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Old 02-07-2008, 03:01 PM #1
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Default Can someone explain the diode in the DDL circuit?

I understand how diodes work and what they do, but looking at the schematic for the DDL driver, the diode appears to just short out the battery if it was hooked up backwards. And it looks like you are basically hoping the diode is ok and shorts the battery out before any voltage/current can continue on the straight path to the LD. Wouldn't 2 diodes, 1 on each LD lead facing the proper way be better protection? Is the single one done just to save space?

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Old 02-07-2008, 03:16 PM #2
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Default Re: Can someone explain the diode in the DDL circu

I think I remember reading that it was to provide a 0.7v drop in voltage to the LD.

Correct me if it I am remembering wrong.

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Old 02-07-2008, 03:28 PM #3
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Default Re: Can someone explain the diode in the DDL circu

It is just there to protect the LD if the supply is connected reverse polarity.

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Old 02-07-2008, 03:42 PM #4
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Default Re: Can someone explain the diode in the DDL circu

[quote author=Daedal link=1185701612/45#53 date=1185914496]

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 properly, there is a drop of 0.7V across the 1N4001. Therefore, 6V - 3V = 3V to the diode. With most cases I have tried this I got a little past 3V and closer to 3.5V to the LD. The diodes we are getting from the SenKat group buy, and that is what this circuit is designed for, and most other laser diodes even, require between 2.5 and 3V to lase. With a 0.7V drop off the 3-3.5V supply form the LM317, you have a margin of 2.3-2.8V going into the diode. Which is perfect! You can take it out if you want, again, this is a personal choice.
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Old 02-07-2008, 03:51 PM #5
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Default Re: Can someone explain the diode in the DDL circu

I have never considered that it would give a voltage drop with correct power connection, I thought with diodes you only got the v drop when current flows through it (not an electronics man so *[smiley=huh.gif]), DDL does say in your quote....
Quote:
You can take it out if you want, again, this is a personal choice.
All my drivers have it fitted on the basis of why not.....

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Old 02-07-2008, 03:58 PM #6
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Default Re: Can someone explain the diode in the DDL circu

If the diode is properly installed it would be reverse biased and not conduct any current. So there would be no affect whatsoever on the circuit. Only when the diode is forward biased would you see the .7V drop.
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Old 02-07-2008, 04:02 PM #7
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Default Re: Can someone explain the diode in the DDL circu

Quote:
If the diode is properly installed it would be reverse biased and not conduct any current. So there would be no affect whatsoever on the circuit. Only when the diode is forward biased would you see the .7V drop.
There you go, what I said..... but using the correct terminology. Thank you for confirming knimrod.

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Old 02-07-2008, 04:41 PM #8
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Default Re: Can someone explain the diode in the DDL circu

Quote:
Originally Posted by knimrod
If the diode is properly installed it would be reverse biased and not conduct any current. *So there would be no affect whatsoever on the circuit. *Only when the diode is forward biased would you see the .7V drop.
What if it's trying to be used as a shunt voltage regulator? I'm not sure of the reverse breakdown voltage of these particular ones.
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Old 02-07-2008, 04:59 PM #9
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Default Re: Can someone explain the diode in the DDL circu

Quote:
Originally Posted by desslok
[quote author=knimrod link=1202400066/0#5 date=1202403509]If the diode is properly installed it would be reverse biased and not conduct any current. *So there would be no affect whatsoever on the circuit. *Only when the diode is forward biased would you see the .7V drop.
What if it's trying to be used as a shunt voltage regulator? *I'm not sure of the reverse breakdown voltage of these particular ones.[/quote]

That's not how it's used in the LM317 current reg. circuit. *Actually, if you look at the schematic for the DIY Power Meter, you'll see a good example of the forward voltage drop of a diode being used for shunt voltage regulation in the zero offset circuit.

The 1N4001 diode often used in our LD power supplies has a max. peak reverse voltage of 50V. And despite popular opinion here, it's forward voltage drop is actually closer to 1V, not .7V
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Old 02-07-2008, 05:32 PM #10
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Default Re: Can someone explain the diode in the DDL circu

Quote:
Originally Posted by knimrod
[quote author=desslok link=1202400066/0#7 date=1202406067][quote author=knimrod link=1202400066/0#5 date=1202403509]If the diode is properly installed it would be reverse biased and not conduct any current. *So there would be no affect whatsoever on the circuit. *Only when the diode is forward biased would you see the .7V drop.
What if it's trying to be used as a shunt voltage regulator? *I'm not sure of the reverse breakdown voltage of these particular ones.[/quote]

That's not how it's used in the LM317 current reg. circuit. *Actually, if you look at the schematic for the DIY Power Meter, you'll see a good example of the forward voltage drop of a diode being used for shunt voltage regulation in the zero offset circuit.

The 1N4001 diode often used in our LD power supplies has a max. peak reverse voltage of 50V. And despite popular opinion here, it's forward voltage drop is actually closer to 1V, not .7V [/quote]

the voltage drop reaches 1V at high currents. if u masure it with the diode test function most DMMs have, it would be aroung 0.55 volts (@1mA). In any case, the forward drop is current- and temperature-dependant. If more current is passed thru the junction, the forward drop is also higher. If the temp increases, the fwd drop (at a fixed current) decreases at a rate of about 2mV/C.
I use 1N5819 against polarity inversion /schottky diode with lower forward drop/. In DDL's regulator the 1n4001 is placed on the output - if the polarity is reversed, the LD will be protected, but LM317 may blow. If the diode is placed on the battery side, the polarity inversion current would flow thru it alone, exposing whatever's behind to its forward drop voltage during that particular critical condition, which, depnding on the power source, may blow the diode itself, in spite of its ruggedness and surge handling capabilities. That's why an in-line fuse rated for a current the diode can safly handle long enough would be advisory for a labby set-up.
...my 2 volts...
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Old 02-07-2008, 09:10 PM #11
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Default Re: Can someone explain the diode in the DDL circu

Quote:
...my 2 volts...
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