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Old 1st May 2003, 01:06 AM   #1
elizard is offline elizard  Canada
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Default Yet another gainclone question

Ok .. so now that I have everything but the transformer, i'm starting to slowly build everything .. including the chassis

and i have a couple questions ..
1) should i just cut the unused pins off of the lm3875's? Also, looking straight at the front of the lm3875, is pin 1 the leftmost one (as seems to be the convention?)

2) I want an LED for the PS case .. eventually anyway ..
say I want to take the +/- for it from the bridges .. can I do that? (provided i use a sufficient resistor to keep the current down ..) Or does the voltage not go down with the current? (still a newbie .. but slowly learning)
i.e. take the + for the LED from the + output from each bridge, and - from the - output from each bridge .. then have 2 LED's ..
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Old 1st May 2003, 11:14 AM   #2
sangram is offline sangram  India
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I would cut off the unused pins from the IC. To go one further, If I was doing P2P, I'd apply some epoxy on top to pot the pins and ensure no risk of shorts.

No big deal to use LED. I used blue LED with a small diode and 10 uF cap from transformer directly as I have separate rects (but single transformer) for both channels, and did not know which to take from.

A small resistor will do fine. You have to calculate the current you want through the LED and figure out resistor value from there. Different LEDs have different voltage drop too. Too small a resistor will mean blowing up the LED, or the resistor, or both, and too large a resistor will result in a dimly lit LED or different colour emission.

Say you have LED with 1.65 V forward V and you want to run it at I of 15 mA. Your input DC V into the circuit is 35V. You are asking the R to drop 33.35V at 15 mA. R=2.223K or about 2.2K. Power dissipation will be about 1/2 Watt, a 1 watt resistor should be good for this application. You can go up or down on all values depending on the LED and the input voltage...

This is however calculated from + to ground. For + to - your dissipation and R value will double. So take your tap from + rail and ground, or have a separate 1N4001 and 10uF circuit to power the LED separately.

To tame the bright torch like glow from a blue LED I use a small square of paper behind the panel, and then the LED shines on the paper. Nice!!!
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Old 1st May 2003, 11:18 AM   #3
sangram is offline sangram  India
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Forgot to mention - 15 mA is a pretty high current. I run my LEDs at 5 to 7 mA and they are good enough for me.

Also IC pins are normally marked left to right in direction of text - at least English text. Specially in power ICs, SIPs etc.

Manufacturer will always put a prominent dot or other mark to delineate pin 1. Usually a dot. Else Pin 1 is leftmost.
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Old 1st May 2003, 05:06 PM   #4
elizard is offline elizard  Canada
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perfect, that's what i wanted to know, thank you!
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Old 1st May 2003, 05:20 PM   #5
elizard is offline elizard  Canada
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btw, what is the formula for calculating the needed resistance for a given voltage/current drop?
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Old 1st May 2003, 05:40 PM   #6
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Quote:
Originally posted by elizard
btw, what is the formula for calculating the needed resistance for a given voltage/current drop?
sangram's explanation, while accurate, was somewhat confusing as it seemed to imply that you need a particular voltage drop across the resitor and that you're calculating the value of the resistor to get that voltage drop.

The voltage drop across the resistor is always going to be your supply voltage minus the forward voltage of the LED regardless of the value of the resistor. Well, provided the value of the resistor isn't so low that it blows your LED but that's a different story.

So instead of thinking about setting the resistor for a particular voltage drop, think of it only in terms of establishing the current you need to draw through the LED.

The voltage drop across the resistor, V<sub>R</sub>, will be V<sub>ps</sub> - V<sub>F</sub> where V<sub>ps</sub> is the power supply voltage and V<sub>F</sub> is the forward voltage of the LED.

Decide how much current you want to run through the LED and the value of the resistor becomes R = V<sub>R</sub> / I<sub>LED</sub> where I<sub>LED</sub> is your LED current.

That help?

se
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Old 1st May 2003, 05:55 PM   #7
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Got this from another thread.


Click the image to open in full size.


"I just read the reference AC supply voltage at the bridge position,
e.g. 24V AC in case of my Zen V2. Meanwhile, my blue
LED lightens lovely when about 5mA currents flow thru it.
Therefore, I calculate the necessary resistor value in this way:
24 x 1.414 / 0.005 = 6787, i.e. about 7.5K ohms (watts?
24 x 1.414 x 0.005 = 0.17w)."

I tried this in my GC and it worked just fine.
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Old 1st May 2003, 07:17 PM   #8
elizard is offline elizard  Canada
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just a small question on that schematic
the +/- for the LED that you do take, is that from the +/- going OUT of the bridge or coming in from the secondaries?
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Old 2nd May 2003, 01:40 AM   #9
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I believe it is refering to the +/- side of the led (diode) itself.
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Old 2nd May 2003, 01:51 AM   #10
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Quote:
Originally posted by elizard
just a small question on that schematic
the +/- for the LED that you do take, is that from the +/- going OUT of the bridge or coming in from the secondaries?

The LED is connected directly to secondaries.
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