Grotesquely Freakish PHR diode...

pianoman2011 · Apr 6, 2009

Just thought this would be interesting for everybody to know.

Everybody, believe it or not, but in an attempt to kill one of my PHR-803T diodes ([smiley=evil.gif]), I ran it at ~650mA down to ~400mA for about 20 minutes straight...

The reason for the decreasing amount of mA's is because my 9v battery was dying. Anyways, my diode did not die after those excruciating 20 minutes... I just got tired of running it for so long and also had to go to sports practice. Before and after this event, I had my diode at around 200mA with extremely frequent use. It continued to run at 200mA (it kept its original brightness) for two days after this until it died a very unnatural death; I tried to extract it from the Aixiz module and failed miserably

. This all happened about a week ago, but I am sure that it would still be alive and running today...

How the heck did I get such a freak diode!!! I expect anything more than 250mA to be an "insta-kill" (as Charlie Bruce puts it), and 650mA was sure to kill it! But it didn't! I am stunned!

-Jakob

P.S. Maybe my voltmeter is faulty???

Grix · Apr 6, 2009

What mw did it output at 650ma? :S

pianoman2011 · Apr 6, 2009

Grix said:
What mw did it output at 650ma? :S

I wish I could tell you but unfortunately I do not have a LPM yet... But it actually not that much brighter than it was while running at 200mA. I guess the efficiency (mA:mW) is a lot less when dealing with such high amperage.

-Jakob

wolfblue · Apr 6, 2009

If you were running off a 9 volt battery, the square ones that us old guys once called a transistor battery( chagre em with a radar range?

). Well you cant get 650mA out of one. I don't know for sure, but doubt you could get more than 200mA out of even a new one. I cant explain what you experienced Jakob, but a PHR at 400mA or more? From what I understand, it would take a freak to hit 200mA. A short lived freak.

How do you measure the current?

--John

leukoplast · Apr 6, 2009

wolfblue said:
If you were running off a 9 volt battery, the square ones that us old guys once called a transistor battery( chagre em with a radar range? ). Well you cant get 650mA out of one. I don't know for sure, but doubt you could get more than 200mA out of even a new one. I cant explain what you experienced Jakob, but a PHR at 400mA or more? From what I understand, it would take a freak to hit 200mA. A short lived freak.

How do you measure the current?

--John

I thinkyou can pull more than the rated mA out of them as you want, it just accelerates the use of the battery? For example, I have some 100mAh 10220 lithiums, and I can pull more than 100mAh from them..but they start losing voltage and dying very fast.

intro74 · Apr 6, 2009

You're sure about your reading :

I think also that your voltmeter is faulty or anything else is wrong :-X

wolfblue · Apr 6, 2009

You would think that your 100 mAh batteries would run something at 100 mA for 1 hour. Tot bad it doesn't work out that way.

I pull 8 amps or more out of my RC cars 2800mAh batteries. Not all full throttle, and it lasts about ten minutes. I just did not think a 9V alkaline could dump a half amp. Or even a quarter amp. I think they are rated around 500 mAh. I did not think it would come out of those little batteries so fast. But thats just what I think

Maybe it will dump that fast. But maybe it was only putting out like 200mA. Jakob did say it was not much brighter @650mA than @200mA. I think it might do that at 650mA( there I go thinking again). But it could be the battery reached its limit.

--John

Krutz · Apr 6, 2009

dont touch it until you get a "real" currentsupply to check with 400mA! if you measured the current directly (currentmeter in series in between the diode and the driver) and it showed those 400mA, then yes, holy cow! i doubt it, but would really love to be proven wrong!

..how many did die in current-lifetime-tests? how many by igor alone? hehe
should it really survive "real" 400mA, send it to someone with an LPM!!!1!!eleven!

manuel

GooeyGus · Apr 6, 2009

a 9v can dump over an AMP when new... I've used them to power c-mount pump diodes with an LM317 circuit and I've been able to turn them all the way up to 1.5A. They dont last long at all, but they can do it!

wolfblue · Apr 6, 2009

Krutz said:
send it to someone with an LPM

But I think he said it died :'( Too bad if it was true, that one may have lasted a long time around 200mA.

And now standing corrected on the 9v battery draw thing....

http://www.powerstream.com/9V-Alkaline-tests.htm

You can even get 1 AMP out of a modern transistor battery, but not for long. And not at all with a LM317 and 405nm diode. Even at 100mA a LM317 starts to dim almost immediately. I know Jakob has other type drivers in hand. Maybe its true. More info please pianoman

EDIT You can run a 808nm diode with a LM317 (as Gus knows) due to the lower voltage that the 808nm uses.

--John

pianoman2011 · Apr 7, 2009

If you were running off a 9 volt battery, the square ones that us old guys once called a transistor battery( chagre em with a radar range? Smiley). Well you cant get 650mA out of one. I don't know for sure, but doubt you could get more than 200mA out of even a new one. I cant explain what you experienced Jakob, but a PHR at 400mA or more? From what I understand, it would take a freak to hit 200mA. A short lived freak.

How do you measure the current?

--John

I was running it off of a square rechargeable nine volt battery (expensive little effer... cost me $11 for one). I think when you pute the battery through the DDL driver, you can get more mA from what the battery delivers... when it's fresh-charged my DDL with deliver 1.1 amps with a nine volt battery and minimum resistance. And I don't know how to explain what I experienced either. I measured the current with a normal voltmeter. I did not use a dummy load though. Might that be the problem?

I thinkyou can pull more than the rated mA out of them as you want, it just accelerates the use of the battery? For example, I have some 100mAh 10220 lithiums, and I can pull more than 100mAh from them..but they start losing voltage and dying very fast.

That's what I thought... my battery barely lasts at all when I am running it through my driver. THat's why I bought rechargeable.

dont touch it until you get a "real" currentsupply to check with 400mA! if you measured the current directly (currentmeter in series in between the diode and the driver) and it showed those 400mA, then yes, holy cow! i doubt it, but would really love to be proven wrong! Smiley
..how many did die in current-lifetime-tests? how many by igor alone? hehe
should it really survive "real" 400mA, send it to someone with an LPM!!!1!!eleven!

manuel

Too late... I destroyed the diode when trying to extract it from the Aixiz

. I measured the current with a voltmeter and then hooked up the diode, both before and after.

a 9v can dump over an AMP when new... I've used them to power c-mount pump diodes with an LM317 circuit and I've been able to turn them all the way up to 1.5A. They dont last long at all, but they can do it!

That's what I though... but mine gives out 1.1 amp max.

-Jakob

pianoman2011 · Apr 7, 2009

wolfblue said:
But I think he said it died :'( Too bad if it was true, that one may have lasted a long time around 200mA.

And now standing corrected on the 9v battery draw thing....

http://www.powerstream.com/9V-Alkaline-tests.htm

You can even get 1 AMP out of a modern transistor battery, but not for long. And not at all with a LM317 and 405nm diode. Even at 100mA a LM317 starts to dim almost immediately. I know Jakob has other type drivers in hand. Maybe its true. More info please pianoman

EDIT You can run a 808nm diode with a LM317 (as Gus knows) due to the lower voltage that the 808nm uses.

--John

Yea it did die. And for a stupid mistake, not because of overpowering it to death. And to get the ~650mA-400mA I used an LM317 DDL driver and a 9v battery. It went from 650mA to 400mA in about 15-20 minutes (I measured before and after that run). So it did lose power pretty quickly. Ask questions and I will answer. It really is too bad that I killed my diode so idiotically.

-Jakob

wolfblue · Apr 7, 2009

So do you measure the current directly, Like with the meter in series with the battery, on amps ?
Or the voltage drop across a resistor ?

But the LM317 needs 3 volts. And the Blu-Ray needs 5, even more as the current goes up. So a 9V can starts to approach 8 volts pretty quick over 100mA. And a rechargeable may only start at 7.2 volts. So our diode now only has 4.2 volts to begin with.

Also, when current draw goes up, the battery's voltage goes down as the internal resistance changes. And the battery gets hot. Watch that you don't damage your battery with heat.

--John

pianoman2011 · Apr 7, 2009

wolfblue said:
So do you measure the current directly, Like with the meter in series with the battery, on amps ?
Or the voltage drop across a resistor ?

But the LM317 needs 3 volts. And the Blu-Ray needs 5, even more as the current goes up. So a 9V can starts to approach 8 volts pretty quick over 100mA. And a rechargeable may only start at 7.2 volts. So our diode now only has 4.2 volts to begin with.

Also, when current draw goes up, the battery's voltage goes down as the internal resistance changes. And the battery gets hot. Watch that you don't damage your battery with heat.

--John

I measured the mA output of the driver by connecting the + and - leads of a voltmeter set on amps to the + and - LD outputs as the driver. The one error that I notice here is that I never used a test-load to measure the milliamps... is that a problem (other than the possibility of damaging the driver)?

My 9V delivers 8.4 volts at full charge so I know that there is no problem with the amount of voltage. I actaully haven't measured the voltage output of the driver though. I should probably do that. And by the way, when I was doing the extreme PHR test, I was using a new normal 9V battery... I just bought my rechargeable 2 days ago.

And my battery is not getting hot... thanks for the warning though!

-Jakob

wolfblue · Apr 9, 2009

pianoman2011 said:
I measured the mA output of the driver by connecting the + and - leads of a voltmeter set on amps to the + and - LD outputs as the driver.

Uh, maybe, uh. If I understand correct, that wont work right if the diode is connected. Your parallel with the diode. Some current going through the diode, some though the meter. You have to flow all the current through the voltmeter to measure current. You can put the voltmeter between the driver and battery . As when you connect one side of the battery clip to one side of the baterry. And then use the voltmeter to connect the other side of the battery. When you hold the testleads the voltmeter becomes the on switch. Flowing all the current though the voltmeter and powering the circuit.

Inside the meter is a known value resistor that is low ohms. They call it a shunt. You have seen where they recommend a 1 ohm resistor inline with, in series with the diode. A resistor between the driver and diode. On the + or - side of the diode. This is a shunt.

I think the best way to check actual current is to use a shunt. The voltage drop across the shunt is proportional to the current flowing through it. Ohms law states that current = volts divided by ohms. Say1/10 Amp= 1/10 volt divided by 1 ohm. Why they say 1 millivolt= 1 milli amp with a 1 ohm shunt. I use .47 ohm resistors because I have a few already laying around. For my 6x the voltage across the .47 ohm shunt is 82mV. So .082/.47=.174 amps. Or 82/.47=174 mA.

Hope this helps

--John

pianoman2011 · Apr 9, 2009

wolfblue said:
Uh, maybe, uh. If I understand correct, that wont work right if the diode is connected. Your parallel with the diode. Some current going through the diode, some though the meter. You have to flow all the current through the voltmeter to measure current. You can put the voltmeter between the driver and battery . As when you connect one side of the battery clip to one side of the baterry. And then use the voltmeter to connect the other side of the battery. When you hold the testleads the voltmeter becomes the on switch. Flowing all the current though the voltmeter and powering the circuit.

Inside the meter is a known value resistor that is low ohms. They call it a shunt. You have seen where they recommend a 1 ohm resistor inline with, in series with the diode. A resistor between the driver and diode. On the + or - side of the diode. This is a shunt.

I think the best way to check actual current is to use a shunt. The voltage drop across the shunt is proportional to the current flowing through it. Ohms law states that current = volts divided by ohms. Say1/10 Amp= 1/10 volt divided by 1 ohm. Why they say 1 millivolt= 1 milli amp with a 1 ohm shunt. I use .47 ohm resistors because I have a few already laying around. For my 6x the voltage across the .47 ohm shunt is 82mV. So .082/.47=.174 amps. Or 82/.47=174 mA.

Hope this helps

--John

John,
I did not have the diode in parallel with the voltmeter… I unsoldered the positive lead each time I tested it. The voltage and current went straight into the voltmeter and back out. But I am worried this may have damaged the driver and/or voltmeter and affected my reading.

And thanks for telling me how to measure mA with a resistor and diode… now I know why people use those “dummy-loads” and how they work. In fact, I have resolved to buy one myself (the one that is built by rckstr I think). Could you give me a link or directions on how to use those? Thanks for the advice!

-Jakob

Grotesquely Freakish PHR diode...

pianoman2011

0

Grix

0

pianoman2011

0

wolfblue

0

leukoplast

0

intro74

0

wolfblue

0

Krutz

0

GooeyGus

0

wolfblue

0

pianoman2011

0

pianoman2011

0

wolfblue

0

pianoman2011

0

wolfblue

0

pianoman2011

0

Online statistics

Forum statistics

Share this page