Hello :) I fight this task in different ways and the results do not match the simulations in Multisim. Let's say I want to calculate the diode voltage in the forward direction for the value E = 0.3 V. In the circuit the current will not flow while multisim shows the voltage drop across the...

For all voltages less than or equal to the diode forward voltage UT = 0.7V, the diode is a gap. So for E = 0.3V UD = 0.3V; for E = 0.5V UD = 0.5V etc. If the voltage exceeds the forward voltage, current begins to flow in the circuit, and the diode can be replaced by an electromotive force of 0.7V. So for all voltages higher than 0.7V, the voltage on the UD diode is constant and amounts to 0.7V. The current flowing in the circuit in this case is: I = U / R = (E-UT) / R = (E-0.7V) / 1k?. After reversing the polarity of the source, the diode is in a reverse state, so in no case will the current flow. So the voltage on the diode will be -0.3V, -0.5V, -0.7V, etc. in turn. Multisim probably takes into account the real diode model.

Have you seen the current-voltage characteristics of the rectifier diode? The voltage drop across the diode depends on the current flowing through it.

The task assumed a constant threshold voltage of 0.7 V.

But a colleague asks why it is different in multisimie.

Maybe you need to improve the multisim - I don't know. Or maybe it will be without this multi-son? Write what you have agreed so far.

Which is not easy, due to the non-linear characteristics of the diode to reach the threshold voltage.

Because the program treats it as a real diode. Seeing the level of the task, I would assume that the author must adopt a model with approximated characteristics.

Your friend Szymon's deductions ... are correct. If you want to calculate the current exactly, I suggest you read the Shockley equation:

Calculating Rectifier Diode Voltage in Forward Direction for E=0.3V: Comparing Multisim Results

elektraelektra 10512 9

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How do I calculate the forward voltage across a rectifier diode when the source is only E = 0.3 V, and why does Multisim show a different result?

For the idealized constant-threshold model, a rectifier diode with UT = 0.7 V does not conduct at E = 0.3 V, so the diode voltage is Ud = 0.3 V and the current is zero [#16425640] Only when the source voltage exceeds 0.7 V does current flow, and then the diode voltage is taken as a constant 0.7 V in that simplified model [#16425640] Multisim likely shows a different result because it uses a real diode model with a non-linear current-voltage characteristic, not the simple threshold approximation [#16425640][#16426116] If you need the current exactly, use the Shockley equation [#16430870]

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#1 16422809 19 Apr 2017 20:42

elektraelektra elektraelektra

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Post #1
16422809 19 Apr 2017 20:42

Hello

I fight this task in different ways and the results do not match the simulations in Multisim.
Let's say I want to calculate the diode voltage in the forward direction for the value E = 0.3 V. In the circuit the current will not flow while multisim shows the voltage drop across the diode. How to calculate for such a task?
From the second Kirchoff's law I make the equation E-IR - Ud = 0, but what next?
Greetings!
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#2 16424668 20 Apr 2017 17:36

Anonymous Anonymous

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Post #2
16424668 20 Apr 2017 17:36

Have you seen the current-voltage characteristics of the rectifier diode? The voltage drop across the diode depends on the current flowing through it.
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#3 16424706 20 Apr 2017 17:56

Anonymous Anonymous

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Post #3
16424706 20 Apr 2017 17:56

Radzio M. wrote:
The voltage drop across the diode depends on the current flowing through it.

The task assumed a constant threshold voltage of 0.7 V.
#4 16424802 20 Apr 2017 18:35

Anonymous Anonymous

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Post #4
16424802 20 Apr 2017 18:35

But a colleague asks why it is different in multisimie.
#5 16425308 20 Apr 2017 21:27

Anonymous Anonymous

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Post #5
16425308 20 Apr 2017 21:27

elektraelektra wrote:
I fight this task in different ways and the results do not match the simulations in Multisim.

Maybe you need to improve the multisim - I don't know.
Or maybe it will be without this multi-son?
Write what you have agreed so far.
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Helpful post

Piecewise constant-voltage model explains Multisim differences
#6 16425640 20 Apr 2017 22:55

Anonymous Anonymous

Anonymous

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Helpful post
Piecewise constant-voltage model explains Multisim differences
Post #6 16425640 20 Apr 2017 22:55

For all voltages less than or equal to the diode forward voltage UT = 0.7V, the diode is a gap. So for E = 0.3V UD = 0.3V; for E = 0.5V UD = 0.5V etc.
If the voltage exceeds the forward voltage, current begins to flow in the circuit, and the diode can be replaced by an electromotive force of 0.7V. So for all voltages higher than 0.7V, the voltage on the UD diode is constant and amounts to 0.7V.
The current flowing in the circuit in this case is: I = U / R = (E-UT) / R = (E-0.7V) / 1k?.

After reversing the polarity of the source, the diode is in a reverse state, so in no case will the current flow. So the voltage on the diode will be -0.3V, -0.5V, -0.7V, etc. in turn.

Multisim probably takes into account the real diode model.
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#7 16426116 21 Apr 2017 10:31

Anonymous Anonymous

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Post #7
16426116 21 Apr 2017 10:31

Szymon095 wrote:
Multisim probably takes into account the real diode model.

Which is not easy, due to the non-linear characteristics of the diode to reach the threshold voltage.
#8 16427180 21 Apr 2017 19:52

Anonymous Anonymous

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Post #8
16427180 21 Apr 2017 19:52

Adamcyn wrote:
But a colleague asks why it is different in multisimie.

Because the program treats it as a real diode. Seeing the level of the task, I would assume that the author must adopt a model with approximated characteristics.
#9 16427330 21 Apr 2017 20:46

Anonymous Anonymous

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Post #9
16427330 21 Apr 2017 20:46

Kraniec_Internetów wrote:
Seeing the level of the task, I would assume that the author must adopt a model with approximated characteristics.

Or Szymon095 counts well.

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#10 16430870 23 Apr 2017 13:08

psooya psooya

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Post #10
16430870 23 Apr 2017 13:08

Your friend Szymon's deductions ... are correct. If you want to calculate the current exactly, I suggest you read the Shockley equation:
https://pl.m.wikipedia.org/wiki/R%C3%B3wnanie_Shockleya
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Topic summary

The discussion revolves around calculating the forward voltage drop across a rectifier diode when the applied voltage (E) is 0.3V. The user notes discrepancies between their calculations and the results from Multisim simulations. Responses highlight that for voltages less than or equal to the diode's forward voltage (typically 0.7V), the diode does not conduct, and thus the voltage drop across it equals the applied voltage (UD = E). When the voltage exceeds 0.7V, the diode behaves as a constant voltage drop of 0.7V. The conversation also touches on the non-linear characteristics of diodes and suggests that Multisim likely uses a real diode model, which may account for the differences observed. The Shockley equation is recommended for precise current calculations.
AI summary based on the discussion. May contain errors.

Calculating Rectifier Diode Voltage in Forward Direction for E=0.3V: Comparing Multisim Results

How do I calculate the forward voltage across a rectifier diode when the source is only E = 0.3 V, and why does Multisim show a different result?

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