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Old 28th September 2008, 09:57 AM   #1
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Default A stupid questions about fuse, please help

I am using a 5AH 12V rechargeable battery for my DIY TA2024, since I mounted the battery inside an external box, I added and extra LED direct to the battery for minitoring the status, plus help keep the battery in health by spend all remain energy as no memory effect, the connections is as shown as attached drawing, however the LED may burn out sometime when plug and unplug the battery from the amp , especailly when the battery in fully charged, I am sure it is about pluse current, I want to have a fuse for protect the LED, then no need to disassembling the chassis fequently when something happen, but how large the fuse I should use? Is there any formular for calculation? Please help, thanks!
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Old 28th September 2008, 11:12 AM   #2
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OK I couldn't find E272 diode, so I can't tell what this part does, and I can't calculate the current.

You don't need to discharge the battery like this, unless it is a NiCd, and memory effect is not common anyway. Better to just deep cycle the battery from time to time.

You probably have too much current thru the LED anyway. Take the E272 diode out of the LED circuit and replace the 100R resistor with a 2k resistor. This will keep the current to ~5mA, which is plenty for an indicator. You shouldn't need a fuse.

w
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Old 28th September 2008, 02:20 PM   #3
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E272 is a CRD (Current Regulator Diode) rated at 2.7ma, the reason I use E272 as a resistor because it never goes hot, but resisitor will generate heat then may burn as time goes by.
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Old 28th September 2008, 05:29 PM   #4
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The device will get warm just like a resistor. Are you sure it hasn't failed short-circuit?

Conventional fuses are not available in the 2.5mA range.

Try bypassing (paralleling) the diode-resistor chain with a 0u1 capacitor.

w
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Old 29th September 2008, 04:53 AM   #5
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No, E272 never get even warm, I use it for all cases which need to have a resistor with LED for power indication, E272 can support up to 60Vinput, plus it is times much expensive then resistor, if it is not stable, always keeping cool, I will not spending money for no reasons.
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Old 29th September 2008, 05:33 AM   #6
Pafi is offline Pafi  Hungary
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goldkenn!

Both CR diode and resistor will dissipate exactly P=I*V amount of power! If you replace CR diode to a 3.3k resistor, then current and voltage will remain the same in normal condition, so obviously the resistor won't be hot either, (if it is as big as the CR diode)!

Fuse is absolutely not neccessary.

What kind of accu do you use? If it is a lead-acid type, then you will surely kill it this way!
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Old 29th September 2008, 08:55 AM   #7
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Not lead acid, I know lead acid will be killed in this way, don't worry, I just not having enough knowledge for some baisc calculation formular, but not really stupid as I told from the topic

By the way, I just don't know why, before I found E272 and use as resistor, I of course stuck with resistor, but no matter what the resistor value and watt, I always experience resistor get hot and even can't be touch, but E272 never even slightly warm, wire.....

Anyway, the LED blow not very fequently, just as if a fuse could add, I never need to disassembling the chassis, since the chassis ius drilled by myself, and the LED is stuck on the panal by hot gule, so if in case replace LED will be a bit complicated.

Thanks for all join discussion.
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Old 29th September 2008, 09:08 PM   #8
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Quote:
Originally posted by Pafi
Both CR diode and resistor will dissipate exactly P=I*V amount of power! If you replace CR diode to a 3.3k resistor, then current and voltage will remain the same
Thanks Pafi, this is exactly what I am saying. If the E272 is TRULY not getting warm, it has failed short circuit. It may be your fingers are mistaken. If it is the same size as a resistor, it should get as hot as a resistor. All diodes have a resistance, the resistance is what controls the current in this case, although the device is not just a simple diode. IT MUST GET WARMER IF IT IS WORKING.

If the E272 is working, you don't need the 100R resistor.

If the E272 is not working, then the 100R may be preventing the diode blowing for a while. Some blue LEDs are very tolerant of overcurrent for a time.

The circuit is overcomplicated for the application. You are spending money for no reason, increasing the likelihood of failure and making the circuit more difficult to diagnose.

Take the E272 out of the circuit and replace the resistor with 2~4k. 1/4 watt will do: @ 15V, 2k the dissipation will be ~70mW, 4k < 40mW. Resistors are supposed to get warm.

There is no real need for a bleeder resistor across the battery.

w
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Old 30th September 2008, 09:23 AM   #9
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Funny.... whatever, I am not selling E272, I no need to lie for how amazing it is, by the way, as you guess, it is no need to use the 100 ohm resistor, it is just what I am going to add with the fuse, as all I need just asking for a calcualtion of fuse value, that's why I put it in the drawing, in general I never need to add any resistor with E272.

Again, thanks for all join the discussion, I found the formular for calculate the fuse value, thanks.
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