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noobie diode question

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shortbus=

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as a noobie i have a lot of dumb questions on how things work. questions;

diodes have a forward voltage drop spec on their data sheets. does that mean that for example ; at say 5VDC a small signal diode with .7Vfd the voltage would read 4.3VDC on the cathode side?

if you put 2 or more diodes in parallel in a circuit do they have a rule like resistors, that for 2 diodes instead of 1.4Vfd the drop would be .35Vfd?

like i said dumb questions to you guys but i can't seem to find the answer or i ask the wrong questions. i do a google search on this stuff before asking.

thank you, cary
 

crutschow

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diodes have a forward voltage drop spec on their data sheets. does that mean that for example ; at say 5VDC a small signal diode with .7Vfd the voltage would read 4.3VDC on the cathode side?
Yes there would be about a 0.7V drop across the diode and you would read 4.3Vdc at the cathode, but that's assuming there is a load on the cathode side (resistor or other circuitry) the controls/limits the current through the diode. If you connect 5Vdc directly across the diode it will blow.
if you put 2 or more diodes in parallel in a circuit do they have a rule like resistors, that for 2 diodes instead of 1.4Vfd the drop would be .35Vfd?
No. Diodes in parallel will still have about a 0.7V drop, but one diode will tend to hog the current because of slight differences in their forward voltage drop. And as that diode heats up more due it's increased current its forward voltage will drop further (diodes have a negative voltage temperature coefficient) which means the hogging will get more pronounced.

Two diodes in series will have about a 1.4V drop.
 
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shortbus=

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thank you crutschow, for the answer.
would a input to a lm339 comparator be considered a load? i don't have a scanner so i can't post a shematic to show what i'm trying to do. but i will try to describe it;

100VDC fed into a voltage divider (100VDC to 430kohm resistor joined to
62kohm resistor to ground)

12vdc from the divider to +pin on first 1/4 of lm339 and -pin of second 1/4 of lm339


12VDC to another volt divider (56k resistor to 50k pot to 30k resistor to ground) wiper on pot to -pin on first 1/4 of lm339. voltage on wiper out is 7V to 2.6V. (equivalent to 58V to 21V)

from -pin on first 1/4 lm339 a 1n5818 schottky diode to +pin on second 1/4of lm339. the diode Vfd is 550mV (equivalent to 4.6V)

this is a single pot window comparator that will give a 4.6VDC null (dead zone) to control direction of a stepper motor driver board. no matter what the voltage setting there should always be a 4.6VDC null.

the questions i have are;

1. am i using high enough resistor values to protect the inputs of the lm339?

2. will the circuit work as i described?

if it won't work could someone help with a circuit that will?

thank you, cary
 
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