50.0 IBV=5.00u CJO=14.8p  M=0.333 N=1.36 TT=2.88u )

Which I picked up from Diodes, Inc..

Another issue I saw is that the datasheet curve you cited uses \$25\:\text{mV}\$ peak. You used \$50\:\text{mV}\$ peak in your schematic. So I changed that, as well, to match the datasheet's choice.

Beyond that, all I did was take the 1N4001 and plot the reverse current vs the voltage across the small sine generator using .TRAN, instead. Here's the plot I got:

enter image description here

Clearly, when the applied sine wave is at its maximum magnitude, the current is zero. Also, when the voltage goes through zero volts (it's maximum slope), then the current peaks. Easily seen in the above graph.

From the larger loop, dark blue, to the inner-most loop, the voltages I used were \$0\:\text{V}\$, \$1.5\:\text{V}\$, \$3\:\text{V}\$, \$6\:\text{V}\$, \$12\:\text{V}\$, and \$24\:\text{V}\$, respectively. The peak currents are, in the same order, \$2.32\:\mu\text{A}\$, \$1.71\:\mu\text{A}\$, \$1.47\:\mu\text{A}\$, \$1.22\:\mu\text{A}\$, \$990\:\text{nA}\$, and \$800\:\text{nA}\$.

The capacitance is \$C_\text{I}=\frac{I}{25\:\text{mV}\,\cdot\, 2\pi\,\cdot \,1\:\text{MHz}}\$. So this works out to \$14.8\:\text{pF}\$@\$0\:\text{V}\$, \$10.9\:\text{pF}\$@\$1.5\:\text{V}\$, \$9.4\:\text{pF}\$@\$3\:\text{V}\$, \$7.8\:\text{pF}\$@\$6\:\text{V}\$, \$6.3\:\text{pF}\$@\$12\:\text{V}\$, and \$5.1\:\text{pF}\$@\$24\:\text{V}\$.

The model I used (as shown above) says CJO is \$14.8\:\text{pF}\$.

It would be possible to use .MEAS cards to pick off the exact values for you, too. I didn't do that, but it's not hard to use.

jonk
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Replace your diode with an actual capacitor of 10pF. Refine the simulation until you can clearly identify the capacitor value from observed voltages and currents. Then, and only then, put the diode back in the circuit.

Remember that SPICE performs calculations using computer arithmetic, and is subject to round-off error. Make sure that you adjust values so that you are not looking for an extremely small change in a relatively large voltage or current.

Elliot Alderson
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Yet another simple test method: is to use an LC oscillator circuit using the diode as a varicap or simply an LC resonator and frequency counter.

You could use any of the following high Q series RLC with diode in reverse bias;

  • 50mV staircase generator stepping every 10us to 30V
  • slow ramp with a 50mV 10 ns pulse (100MHz) every 10us (100kHz) as below.

enter image description here

Tony Stewart EE75
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