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So I'm confused about the constant voltage model (which assumes diode has .7 volt drop).

In the pic below, I'm confused about the third case. What if the voltage source has voltage less than .7 V but the diode is in forward bias? Does that mean it's open circuit and hence the voltage drop across the diode will equal that of the voltage source? (In the last two cases---in the circuits to the right--I just made up numbers to illustrate the concept)

enter image description here

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If you are modelling the diode as a constant voltage, then yes, if the source voltage is less than 0.7v, that will be the voltage across the diode. It looks like you understand it fine. With reverse bias, and with forward bias < 0.7v, it's open circuit. Otherwise, it looks like a 0.7v battery.

A model as simple as this is adequate for some purposes, and not for others.

Remember, all models are wrong, but some models are useful George Box

If a constant 0.7v is too wrong for your purposes, let's say you want to estimate the diode voltage drop at 1nA, then you would use a better model.

A popular one is the Shockley Diode Equation. Watch out for the 'ideality factor', a fiddle factor which allows you to align the model with measurement over a wide range of parameters. At high currents, you'd need to add some series resistance. At high frequencies you'd want some parallel capacitance (voltage dependent), and perhaps some series inductance as well. And you still haven't taken account of the charge storage, which is as important for stopping 1N4004s from being used in SMPS as it is for making GHz step recovery diodes work.

You stop elaborating the model when it's good enough for your purposes.

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  • \$\begingroup\$ Also why does it say you can replace the diode with voltage source of .7 in the forward bias \$\endgroup\$
    – David
    Sep 16 '18 at 6:32
  • \$\begingroup\$ @David because that's the forward voltage drop the model assumes. When a current is flowing, the voltage drop is always 0.7v. Which is approximately true enough for simple modelling. \$\endgroup\$
    – Neil_UK
    Sep 16 '18 at 6:35
  • \$\begingroup\$ Also does that mean if they ask for the current in the circuit when voltage source is less than .7 and diode is connected in forward bias, it would be better to model as ideal diode? \$\endgroup\$
    – David
    Sep 16 '18 at 6:35
  • \$\begingroup\$ @David it depends who 'they' is. In an academic setting, you're usually told which model to assume. Otherwise, you always use the simplest model you can that meets your requirements. If you assume a particular model, and you're not certain whether it's the 'right' one, state explicitly what model you're assuming. That way, if your error is not guessing what model 'they' wanted you to use, your answer may still be correct in the sense of consistent with the model you've assumed. \$\endgroup\$
    – Neil_UK
    Sep 16 '18 at 11:27
  • \$\begingroup\$ A shout out to George Box who first said "All models are wrong; some are useful". Words to live by. \$\endgroup\$ Sep 16 '18 at 12:40
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I think you might be confused about a basic concept with how these simple diode models are used. You must use a different model for the diode depending on whether it is forward biased or reverse biased. In forward bias the diode looks roughly like an ideal voltage source of 0.7V and in reverse bias the diode looks roughly like an ideal current source of 0A (an open circuit).

If you are not sure which to use, then just pick one. If you assume that the diode is forward biased, analyze the circuit, and determine that the diode current is negative then your assumption was incorrect and the diode is in fact reverse biased. If you assume that the diode is reverse biased, analyze the circuit, and determine that the voltage across the diode is greater then 0.7V then your assumption was incorrect and the diode is actually forward biased.

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Consider this, where the plot is a log-vertical-diode-current, and linear-horizontal-diode-voltage

enter image description here

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Simple answer is that diode can't act as a voltage source. If external voltage (Vext) is greater than 0.7V then drop across diode is 0.7V and if Vext < 0.7V then the drop across the diode can't be greater than Vext. So, if you see the I-V chart of this approximation you can see that before cut-in voltage(0.7V) current(Id) is zero.

Similar Analogy: Assume a box on ground (with "MAX" frictional force say 10N)

Case 1: If we push it with 20N then frictional force acting is 10N to oppose motion.
Case 2: If we push it with 3N force then frictional force acting is 3N only else if its greater than 3N then box will be in motion.

So frictional force is for opposing the motion, it can't make box move (and similarly a diode can't be a source).

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