Diodes - General Question

BobTPH

Joined Jun 5, 2013
11,522
Hi,

Read some of the other replies he is not using a 1N4000 series diode.
The leakage current being seen here is reasonable for his actual diode, but these specs are always just rough estimates anyway.
I must have missed where he identified the diode. I only gave the 1N4000 as an example to show the leakage current to get the stated reading was reasonable.
 

Thread Starter

NewToElectricity

Joined Jul 6, 2023
7
Hello again,

The anode is only theoretically at 0v because there is nothing connected to it and the cathode is at ground. In reality, the slightest current will make the anode slightly positive with respect to ground (the negative terminal of the battery) and see figure 3A. That means that when you go to actually measure it in real life you will see something other than a purely theoretical result (as in 3B).

If you connect almost any meter from +24v to the anode, there will be a small forward bias current and that will raise the anode voltage slightly. It can be almost anything between just above 0.0 volts to around 0.7 volts. In figure 3B we see the meter itself biases the diode slightly and so we will see a small voltage drop. In your case you seem to have measured 23.7 volts, but you can expect just about anything up to about 0.7 volts or so. It usually won't go any higher than that unless the current is fairly large, like 100ma or maybe 500ma or 1 amp or higher. Since 24/100000=24ua you will just see a small voltage drop.

View attachment 298064
Thank you so very much, MrAI, for your thorough explanations & patience! And thank you very much to everyone else who chimed in.
 

AnalogKid

Joined Aug 1, 2013
12,142
In case no one has said this, the same thing applies to capacitors and inductors. A real world capacitor can be thought of as a theoretically perfect capacitor in parallel with a high-value resistor (the source of leakage current) and also in series with a low-value resistor (the intrinsic resistance of the materials). This is called the ESR (equivalent series resistance) on datasheets.

Likewise, a real inductor is a perfect inductor in series with a small resistor.

ak
 

MrAl

Joined Jun 17, 2014
13,707
In case no one has said this, the same thing applies to capacitors and inductors. A real world capacitor can be thought of as a theoretically perfect capacitor in parallel with a high-value resistor (the source of leakage current) and also in series with a low-value resistor (the intrinsic resistance of the materials). This is called the ESR (equivalent series resistance) on datasheets.

Likewise, a real inductor is a perfect inductor in series with a small resistor.

ak
Hi,

Yes, and of course a resistor in parallel to represent internal losses.
 

AnalogKid

Joined Aug 1, 2013
12,142
Yes, and of course a resistor in parallel to represent internal losses.
Keeping in mind the experience level of the TS, I thought it best to start with the basics.

In the most simple model, the perfect inductor has zero resistance, so a parallel resistance does nothing. All losses are combined into one ESR element. In the next layer more complex model, the total inductor is represented as a series of small inductances, one per winding turn, with a small capacitor in parallel with each turn and a small resistor in series between turns. This still is not a perfect model, but there is a point beyond which increasing layers of model complexity do not yield significant changes in the quality of the analysis or simulation.

ak
 
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WBahn

Joined Mar 31, 2012
32,871
This still is not a perfect model, but there is a point beyond which increasing layers of model complexity do not yield significant changes in the quality of the analysis or simulation.
And where that point lies depends on what level of quality is "good enough". Different models have different things -- in some situations, the lead wire inductance of a surface mount resistor is important enough that it has to be modeled, so too might the difference in capacitance to the PCB ground plane depending on whether a capacitor is mounted on it's front or on its side. But, in most situations, these factors have no discernible effect. Similarly, there are all kinds of non-linear effects that can be modeled. We usually ignore them because usually we can ignore them without it impacting the fidelity of the circuit model to the actual circuit too much, at least as far as what is important to us happens to be. Other times we have to account for them otherwise our analysis/simulations will be off more than we can tolerate. Yet other times we have to account for them very carefully and faithfully because that behavior is what the circuit's entire purpose centers on.

When I was working with an IBM process, each transistor was modeled using a 300+ component subcircuit because that was the level of detail they needed to do in order to model all of the various behaviors that designers cared about. My guess is that any given designer could have gotten by just fine with a much simpler model, but then you would have thousands of models and the difficult task of choosing which one was appropriate, along with a good probability that the one you really needed didn't exist. So, instead, they took a sledgehammer approach. The result was incredibly slow simulations that were phenomenally faithful to how the fabbed chip actually behaved.
 

MrAl

Joined Jun 17, 2014
13,707
Keeping in mind the experience level of the TS, I thought it best to start with the basics.

In the most simple model, the perfect inductor has zero resistance, so a parallel resistance does nothing. All losses are combined into one ESR element. In the next layer more complex model, the total inductor is represented as a series of small inductances, one per winding turn, with a small capacitor in parallel with each turn and a small resistor in series between turns. This still is not a perfect model, but there is a point beyond which increasing layers of model complexity do not yield significant changes in the quality of the analysis or simulation.

ak
Hi,

I'm not so sure i can agree with that, that the perfect inductor has zero resistance, so the parallel resistance does nothing.
The secondary issue with inductors (magnetic devices like that) is the core losses, which comes into play with almost every application. That's because the AC performance is affected by the core losses as well as other losses. That's what the parallel resistance is there to mimic.
I mainly mentioned this because you mentioned the parallel resistance in the capacitor so I thought it was odd that you didn't mention it with the inductor. It's not like we will always use this just like we don't always use the parallel component with the capacitor.
Simulators have a setting for this also although they go by conductance not resistance.

A lot of simulations require going to a higher-level model anyway such as the coupled inductor models.

As WBahn was pointing out, the application is what dictates the model, not the other way around. If we need parallel resistance, we use it, if not we don't use it, same with capacitors. Sometimes we even need to see the effects of the core material, then we might move to Jiles Atherton. All depends what we need to see. I just thought it would be more consistent to mention the parallel resistance that's all.
 

WBahn

Joined Mar 31, 2012
32,871
Keeping in mind the experience level of the TS, I thought it best to start with the basics.

In the most simple model, the perfect inductor has zero resistance, so a parallel resistance does nothing.
Even with zero series resistance, a parallel resistance has an effect, because a voltage across the inductor exists whenever the current in the conductor is changing, and thus there will be current flowing in the parallel resistor at those same times. The issue is whether such behavior models actual inductor behavior. I can see that it might, but I haven't done much with inductor modeling so I don't know.
 

AnalogKid

Joined Aug 1, 2013
12,142
Even with zero series resistance, a parallel resistance has an effect, because a voltage across the inductor exists whenever the current in the conductor is changing, and thus there will be current flowing in the parallel resistor at those same times.
Agree. The original question was about metering a steady state condition, so that was what I was addressing.

ak
 
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