led
How would I wire 4 LEDs so they would serve as their own bridge rectifier? I have a limited space (the interior of a bicycle taillight fixture) so I would like to use the LEDs themselves to rectify the A.C. coming from a bicycle wheel generator.
Also, to limit the current so as not to exceed the forward current to each LED, how would I calculate the resistor(s) value?
A bridge design is odd for this use, however it'll work if you use a single resistor as the load of the bridge. That resistor will determine the current through the LEDs, and thus the brightness.
What's the reason for not having a common cathode, instead having +/- for each channel? Can the cathode go to a common ground in my circuit, while each of the anodes go to the emitter of each Q1?
But the cathode is common (assuming the led resistor is connected to the anode side), all cathodes are connected to the ground so they are connected to each other too.
If you have connected the resistors to the cathode side of the leds then you can't connect them together.
I'm not sure why you mention the emitter, the leds are connected between the collector and ground
Would the schematic above be sufficient to drive these LEDs? (remember the schematic represents a single channel)
If you increase the base current to be sufficient then it can drive higher power leds too.
How do I calculate the resistor values for each channel for a 12VDC power supply? Is the 350mA current for all 3 channels or is it per channel?
The base current should be about 1/10 to 1/20 of the output current to saturate the transistor and have a low voltage drop across the emitter-collector.
When you are using one color (the others are off), you can use 350mA for sure but I'm in doubt of the total max current when all three are on.
This doesn't seem right to me at all, because the total power comes to 4.4 Watts.
There is nothing wrong in the calculation. What you get is the dissipated power in each resistor for the given input voltage/output current.
Best Answer
LEDs typically have much lower reverse breakdown voltage limits (Vrrm = 5 to 15 Volts) than even the cheapest, smallest silicon diodes (1n4001 = 50 Volts , 1n4007 = 1000 Volts, 1n4148 = 100 Volts).
Forward voltage Vf on the other hand is anywhere from 1.7 Volts for some red LEDs, to 3.5 Volts or higher for some blue and white ones. Compare this to a typical 0.7 Volt Vf for a standard silicon rectification diode.
In a full bridge rectifier configuration, the voltage is dropped by 2 x Vf during the forward conduction part of the cycle.
Thus if you were to build a bridge rectifier with LEDs, the output voltage would drop anywhere from 3.4 to 7 Volts or more, compared to the input. The conduction would start much later into the positive part of the cycle (once the voltage rises above the LED Vf), compared to a silicon diode. It would also end earlier. Also, during the reverse part of the cycle, depending on the specific LED used, the "diode" is likely to enter conduction well within the input voltage range.
In other words, the bridge will provide voltage during a much smaller part of the AC cycle than with silicon diodes, and will provide a lower voltage even then.
As your purpose seems to be actually lighting the LEDs rather than just rectifying the voltage, the output of the bridge may not matter, but it's useful to know.
A suggested alternative if you are keen to use LEDs in this particular way, is to hook up the LEDs as you desire, but to add one inexpensive and tiny 1n4001 or similar, in series the current limit resistor and each LED. The diodes are not much bigger than the discrete resistors you will probably use.
Why this works:
The silicon diodes will block reverse voltage better than an LED can, i.e. to a much higher voltage.
However, the recommendation remains to use just an integrated 4-lead bridge IC (35 cents single unit, 800 Volts/1 Ampere), and use the LEDs as in any DC circuit configuration. Space requirement will not go up significantly.
Regarding current limiting resistor calculation:
Obtain the peak voltage of the bicycle dynamo's AC signal, using a multimeter with either peak AC voltage mode if available, or RMS AC voltage (more common in basic multimeters) and multiplying by 1.4142 to estimate the peak voltage. All this while pedalling as fast as you can with the cycle on its stand.
Add in a safety factor by doubling this peak voltage if you are concerned about blown LEDs due to high voltage when speeding downhill, for instance. LEDs are somewhat forgiving of minor current spikes if they are for brief durations - It isn't over-current that kills an LED usually, it is inability to get rid of generated heat.
Now, if Vmax is your safety-adjusted peak voltage, subtract your LED Vf from it (2 x Vf if you are going with the bridge, 2 x (Vf + 0.7) if you add in the silicon diodes), to obtain the voltage Vres each resistor needs to drop.
R = Vres / I,
So plug in the recommended current I for your LED of choice, and you have your numbers.
LEDs commonly come in 5 mA (SMD and some 3mm ones), 20 mA, 25 mA or in the region of amperes (e.g. Osram "Golden Dragon" LEDs). Refer to appropriate datasheets.