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Current/Voltage Regulator

Benm

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An AMC7135 is interesting, but it has a preset current of 350 mA and no means to control that. In my opinion you'd be driving the diode pretty hard if you use one.

I understand many people are looking for a low-drop 317 that you can just fit in there, but such chips are rare if available at all. Obviously the 1.25 volt reference is a problem in itself, as that is effectively in series with your diode. This is a problem common to all adjustable voltage regulators - they were never -really- designed as current sources with good performace.

Even modern low-drop voltage regulators have references in the order of 0,8 volt, despite their dropout of 200 mV or less.
 



Gazoo

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Benm said:
An AMC7135 is interesting, but it has a preset current of 350 mA and no means to control that. In my opinion you'd be driving the diode pretty hard if you use one.
Why can't a resistor be placed series with the diode to lower current? Or maybe a silicon diode would drop the current to a comfortable level.
 

o_DEATH_ANG3L_o

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Yeah Gazoo those chips from DX look really cool. You get 20 of them for $14.00. Not bad at all. But if that is not adjustable from the 350 mA, I am sure you can just put a current limiting resistor in series as well and just play with the numbers in ohms till you get a comfortable current. Well thank you guys for all your input. For now I am just gonna stick with Daedals circuit as is.
 

chimo

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Gazoo said:
[quote author=Benm link=1191266847/15#16 date=1191322414]An AMC7135 is interesting, but it has a preset current of 350 mA and no means to control that. In my opinion you'd be driving the diode pretty hard if you use one.
Why can't a resistor be placed series with the diode to lower current? Or maybe a silicon diode would drop the current to a comfortable level.
[/quote]

A resistor in parallel with the LD would be a rough fix but anything in series with the LD will not affect a current source (just waste power :)).

In this case you would have to start with a low resistance (gets a larger share of the AMC7135's current) and gradually increase it until the LD gets the share you want it to get.  Recall that the LD's voltage will drop as it heats up so it will grab more of the current when it gets warm.
 

Gazoo

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Ok, I think I get it but what about placing a silicon diode in parallel with the LD? And I do believe this is a linear regulator? So the input voltage will effect the amount of current going to the LD...right? Thanks for your help.
 

chimo

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Gazoo, once the silicone diode goes into conduction, the LD will only see the diode's forward voltage and it will be too low for the LD to operate. You could switch to a zener configuration but that would be equivalent to going back to a voltage source. The easiest way to use the AMC7135 would be to shunt some of the current with a resistor parallel to the LD. Cheers,

Paul
 

Gazoo

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Ok Paul, thanks... :) I totally understand now. It will be interesting to play with. It would be perfect for using with an open can diode, or the phasor diode as is. But I really want to get it to work with the GB diode.

According to my calculations, and please correct me if I am wrong, I would need at least a 1.5 watt resistor assuming a power supply of 4.5 volts. Therefore I should be able to use a 3 watt 25 ohm rheostat in parallel with the laser diode to calculate the resistance needed. Do you concur.?..thanks again.

Oops...never mind...I just realized I need to use the following formula..P = E 2 / R. This changes the whole picture because I think it will require a hefty resistor as far as the wattage rating.
 

chimo

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Gazoo, according to the spec sheet keep in mind the output of the AMC7135 may vary in output from 300-380mA so you will want to test it to be sure.

If I was doing this, here are the steps I would follow.

1. Determine the actual AMC7135 output on a dummy resistive load.
2. Choose a current you would like to run the LD at.
3. Use an adjustable LM317 based current source to determine what voltage range the LD you will be using has at the desired current.
4. Take the actual current output of the AMC7135 and subtract the desired current through the LD. That's the current you will have to shunt through the resistor.
5. Determine the required resistance value by taking the LD operating voltage and divide it by the shunt current.
6. Determine the resistor wattage value by P=VI or I^2R

Example of above (made up values!!!!!!)
1. Let's say the AMC7135 came in bang on at 350mA
2. Let's say you want 200mA through the LD
3. Let's say you measured the LD's voltage to be 2.5V at 200mA
4. You will need to shunt 150mA (350-200)
5. Resistor value = 2.5V/0.15A = 16.6666667ohms
6. Resistor wattage = 2.5*0.15 = 0.375W (for short durations a 1/4W should be OK but the the design would call for a 1/2W)

All clear??? :)
 

Gazoo

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Yup...all crystal clear now...thanks. ;) I have a couple of LD's that turned into LED's. One is an open can and the other is a GB diode...so I think I have the perfect load to begin testing this with. I plan to test this with a range of 6 volts all the way down to 2 volts. I will report back when I have completed my testing. Thanks for all your help..
 

o_DEATH_ANG3L_o

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Gazoo said:
Yup...all crystal clear now...thanks. ;) I have a couple of LD's that turned into LED's. One is an open can and the other is a GB diode...so I think I have the perfect load to begin testing this with. I plan to test this with a range of 6 volts all the way down to 2 volts. I will report back when I have completed my testing. Thanks for all your help..
Yeah please let us know. Thank you. ;)
 

Benm

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This isnt very efficient, but it would probably work okay to some extent.

Bear in mind that the paralel resistor DOES mess up regulation though, since the current it sinks is proportional to the voltage across it. If you work the numbers, you'll see that the current through the LD will also depend on the voltage across it.

On the positive side, the current through the LD will drop with rising diode voltage, so there is little chance this running out of control on its own. On the downside, the voltage across the diode drops with rising temperature, which leads to the diode taking even more current.
 

catdog

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How much current are you actually aiming for with these diodes? The same manufacturer makes a different part with a set value of 220mA:
micro-bridge.com/A705.asp
 

Gazoo

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It may not be very efficient, but I would think it is a heck of a lot more efficient then the 317. :p

catdog...thanks. I will see if I can find some...that would be perfect for the GB diodes.
 







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