How Can I Understand the Solution for Diodes Homework in Electronics?

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Discussion Overview

The discussion revolves around understanding a homework problem related to diodes in an electronics context. Participants are attempting to clarify the solution provided in a manual, particularly focusing on the diode equation and the expected output voltages in different scenarios involving diode conduction.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses confusion over the solution manual's output, particularly regarding the I2 equation and the values obtained when substituting for VD.
  • Another participant questions the clarity of the problem description and suggests that the diode equation appears correct, noting that the expected current should not exceed 1 mA with a 5 kΩ resistor.
  • A participant clarifies that diodes 3 and 4 will only conduct if the input voltage VA is positive, while diodes 1 and 2 will conduct for negative voltages, leading to different output voltages at points C and B.
  • It is mentioned that the output voltage at B cannot go more negative than -2V due to the voltage drop across the diodes and the resistor, assuming the diodes operate within their current ratings.
  • There is a suggestion that if the images provided are unclear, the participant should type out the relevant equations and text for better understanding.

Areas of Agreement / Disagreement

Participants do not reach a consensus; there are competing views on the clarity of the problem and the correctness of the solution. Some participants express skepticism about the provided images and the understanding of the diode behavior.

Contextual Notes

Limitations include unclear images of the problem and solution, which may hinder understanding. There are also assumptions regarding the diode characteristics and current ratings that are not explicitly stated.

Alkozy
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Homework Statement


A picture of the problem is provided in the attachment

Homework Equations


there is no special relations except
I= Is x e^(vd/Vt)

The Attempt at a Solution


i have a solution for from the solution manual but i do not understand it
like the I2 equation i don't get reasonable values when i substitute VD for any value > 1 or even 1
i attached the solution that i do not understand
if someone can help please
 

Attachments

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  • diodes .png
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Can you be bothered to provide a clear description ? Of the problem and also of the part you don't understand.
Your pictures are so fuzzy my eyes hurt after discerning 0.025 which is 1/40 which I recognize as kT for room temp, so I don't think there is much wrong with that diode equation. What is it that goes wrong for you ? (Note that with 5 k##\Omega## you don't expect currents exceeding 1 mA anyway).
 
BvU said:
Can you be bothered to provide a clear description ? Of the problem and also of the part you don't understand.
Your pictures are so fuzzy my eyes hurt after discerning 0.025 which is 1/40 which I recognize as kT for room temp, so I don't think there is much wrong with that diode equation. What is it that goes wrong for you ? (Note that with 5 k##\Omega## you don't expect currents exceeding 1 mA anyway).
i am sorry for the fuzzy picture. it is the pic of the solution which i don't understand. leave it alone
the first pic is about the question and it is clear
what i know about the problem: that the diodes have 0.7 voltage each as VD
diodes 3 and 4 only will conduct if the input voltage VA is positive. in this case there will be an output voltage at both C and B ( C is that output before the 1k ohm R )
for a negative voltage diodes 1 and 2 only will conduct and for this case there will be output voltage at B only

how much?
this is what i am asking for

note: these diodes are working as voltage regulators
 
Last edited:
The output voltage at B is unable to go more negative than -2V. If you try to send it more negative than this by giving A a large negative voltage, then increased current flows through the 5k resistor via the diodes, and this results in more voltage dropped across the 5k resistor—leaving still 2V across the pair of diodes.

This assumes the diodes are within their current rating and don't overheat or burn out, and the voltage across each when conducting can be approximated as 1V.

Note: if it is not possible to provide a clear image, we expect you to remedy this by typing out the relevant equations and text yourself, preferably using LaTex. Blurred images help no one.
 
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