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I'm new to LTspice so I would like to have a few ideas. Here is a circuit:

enter image description here

Basically, the output of BLOCK_1 is a small triangular waveform. I need to pass this output through a 1N4148 diode which is a test device. The current which I get through it is:

enter image description here

Essentially, this is what I obtain after using the current probe of LTspice.

I've been thinking of a situation where I am not allowed to use the current probe, for instance.

In that case, I need the I/V characteristic for the diode D3 using the voltage probe only.

In that case, should I just add a capacitor in parallel with D3? I just need to add a single component to get both I and V characteristics of D3, I have a hunch.

I would probably need to probe two node voltages to make that work. I need to figure out those 2 nodes and plot one against the other in LTSpice to be sure, but I've been unable to do so.

Somehow, I need to know the Shockley characteristic current for D3 without using the current probe. Can someone please help me out in this with a nice explanation?

IMPORTANT: I would be building the circuit on a circuit board and probing it with an oscilloscope, so I won't be able to probe the current. That's why I need a way around.

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    \$\begingroup\$ I am not allowed to use the current probe in the simulator a current probe just make the current available for plotting and calculations. There is no way to "not allow" that. Do you mean not available when measuring a diode on a bench? I do not understand how you could determine a proper I/V curve of a diode without knowing the current. That can be done by forcing the current (using a current source) or using a current probe. \$\endgroup\$
    – Bimpelrekkie
    Aug 18, 2021 at 13:44
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    \$\begingroup\$ I need to know the Shockley characteristic current for D3 without using the current probe. Can someone please help me out in this with a nice explanation? How about using a small value resistor in series with the diode and measuring the voltage across that resistor? That is basically a current probe with some voltage drop. \$\endgroup\$
    – Bimpelrekkie
    Aug 18, 2021 at 13:48
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    \$\begingroup\$ where only measuring certain nodal voltages could give me the I/V characteristics for D3 As I wrote: resistor in series with diode. Use a 1 ohm resistor. If 1 mA flows you get 1 mV across the resistor. But all this isn't needed in a simulator as the current is directly available. It is unclear to me why you're making things more difficult for yourself. \$\endgroup\$
    – Bimpelrekkie
    Aug 18, 2021 at 13:50
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    \$\begingroup\$ @electrovolt A current probe does not need to be strictly in the circuit, it can use the wires making up the circuit, too. So use a clamp meter through a wire, since if that diode is to be added, it can be added with wires. I don't understand what this or the DSO have anything to do with LTspice, or simulation, in general. \$\endgroup\$
    – a concerned citizen
    Aug 18, 2021 at 13:55
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    \$\begingroup\$ @electrovolt said "LTSpice doesn't allow probing for the voltage drop"...that is not correct. Look here for help. \$\endgroup\$
    – evildemonic
    Aug 18, 2021 at 14:04

1 Answer 1

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How do I do that if I connect a resistor in series to D3?

Like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Then Channel 1 will give you the voltage across the diode

Channel 2 will give you the current through the diode

As Rsense has a small value, the voltage across it will be very small. So much smaller than the voltage across the diode that you can ignore it (assume it is zero).

If you're nit-picking then use the subtract function on the oscilloscope and subtract the voltages: Vdiode = Vch1 - Vch2

Maybe the current through the diode is small making the voltage across Rsense also small. Then simply use a larger value for Rsense, like 10 ohms or even 100 ohms.

Aim to keep the voltage across Rsense below 100 mV so that the diode voltage will not be affected much by that voltage.

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  • \$\begingroup\$ Thanks a lot for your effort. I just figured out now. Sorry for the disturbance caused. \$\endgroup\$
    – electrovolt
    Aug 18, 2021 at 16:39
  • \$\begingroup\$ 1 - The smaller the resistor, the better 2 - with small resistors, make sure you measure the voltage on both sides of the resistor, and make a differential measurement. Circuit boards can have significant resistance between "ground" points, and this will provide a spurious voltage at the bottom of the resistor. \$\endgroup\$
    – WhatRoughBeast
    Aug 18, 2021 at 17:29

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