Questions about voltage drop across 1N4001 diode

These generic silicon diodes are known to bring a 0.7v voltage drop.

For design purposes, should one anticipate the .7v drop regardless of diode orientation, or regardless of whether the circuit is open behind the diode?

Example circuit (I took me 10 minutes to draw this, so don't make fun of it! ):


The + supply goes to a 4-pole switch: OFF, Pole 1 which powers Light 1 only, Pole 2 which powers both lights, and pole 3 which powers Light 2 only.

Questions:

In Mode 1 (Light 1 only), will I see a 0.7 drop from diode #1 ? I'm assuming no, because the diode is only acting as a barrier, and the circuit is open behind diode 1. Am I correct? If not, care to explain?

In Mode 2 (both lights powered), Will I see a cumulative 1.4 volt drop due to the two diodes, or ???

Another scenario: the DDL circuit (image courtesy of that thread):


According to the DDL thread, the 1N4001 diode in this version of the circuit contributes a 0.7v drop.

I'm aware that in these circuits, we regulate current, and the voltage sorts itself out. However, in a situation where the supply voltage is low and cumulative dropouts can push the available voltage below the minimum threshhold required by the light, then this diode could become an issue.

I'm also aware that the diode isn't essential to the circuit (it's for reverse polarity protection) and therefore can be removed.

What's not clear to me, is exactly when you will or will not see the .7 drop due to the diode. That's why I'm asking.

In the DDL circuit, you'd think that the .7 drop might only occur when the mistake of backwards batteries was made, thereby enabling the diode to do its job. However, there's no open circuit behind the diode in the DDL circuit, so maybe that makes a difference.

Yep - I'm scratching my head.

Any diode experts care to shed some light?

Thanks!

Re: Questions about voltage drop across 1N4001 dio

I´am no diode expert, but a voltage drop over a diode only occurs if there is current floating throuh it.
So in the lm317 schematic you posted, there would only be a voltage drop if the diode was connected in series with the ld rather than anti-parallel. That considered, I would say that your right with what you guessed about your cuircuit. However, I´am not _that_ sure about the 1.4 V drop or how 2 diodes will behave when connected in parallel, but it should be right.

ArRaY


*edit*
Looking at your picture again, im now sure that there will be a .7 V drop for each driver.

Re: Questions about voltage drop across 1N4001 dio

The voltage across a diode IS directionally dependent.  They do not work the same forward as they do in reverse.  You MUST connect them in the proper orientation. Also, the .7V is an ABOUT .7V, it changes with how much current is flowing through the diode, but .7V is a good guess.

Voltage across the diode is when it is in the closed circuit with current flowing through it.  In mode 1, you have a short across the diode, the diode can be considered to not even be part of that circuit.  It will do nothing.  The same goes for mode 3, in both of those, consider it as though there is nothing in the circuit except the driver and the LD.

In mode 2, voltage is equal in parallel, it doesn't divide.  The voltage across each diode is still .7 (as long as it is oriented properly).  Things get funky with drivers, because there are more complicated things in play there.  But look at it this way, assuming you're using the LM317 (which is a linear driver): There will be .7V across your 1N4001, then whatever voltage the driver takes, then the voltage of the LD.

As far as the DDL circuit, when in proper polarity, the 1N4001 does NOTHING.  It is off, no current flows through it, the voltage across it is the same as the voltage across the laser diode, but since the diode is in reverse polarity, nothing flows.  It only does anything if the battery is in reverse polarity, the diode will be on, with its .7V drop and the driver taking all the rest of the excess energy/voltage, and all the current will be flowing through the 1N4001 instead of the laser diode.  Properly configured, the 1N4001 and the laser diode will never have current flowing through them at the same time.

Why do you need the 1N4001 in your original image?

Re: Questions about voltage drop across 1N4001 dio

Great explanations! Thanks to both! Wow!

To answer PBD's last question, the diodes in my drawing prevent supply current for mode 1 or mode 3 from spilling across the unused "Y" to the the other side. If there's a better way, lemme know!

BTW - DDL afficionados should take note:

1st - thanks again to Daedal for posting that guide and taking the time to explain the CCR circuit to everyone. That thread singlehandedly is what got me started modding lights and lasers. ;D

Having said that, it seems there is a mistake in his comments on page 3 of the guide:

Fourth, the 1N4001 is being used for two reasons. One is to stop the diode from overkilling itself when you connect the battery the wrong way around by passing all the current through the 1N4001 and not through the LD. Second, when the battery is connected [highlight]properly[/highlight], there is a drop of 0.7V across the 1N4001.

Click to expand...

As explained above, the diode would only contribute a .7v drop to the circuit when the battery was connected INCORRECTLY, in which case the .7v drop would be irrelevant, because the diode is only serving to save your LD by dumping the "backwards" current before it hits the LD.

Anyway, thanks again! Awesome!

Re: Questions about voltage drop across 1N4001 dio

1N4001 actually have a .6v forward voltage drop. says so right on the package

Re: Questions about voltage drop across 1N4001 dio

HumanSymphony said:

1N4001 actually have a .6v forward voltage drop. says so right on the package  

Click to expand...

Check my 1st paragraph, the voltage across a diode changes with the amount of current flowing through it (or, the current changes with the voltage applied, whichever you prefer).  That's just how diodes work, it's variable.  .6 to .7 to .8 is a very good guess, but don't think it's exact.  On the 1N4001 datasheet from Fairchild (one manufacturer of them, there are tons of semiconductors companies that make these simple components, it's jsut one example), it's about .6 V at 10mA, .7V at 50mA, and .8V at 200mA.  But they're going to vary slightly from one to the next, as well.  



To the OP: That makes sense, and I can't think of a better way to do that off the top of my head with a switch like that. There might be a kind of switch or relay that would make the diodes unnecessary, but that looks like it will work as long as you supply enough voltage to account for the drop of the 1N4001 diodes when on setting 2, such that the drivers and laser diodes will still have all the voltage they need. Also, for something like that, it doesn't have to be 1N4001, it could be any number of simple, standard diodes. 1N4002, 1N4004, etc. Since all you need is the rectifying behavior, most every simple diode will give you that.

Re: Questions about voltage drop across 1N4001 dio

Thanks again!

I guess one solution to ensuring equivalent voltage supply to drivers regardles of mode is to install diodes on the #1 and #3 legs as well. ;D Crude, but effective.

Re: Questions about voltage drop across 1N4001 dio

Jess said:

Thanks again!

I guess one solution to ensuring equivalent voltage supply to drivers regardles of mode is to install diodes on the #1 and #3 legs as well.   ;D  Crude, but effective.

Click to expand...

yeah that would work. the lasers will work best with single operation, sharing the voltage to both the drivers will result in less power to it, but it should still work... perhaps you could have 2 power sources, using 1 for single mode and use them both for dual mode. if you put the batteries in parallel the voltage will stay the same, series it doubles.

good luck w/ ur project!
-Kendall

Re: Questions about voltage drop across 1N4001 dio

it's about .6 V at 10mA, .7V at 50mA, and .8V at 200mA. But they're going to vary slightly from one to the next, as well.

Click to expand...

That sounds about the right, the 0.7v is just an indication, it depends on forward current and temperature. In real-world applications, expect 1.0 to 1.1 volt drop at full load (1 amp). Note that the diode does get warm at this current, which in fact lowers its voltage drop a bit.

With 7.2v supply and the lm317 driving a blu-ray, this drop is something to consider.