Warning light is dim after fitting led indicators.

[spuggy]

Incandescent bulbs are just a coil of wire. A little heater coil that glows white-hot. They don't care which direction the current is flowing, they work either way round. They have a fairly high resistance - which is converted into heat as well as light.

Many old indicator circuits are designed around using bulbs; as these aren't polarity-sensitive at all, as soon as you provide current with a path to ground, it's "job done"...

An LED is just a diode that happens to emit light (they actually generate small amount of electricity if you expose them to light as well, but usually not enough to be useful)

The voltage the LED itself wants is determined by the doping of the substrate - and thus the colour they produce. White LEDs only work with 3V, red ones want 1.9V, UV ones 4.1V. They either have a current-limiting (aka "ballast") resistor built-in to the package to allow it to be directly connected to a 12V circuit, or sometimes multiple - eg 4 - LEDs are wired internally in series in the package.

LEDs only allow current to flow in one direction. Even with the ballast resistor, they have a very low resistance to current flow in this direction - which is why they draw so little power.

(A 2W bulb, at 12V, draws 160ma, and a 3W bulb 250ma. A typical white miniature LED uses 10ma - and probably brighter than a 10W bulb. It'll generate a little warmth in the ballast resistor - but still a lot less power overall than a bulb.)

However, current flow in the other, "wrong", direction, with an LED has a very, very, high resistance - think of it as infinite. Much, much more resistance than a bulb. That's the "diode" part of "light emitting diode".

With non-LED flasher relays and no load resistors, the relay hyper-flashes because the the forward resistance of the LED - even with the ballast resistor - is nothing compared to a bulb. So it behaves as though a bulb is out.

If you resort to load resistors, you pretty much throw away the "low current draw" advantage of LEDs (the load resistor just turns the extra current into heat), but you do get to keep the "I don't blow bulbs", "they're brighter" and "they respond almost instantly" benefits.

[utopia]

Originally Posted by spuggy

With non-LED flasher relays and no load resistors, the relay hyper-flashes because the forward resistance of the LED - even with the ballast resistor - is nothing compared to a bulb. So it behaves as though a bulb is out.

That's the bit I've never really understood.
I've always reasoned that if the system behaves as if a bulb is out, that would tally with an ultra high or infinite resistance, not the ultra low resistance of an LED.
In other words, it would behave as if the bulb had been short circuited (low resistance path) rather than blown (infinite resistance path).

[spuggy]

Originally Posted by utopia

That's the bit I've never really understood.
I've always reasoned that if the system behaves as if a bulb is out, that would tally with an ultra high or infinite resistance, not the ultra low resistance of an LED.
In other words, it would behave as if the bulb had been short circuited (low resistance path) rather than blown (infinite resistance path).

Good lord, I'm too young to understand all that analog stuff!

Heh. As-if...

https://auto.howstuffworks.com/turn-...he%20resistor.

When you push the turn-signal stalk down, the thermal flasher connects to the turn-signal bulbs by way of the turn-signal switch. This completes the circuit, allowing current to flow. Initially, the spring steel does not touch the contact, so the only thing that draws power is the resistor. Current flows through the resistive wire, heating up the smaller piece of spring steel and then continuing on to the turn-signal lights. At this point, the current is so small that the lights won't even glow dimly.

After less than a second, the small piece of spring steel heats up enough that it expands and straightens out the larger, curved piece of spring steel. This forces the curved spring steel into the contact so that current flows to the signal lights unimpeded by the resistor. With almost no current passing through the resistor, the spring steel quickly cools, bending back away from the contact and breaking the circuit. The cycle then starts over. This happens at a rate of one to two times per second.

Typical spec for these relays is 90 +/- 5 cycles/minute - when the load is 2x 21W bulbs wired in parallel, as per this example diagram:



So the resistance from the bulbs in parallel is lower than two bulbs in series (parallel resistance is calculated as 1/Rt = 1/R1 + 1/R2 + 1/R3 + ...).

A modern one will still operate with one bulb out - but if you've blown a 21W bulb, it'll hyperflash - because a single bulb means more resistance, so with less current flowing, bi-metallic strip doesn't get as hot - and cools down faster.

LEDs draw so little current without load resistors that an electro-mechanical relay might not operate at all - but if it does, it hyperflashes, same as if you've got a bulb out.