LED power supply

[chunkitup]

I am adding a string of led's to my amp so it will light up the inside of the chassis. This will probably be a dumb question to many but I am going to ask it anyway. In the attached pic I was going to wire up the first scheme but then I thought why when I could just use one dropping resistor instead of one for each string. Am I correct? There will be four led's in all.

[chunkitup]

Pic

whoops

[defaced]

Generally spec sheets want a single current limiting resistor for each LED. This is probably to force each LED to operate at a given current, because if left to their own devises with a single combined resistor things can get unbalanced - but that's just a guess on my part. Similar idea to using balancing resistors with series caps and such.

[Enzo]

The upper one is the better of the two shown, but you left out the one I would use - series. If the LEDs are to be on at the same time, just run them in series. Same current runs through them all then. The resistor would be reduced since it has to drop less voltage. I mean if you wanted, you could run 10 LEDs in series across a 24v DC supply, with an appropriate resistor to take up the slack. Or 20 across 48v or whatever.

[Phostenix]

What if you had 12VDC & 2V LEDs? Would you put 6 in series with no resistor or 5 with a resistor to pick up 2V?

[Phostenix]

Looks like the answer to my question is 5 with a resistor.
L.E.D Basics; gaining an understanding of how to work with L.E.D.s

[Enzo]

You need a resistor to limit current.

[Burnt Coil]

Number 1 is correct. Number two could work but you could burn an LED; they are very sensitive to current.

Another approach: if you have any diodes in the circuit you could replace them with LEDs.

[chunkitup]

Phostenix, thanks for that site. It is very informative. I would definitely go with the series with one limiting resistor but I am running this off the heaters. I am using green leds at 2.1v per. So I would not have enough juice. On that site it says not to run leds in parallel.
Quote:

Do not put LEDs in parallel with each other. Although this usually works, it is not reliable. LEDs become more conductive as they warm up, which may lead to unstable current distribution through paralleled LEDs. LEDs in parallel need their own individual dropping resistors. Series strings can be paralleled if each string has its own dropping resistor.

So drawing one would be the most fitting or should I use a 12v tranny, which I have.

[Don Moose]

Drawing one is the reliable one. Rectified heater voltage is fine, if you have the spare current. DO NOT GROUND the bridge or the LEDs unless the heater winding is also grounded. Just replace the grounds you've drawn with the minus connection, and you're good to go.

Install recto and cap, measure voltage.
Calculate (nearest standard) resistor to load that voltage to (numberofLEDs*LED_current)mA; add resistor load.
Measure voltage - this is your supply V.

IF your V is, say, 2.25*Vf, you can put 2 LEDs in series - you want some headroom, but you don't need much.

You have a Vf of 2.1 @ LED_current. Find the next-higher standard resistor that gets you LED_current for V - Vf. You might want to verify the Vf and adjust R slightly ... maybe.

Remove your test load, add LED+R assemblies and go.

Hope this helps!

[mooreamps]

Generally spec sheets want a single current limiting resistor for each LED. This is probably to force each LED to operate at a given current, because if left to their own devises with a single combined resistor things can get unbalanced - but that's just a guess on my part. Similar idea to using balancing resistors with series caps and such.

that is sort of generally true. However, you can have one large value dropping resistor, for total combined load current ; figuring lets say 20 mills per device, and then a singular 1 ohm series resistor for each device to keep individual led currents in check.


-g

[Old Tele man]

...only one SERIES resistor is needed with a SERIES string of connected LEDS because current is constant through a series path.

...conversely, only a single VOLTAGE source would be needed if a PARALLEL connection of all the LEDS were used.

...one resistor with all the LEDS in series is the answer.

[raiken]

If your total LED forward voltage drop doesn't exceed your supply voltage, you can connect them in series with a single resistor UNLESS you have a large number of LEDs where the normal lot-to-lot variation in Vf can result in a wide range of voltage drop across the series string, making it impossible to pick a resistor value that will work in all cases for production, without too wide a variation in LED current. If you are building a one-off, however, you can simply select a resistor that gives you the current you want and leave it at that.

In these cases is is best to either use a parallel string with individual current-limiting resistors, or better yet, use a current source to drive them. There are lots of nice buck and boost mode high-efficiency switchmode LED drivers on the market today from vendors like Linear Technology and others.

Since you are running 4 LEDs, if they are green or red, you'll have a series drop of anywhere from 8V to 12V, and if they are white, you'll have a series drop of around 12V to 16V. You will need a voltage supply a few volts higher than that to drive them and account for the drop across the series current-limiting resistor. If you don't have that voltage available, and you are planning on rectifying and filtering the filament supply and using that instead, you'll have to use a parallel arrangement with a separate dropping resistor for each LED.

Never connect them in parallel with one dropping resistor like you show in your second drawing. This is a bad idea.

[Enzo]

Here is the deal with straight parallel LEDs and why it is bad. This is really the same idea behind ballast resistors on parallel power transistors.

Assume you have a power source and a series resistor that makes sens and an LED. The LED will have some natural voltage drop across it, let us say 2.00 volts for example.

Now slap another LED in parallel with the first. You can change the resistor to reflect the extra current. But lets say the new LED has a natural drop of 1.80v. The circuit will turn that LED on and there will be almost zener like a steady 1.80v there across it, across them. Well that isn;t enough to light the other 2.00v LED. SO the lower drop LED will hog the circuit.

[chunkitup]

Alright here is my plan. This is a gutted PV Butcher and I was going to wrap up and hide the orange taps on the tranny that was origially used for the bias since my new bias is using the HV sec tap. I believe and please someone varify this that the voltage is about 57v on the orange taps. Here is the pic of my scheme. Can I use a 1.8k limiting resistor instead of the 1.62k or will that drop too much voltage? And from one of the replies, I should keep the circuit closed and not grounded to the chassis?