LED replacement resistor/diode calculation

[Scott75s1974]

I'm replacing the tuner pointer illuminator lamp on a Scott R75s.
It's a rice-grain bulb run off of a 6 VAC line with a 33 ohm resistor.
Not having a suitable replacement, I used a white led out of a penlight. It runs on 3 button cells, so about 4.5 volts. (I turned it down lathe-like in a drill chuck. Took less time than placing an order, let alone waiting for it to arrive.)

Using http://ledcalculator.net, entering a suggested voltage drop of 3 v for a white LED. I got a suggested resistor of 150ohms. Does this sound right?

I've also heard it's best to install a rectifying diode to extend life of the LED. Do I install this anode to cathode? E.g. as a series double diode setup both in same orientation?

 

[Scott75s1974]

Ok, googling around, I see for AC circuits, the LED should be used with another LED or a diode in inverse parallel configuration, to protect it from reverse voltage breakdown. Easy enough to do if you have a string of them. But this is only one. So I'll prob go with a regular diode.

 

[elnaldo]

I'd install a diode in series and a small filter capacitor to reduce flickering. Dc will be higher than 6V once rectified, so redo your math. Or, install a 1K trimpot and adjust to the desired brightness. You can also connect alligator clips for your tests and swap resistors, so you choose the one you like better.

Start around 470 ohm

 

[Scott75s1974]

I'd install a diode in series and a small filter capacitor to reduce flickering. Dc will be higher than 6V once rectified, so redo your math. Or, install a 1K trimpot and adjust to the desired brightness. You can also connect alligator clips for your tests and swap resistors, so you choose the one you like better.

Ok, say 5-20 uf?

 

[petehall347]

with led i look up the power consumption of it and use my meter to work out the resistor value after adding a resistor that works out in the ball park .

 

[Scott75s1974]

So I wired it up. Ended up using a 680k ohm with a diode in inverse parallel to the LED--safely below peak, and bright enough for pointer. I don't want to replace this ever again (folded brass clips hold the pointer in, and have a limited number of cycles).

I thought this would be simple, but maybe it's not. The pointer shares the same AC circuit with the other panel lights.
So, if I mount this LED with another diode in inverse parallel, that voltage drop will affect that whole circuit for the other AC lamps. I don't notice any ill effects. Should I care? (Of course the LED voltage drop will be doing the same on the other side of the sine wave.)

Or should I just wire up a small full bridge rectifier board with filter caps? That seems overkill. (If so, what audio-grade caps do you recommend -- joke!)

I don't think a capacitor is going to help if I'm using the diode in inverse parallel. A 10 uf electrolytic just dims it quite a bit--either polarity.

 

[Scott75s1974]

with led i look up the power consumption of it and use my meter to work out the resistor value after adding a resistor that works out in the ball park .

Alas, this LED is removed from a bright penlight I found crushed on the side of the road. So I'm not sure on the specs.

 

[Scott75s1974]

I'd install a diode in series[/QUOTE

What I read somewhere online (sorry, I lost the link) is that a series rectifier doesn't save the LED from reverse voltage damage. The highest voltage drop device will take the reverse voltage hit, which is likely the LED. Does that make sense?

 

[BinaryMike]

A series diode should protect the LED from reverse voltage damage, even though it won't necessarily prevent reverse voltage breakdown. Damage results from excessive current flow during breakdown, not from breakdown alone. I've seen deliberately induced breakdown phenomena many times on a curve tracer --- when current is adequately limited, no failure results. That said, I would still prefer an anti-parallel diode.

 

[pustelniakr]

LEDs are simple. If you have AC to drive it, put a regular diode in the opposite direction across the LED. Silicon diode forward voltage drop is lower than what will damage an LED in reverse bias. Next, find what the forward voltage drop is across the LED at the desired forward current. The rest of the voltage must be dropped across the series limiting resistor, at the desired LED driving current.: E = I * R. Hence, the resistance must be selected to drop the desired voltage (beyond the forward voltage of the LED): R = E / I. This is not exactly rocket science folks.

Rich P

 

[elnaldo]

I usually connect a diode in series + one filter capacitor to rectify the AC, then a resistor and the led. No problem yet in years. We have 50Hz down here, so the flickering is noticeable without the smoothing capacitor.

 

[Scott75s1974]

Well the inverse parallel diode with the resistor works. And I trust series diode/capacitor should work as well too.

But as I overthink this minor issue, it occurred to me that this AC light circuit is probably the source of the low level 60hz hum I can hear at high volume/low input on the left channel. It's definitely 60hz not 120, so I've been stymied to figure out what would cause it. The AC to rails and regulated supplies are isolated off in their own well designed corner. It isn't an issue with the driver boards (which have no ac anyways, noisy rails would be 120hz)--stays constant when I switch them.

The 6V AC buss wiring for the lights goes across the panel, with some leads running right along the tone circuit board. That seems a very likely culprit.

So... what if I just rig a small full bridge rectifier circuit w/ small filter caps and put ALL the lamps on DC? I don't really understand RMS at all, but the DC output of that 6v secondary should't be blow-the-incandescents different. (Power wise it has to be the same or less right?)

Thanks for your patience with a trivially small issue that got slightly bigger.

 

[elnaldo]

You can. DC will be around 8- 8.5V. You can drop a couple of volts with some diodes in series or-and some resistor in series.

But I'd twist the AC wiring and run in further from the preamp, even if you have to solder some extension wiring.

 

[w1jim]

The resistor in series is for current limiting, not voltage drop - otherwise the diode alone would draw way too much current (it would be like a short circuit).

 

[pustelniakr]

For the OP...

The voltage across a forward-biased diode, including LEDs, will approximate a constant voltage across a wide range of driving current, once the diode has been fully forward-biased. The spec sheet for the diode will tell you what that voltage will be.

The resistor is selected to drop the rest of the voltage at the desired current through the LED. The formuli above will lead you to the right value for the resistor. You need to make sure that your resistor is able to handle the power it will dissipate. Otherwise it will get too hot and eventually fail. Since it is already installed, you could feel the resistor with your fingers. A good, simple rule of thumb is if you can hold your fingers there, the resistor will be fine.

Otherwise, determine how much the resistor is dissipating, by measuring the voltage across the resistor. The power being dissipated is indicated by the square of that voltage divided by the resistance value. I like to choose a resistor that can handle twice the indicated dissipation (ex: a 1/4W resistor for 1/8W of dissipation). Just make sure that the resistor has access to the air (not covered with insulation), so it can cool normally.

Rich P