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Old 5th September 2003, 02:42 PM   #1
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Default Adding LED to gainclone

hi, finished my lm3975 gainclone and just putting the finishing touches to it. (sounds great btw, with just a few hours burn in)

Want to add a blue led in the amp case (one case is psu, one case is amp).

Can i just connect one leg the led to one of the dc power lines going into the chip and the other to the power ground? (with suitable resistors ofcourse)

Is it gonna effect the output of the amp, if its only connected to one of the channels?

thanks
Amit
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Old 5th September 2003, 03:06 PM   #2
pete.a is offline pete.a  United Kingdom
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Default Adding an LED

Hi,

No it shouldn't make any difference to the sound of the amp,
just make sure you get the polarity correct.
I guess you know how to calaculate the series resistor?

LED SERIES Supply voltage = 24 - operating led voltage ( normally around 3.3 v) Divided by the forward current of the led(50Ma)

=24-3.3 /.05= 414 Ohms

I used a 1k 2w resistor on mine, as that's what i to hand and it works fine.
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Old 5th September 2003, 03:15 PM   #3
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yep, thanks.

just tryin to figure out if i have a high enough wattage resistor in the parts box.

could i just use p=iv?

p = (supply voltage-working voltage) * current draw

p = 26-4.5 * 0.02
p= 0.43 W

hmm, think i only have 1/4 W resistors about..
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Old 5th September 2003, 03:32 PM   #4
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just keep some old pcb's of whatever in the nailbox, should find what you need on them.
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Old 5th September 2003, 11:20 PM   #5
indoubt is offline indoubt  Netherlands
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And for those who are afraid to put a led on the secondary power, there is a small pcb available for a couple of and you can feed the led from the main power.

if that will affect the sound than everything op to the generator will so it will be out of your control anyway

(example conrad partno:184985-44)
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Old 5th September 2003, 11:24 PM   #6
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Question Another option?

I recall a while back that Peter Daniel and others suggested wrapping additional windings around the outside of a torroidal transformer to create a separate power supply. If this is a viable solution, it would be practically free and unrelated to the circuit.

Just trying to throw something else out there.

Sandy.
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Old 6th September 2003, 02:14 AM   #7
ir is offline ir  New Zealand
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the resistor wattage is irrelevant as the LED will only be drawing typically 20mA.

the formula is R=(E-Vf)x1000/I where R is the series resistance in Ohms, E is the supply voltage, Vf is the LED's forward voltage (typically 2V), I is the LED current in mA

so for a normal, 5mm LED for example, running of +22V you need 1KOhm

for an ultra bright blue LED, Vf=3.8V, you need 910Ohms, using 1K is fine.

25V would give 1060 use 1.1 or 1.2K
32V would give 1410 use 1.5K

simply use the formula and then the nearest, HIGHER, common value

i.e. 800=820, 910=1K, 1100=1200 etc (@10%tol, less tol will provide more divisions and a closer match - but again, that's totally unnecessary)
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Old 6th September 2003, 03:02 AM   #8
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Default Then tell me, "future boy", who is president in the United States in 1985?

Quote:
Originally posted by ir
OC4Free - Performance on the cheap.
A bit off topic- sorry- but I couldn't help notice the date format on your website e.g. 20-08-2k3. We ain't there yet....
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Old 6th September 2003, 03:10 AM   #9
ir is offline ir  New Zealand
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ummmm... okay.
20-08-2k3=
20th of the 8th (august)=2003

2k=2000 and 3 makes 2003 = 2k3

just like roman numerals really MMI = 1000+1000+1

logical eh?
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Old 6th September 2003, 09:10 AM   #10
Nuuk is offline Nuuk  United Kingdom
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Quote:
2k=2000 and 3 makes 2003 = 2k3
Except that in 2K3, the 3 is 300! I'm not nit picking here but many less experienced may get confused when selecting resistors or capacitors specified like that.

Think of the letter like a decimal point . So 2.3 (k=thousand) = 2,300.
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