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Odicforce Boost Driver?

paul1598419

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Yes, you can set up a voltage regulator to adjust the current to a laser diode because it is a nonlinear device and as long as you have reached the lasing threshold, any increase in the output will "look" like an increase in current. The problem arises when the diode heats up and the effect resistance decreases. This acts like using fewer diodes in your dummy load and the current will increase. This will then lead to further heating with more current increases and.... thermal runaway.
 



astralist

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These are what makes me confused about all this driver's thing and the method you are all using all these time, since the emerging of custom driver.

  1. First of all, i usually don't use a dummy load to adjust the current, i simply check if there is an overshoot or not.
  2. If there is an overshoot, simply throw away the driver, OR mod it.
  3. I will then connect the driver, laser diode, and the multimeter (ammeter) in series, but the driver should set at lowest current setting.
  4. Then connect the battery and adjust the pot little by little

The method above is of course quite dangerous for the LD, if the dynamic range of the pot gives you a high current increase/decrease with little adjustment of the pot.

And here is the background story:

I do the above ever since i got the laser module with the driver from member here, but when i measure current between the driver and diode, the current set are different by order of 1A!!!
It should be 4.5A but the current to diode is only 3.5A.
The driver is using "a buck driver which many members here are using" which should be a constant current,
The LD is NUBM07E.

There should be no mistake with the seller because he's already selling so many units to many of member here.
Then i decided to disasemble the module and retest the driver,
As i tested the driver of its short circuit current, the current are correctly set at 4.5A
But as i added more diode, the current gradually decreased.

---End of the story

So given that fact, what's the point we test/simulate the current using accurate Vf?
If the driver are constant current, then it will give the same current throughout every kinds load (provided the voltage is adequate)

Although i only have one kind of custom driver from the forum,
You can try testing all the driver that have been exist
Especially the one that has been used by many people
And i'm sure the result will surprise you :crackup:



And here is my analysis, please CMIIW:

For the "true" current regulated driver, such as the one with dedicated current sense IC and current sense resistor (CSR in short),
They are unable to get a constant current with changing Vf because the load itself (diode) are not linear
so the if we think a diode and CSR form a voltage divider, the voltage across CSR are no longer linear
Furthermore the voltage feedback itself are translated to buck/boost converter's PWM, which might be non linear.


Analysis for the OdicForce/Chineese boost driver we are talking in this thread:

As i said before, those driver's current are limited by its switch current.
The pot you adjust are actually for adjusting the limit of input switch current.
So if the input are constant on 0.5A and 3.7V (as limited with switch current), the output would be vary depending on the voltage, with maximum Pout = 0.5 * 3.7 * 0.9 (efficiency from datasheet)

Example output if the load are a set of diodes, with calculation above:
476 mA @ 3.5V
396 mA @ 4.2V
340 mA @ 4.9V
297 mA @ 5.6V
264 mA @ 6.3V
238 mA @ 7.0V
216 mA @ 7.7V


My conclusion:

AFAIK nobody has mentioned about all these above before, so i think nobody noticed it before, or maybe just "my laser do just fine, so there is no problem with it"

I have a great confidence that every "small driver" out there behave this way,
The good things is, when current set at certain Vf, the current will be a constant current.
And AFAIK the thermal changing of LD doesn't decrease the Vf, so as long as the Vf stays, so does the current.

Please CMIIW
 

jnrpop

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[*]I will then connect the driver, laser diode, and the multimeter (ammeter) in series, but the driver should set at lowest current setting.
[*]Then connect the battery and adjust the pot little by little
[/LIST]

This is the only way i believe you can accuratley measure the current to the diode.....I never had much confidence with these Testloads, setting the Vf using Diodes which im my opinion all provide different Vf drops anyway with Heat change, current draw and even thier construction.

So given that fact, what's the point we test/simulate the current using accurate Vf?
If the driver are constant current, then it will give the same current throughout every kinds load (provided the voltage is adequate)
You can try testing all the driver that have been exist
Especially the one that has been used by many people
And i'm sure the result will surprise you :crackup:

Exactly!, i dont think many Laser diode drivers we have are indeed truely current regulating, i wonder how the big sellers set there drivers if they use a guinea pig diode to set all drivers at once or using a Testload to simulate Vf?

The good things is, when current set at certain Vf, the current will be a constant current.
And AFAIK the thermal changing of LD doesn't decrease the Vf, so as long as the Vf stays, so does the current.

Paul would know more about the affect temperature would have on the Vf drop of a diode, AFAIK the current draw increases with a rise in Temperature, i believe its due to the internal resistance of the Diode decreasing with an increase in temp?
 
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Atomicrox

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astralist, I'm not so sure about that. If it were true for most driver's we'd see a much higher diode failure rate.
I remeber setting the blackbuck 8M and it regulated current correctly when tested with a few different Vf's after being set and also with changing Vin. I wasn't trying to test it, though, didn't write down any numbers.
 

astralist

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Most likely they will not kill your diode,
As i said before, once the current has been set at certain Vf, they're able to keep it constant at those certain Vf.

Never own the blackbuck, but i have the other driver with violet PCB and it behaves that way,
still i'm curious, that's good to hear about blackbuck :D
And because you said that, I just bought one from ebay minutes ago.
 

jnrpop

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Most likely they will not kill your diode,
As i said before, once the current has been set at certain Vf, they're able to keep it constant at those certain Vf.

Never own the blackbuck, but i have the other driver with violet PCB and it behaves that way,
still i'm curious, that's good to hear about blackbuck :D
And because you said that, I just bought one from ebay minutes ago.

Another test i ran with the Odicforce driver, when i adjusted the input voltage
to the driver on my benchtop from 3.7V in to 4V in (freshly charged 18650),
the current and the voltage out of the driver also changed :thinking:

I think im digging my hole deeper here :crackup:
 

paul1598419

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The drivers I have been using all this time do regulate current as can be seen when you change the load while measuring with a dummy load. These diode all heat up especially since few here stick to the suggested maximum forward current and push their diodes harder. When this happens, if the Vf is held constant, the current will increase and cause the diode to run hotter. You can see where this will lead. As long as Vf is held at a certain point, the diode will go into thermal runaway and fail. Try running a diode at a constant Vf with a power supply with the current set to max. It will fail every time. At first the diode will only pull enough current to keep lasing. But, as it heats up it will draw more and more current. It can only fail at that point.
 

honeyx

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Another test i ran with the Odicforce driver, when i adjusted the input voltage
to the driver on my benchtop from 3.7V in to 4V in (freshly charged 18650),
the current and the voltage out of the driver also changed :thinking:

I think im digging my hole deeper here :crackup:

I did a closer look to the pics at OdicForce and it looks to me like it is a simple CV boost driver based on the SX1308. By trying to reverse engeenering the circuit, I only found and additional resistor going from the pot to Vin.

So if this is really a SX1308 based driver, there is no soft start and reverse polarity protection and the resistors are simple voltage dividers to set the output voltage.

However you can modify a SX1308 based driver to output constant current. I have done this already sucessfully. You need to do so to use the FB pin as the V(out)- and a shunt resistor between FB and GND. Where A=0.6/R

led.png
 

astralist

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After lurking around, i found that the Vf (of diodes) are indeed went lower as the temperature goes higher.

3500Fig03.gif


Fig-3-Measured-diode-forward-bias-voltage-vs-heater-temperature-of-a-thermodi.png


Then still, there is no point using an accurate Vf or voltage drop of dummy load if the driver itself is a real constant current.
Altough we need a minimum of 2.5V of Vf/voltage drop to make the current sense IC do it's job (if any).

I'll try to disassemble several of my custom build laser to investigate further about all this.
Because what i remember are indeed the opposite of your experience, the driver I've been using are in fact changing its current when introduced to different Vf.


Sorry for all the OOT,
Back to the topic, about this OdicForce/Chinese boost driver.
My conclusion is that those driver are an APC driver
I believe we can still use it, with several condition:

  1. Use a lower Vf (than the actual Vf of LD at certain current) when simulating with dummy load, to accommodate the LD's Vf which decreased as the temperature rises.
  2. Use full battery when simulating using dummy load
 
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Atomicrox

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IMHO the dummy load doesn't have to be accurate, the point is having a ballpark Vf to make sure the driver behaves properly around that voltage. Many drivers will be well behaved only within a relatively small range of voltages, falling out of regulation or producing a lot of ripple outside that range.
 

jnrpop

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the driver I've been using are in fact changing its current when introduced to different Vf.

I agree A,
I'm pretty sure in all cases it will change,

Using V=IxR
The Resisitance of the dummy load does not change, so if the Vf changes (changing the number of diodes), the current must change as the Law of V=IxR is :thinking:

IMHO the dummy load doesn't have to be accurate, the point is having a ballpark Vf to make sure the driver behaves properly around that voltage. Many drivers will be well behaved only within a relatively small range of voltages, falling out of regulation or producing a lot of ripple outside that range.

From the tests we have done with this driver Atom, even a 0.5-0.7Vf difference can create a change of around ~50mA with these drivers :(

Here's two tables i've put together using the two 515-520 drivers i got from Odicforce. These look alot like Astralists table posted above.

Driver1-started with 149mA on the Pot
10. ? @ 8V
9. 149mA @ 6.96V
8. 182mA @ 6.3V
7. 227mA @ 5.64V
6. 292mA @ 5V
5. 389mA @ 4.36V
4. 535mA @ 3.78V
3. 941mA @ 3.45V
2. 1028mA @ 2.71V
1. 1066mA @ 1.92V

Driver2-started with 267mA on the Pot
10: ? @ 8V
9: 267mA @ 7.32V
8: 313mA @ 6.63V
7: 370mA @ 5.94V
6: 448mA @ 5.27V
5: 561mA @ 4.63V
4: 710mA @ 4V
3: 952mA @ 3.46V
2: 1014mA @ 2.7V
1: 1074mA @ 1.93V
 
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Atomicrox

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Ohm's law only applies to linear "ohmic" resistance. It does not apply to diodes at all, and cannot be used for dummy load calculations.
 

jnrpop

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Ohm's law only applies to linear "ohmic" resistance. It does not apply to diodes at all, and cannot be used for dummy load calculations.

Its the only Law we use with the Testloads.....across the resisitor :D

Edit: i shouldn't say the only one.. :crackup:
 
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astralist

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@Atomicrox:
Yeah you are right, the diode are non linear system.

@jnrpop:
V=I*R still applies, only the difference is:
If the driver are a real constant current, hence the I are constant while the V generated will ramp up to whatever voltage (total Vf/ total Voltage across + and - of driver output), up until the I reach what we've set.
As for the R, couldn't insert the R directly to the equation because of the drop voltage generated from the diode are cannot be translated by "resistance", please CMIIW.

EDIT:
do you have another driver to test?
like the driver that have been used by many member here.
you could also test them out as a proof if this happened to all drivers
 
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jnrpop

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@Atomicrox:
Yeah you are right, the diode are non linear system.

@jnrpop:
V=I*R still applies, only the difference is:
If the driver are a real constant current, hence the I are constant while the V generated will ramp up to whatever voltage (total Vf/ total Voltage across + and - of driver output), up until the I reach what we've set.
As for the R, couldn't insert the R directly to the equation because of the drop voltage generated from the diode are cannot be translated by "resistance", please CMIIW.

Hmm,
The resistance of the Testload is the 1 Ohm resistor we use, this shouldn't change right?
It won't even change when the Testload is operating and heats up as these Resisitors and Diodes have a very high tolerance to heat before their operational values change.
 
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astralist

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Yeah if you are seeing only at the 1 Ohm resistor, then it's okay the V=I*R is of course applies :D

What i thought about resistance on the post above is an R generated from the Vt/I
where the Vt = diode's Vf + 1Ω.
but it seems the formula isn't as simple as that :crackup:
 




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