Hi,
Ive read many articles that state that the voltage drop across a diode is 0.7v.
However when i make this simple circuit with 1N4148 Diodes the voltage drop is 7V!

For some reason i cant measure the voltage drop across resistor and led!


Any ideas?

thanks

Attached Images

Diagram.png (2.1 KB, 45 views)

 

hi bite,

Sorry if these are obvious questions.

Which end of the battery are you using as the 0V connection on the DVM?.

Does the LED glow?.

Is the LED/diode the correct way around ?

Check the resistor value.

Like to help

EricG

 

First of all, is the circuit working? Does the LED glow brightly? By your values, you should be putting about 20 mA through the LED which is correct.

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Writing while you were, Eric. Good morning.

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Sorry to not include crucial info.
No led does not glow as it must only be receiving pitiful voltage, As far as i can tell i have connected everything right. If i bypass the diode and connect v+ to resistor, led glows.

Have tried the diode both ways, The resistor heats up violently if i change the direction.

thnx

 

hi,

If i bypass the diode and connect v+ to resistor, led glows.

If I'm reading your diagram correctly, how can you do that ?.

Eric

hi AllVol

 

Are we sure the LED is in circuit correctly? The long leg (anode) should be on battery side.

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Sorry i lazily threw together that diagram. Just to illustrate what was connected in series.

Here is how it is physically connected.

Attached Images

Diagram.png (2.1 KB, 35 views)

 

hi bite,

The 1N4148 diode and diodes of a similair technology, have forward voltage drops 'around' 0.7Vfwd.

If you slowly increased the forward voltage across the diode, [with a limiting resistor], you find the diode starts to conduct at about 0.6Vfwd.

As you increase the applied voltage, the current will increase thru the diode and the Vfwd go to 0.7V, if you continue to increase the applied voltage, the Vfwd increase to 1.0v. If didn't limit the current, the diode would be destroyed.

The diode has whats referred to as a 'knee' on its applied voltage/conduction curve, were the Vfwd increases with current.
Also the Vfwd is dependant upon its body temperature. check the datasheet.

So don't take for granted you will get 0.7Vfwd across the diode.

Eric

 

hi bite,

If you jumper across the diode and the LED glows, then either the
diode is the wrong way round, or its faulty.

The resistor is carrying 25-30mA, so if its about 0.25W rating, it will get warm.


EDIT: if its 0.125W it will get hot!!.

Eric

 

Hi Eric,

By your latest diagram, if you are reading, with a multimeter, 9volts across the diode (positive probe on battery side, neg probe at cathode [striped] end of diode) that means no current is flowing through your circuit.

Possible causes:

Battery weak or dead
LED backwards
Diode backwards
An open connection in your circuit
Resistor open (probably not)

Check, then double-check everything. The circuit as you have drawn it should work.

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i agree with with you only up to certain extent.

even if the battery is weak, 7V is still enough to power up the entire circuit isnt it?

another thing is that, he said that he read 7V from the multimeter accross the diode. If the diode is not conducting (due to faulty diode, resistor open, faulty LED,and whatever), then isnt the reading should be 9V instead of 7? even if you consider the leakage, it would not be up to 2V right?

correct me if i am wrong

 

Are you doing this in real life or are you just simulating it in Crocadile Craps?

If it's the latter then I recommend upgrading to something less Mickey Mouse like LTSpice.

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It would appear that I have reached the dizzy heights of idiocy; I’ve only been connecting the diode wrong way! as suggested by many people!

Sorry all.

I just use "Crocodile Craps" for schematic design (I realise its pretty rubbish) but I Like the windows interface the most out of all the progs.


Thanks very much everyone.