Martin's Dynamo Circuit #7 - cap & tail-light query...

Hi,

In Martin Schmidt's #7 circuit (http://pilom.com/BicycleElectronics/DynamoCircuits.htm), for the 4-led version he specifies a 470uf 100V value for the 2 tuning capacitors. I would be grateful for any thoughts as to why Martin has chosen 100V caps here rather than say, 50V? I have both to hand and the 50V parts are much smaller and would certainly better suit what I am trying to do. Any reason why I can't use 50V?

I'm actually going to use 3 x white XR-Es in one of Cutter's mpcb + lense packages along with a generic, Luxeon-type 350ma red led in series, as Martin shows in circuit #9 (with the red led at the 'top' of the string). The vf of the red led will be around 1V less than that of the white leds. Will that be a problem at all? Will I be overdriving it? I have several of these red leds, some rated at 350ma and some at 500ma. Would it be better to use 2 of them in parallel in the series string?

Thanks,

Sam.

hi Sam,

have you considered that this is a perfect excuse to play with a circuit simulator such as LT Spice?? I've got a model for a Schmidt dynamo, if you want to fiddle around with it.

Without the benefit of doing a proper circuit analysis or circuit simulation, let me throw out a couple of ideas.....

the first is that the voltage rating is just a proxy for selecting a cap with a low ESR and a high ripple current rating. If that really is the intent, then it would have been better to specify these parameters/characteristics, and perhaps recommend a couple of part numbers from specific capacitor families from the major suppliers.

the second, and somewhat lousier, reason would be the desire to have a cap that would withstand the dynamo voltage in the event that the LED load was disconnected. It doesn't explain why different voltage ratings are required for different numbers of LEDs.

As far as the LEDs go... the Vf of the red LEDs shouldn't make much difference. I'd make sure that the LEDs are rated for over 500mA, though... preferably 1 amp? I was thinking that Martin had a graph of dynamo current, but I'm not seeing it. If you want to parallel red LEDs, you'll want them to be the same type so that they share current equally.

The way I remember the effect of wiring LEDs is - in series voltage is cumulative, current is constant, in parallel voltage is constant, current is cumulative.

What you describe is the basis of the light for one of my bikes. The only differences are that my red led is an XP-E which is rated at 700 mA and I have added 2 x 1.5F capacitors for a standlights. With your dynamo producing 500 mA, the 3 XR=E's will have a vf of ~10.3 + ~2.4v for the luxeon giving a vf of ~10.7.

In practice, the only thing noticeable about the different voltages is that the red led lights up first

Will this circuit work?
White led headlight and a red led taillight on a current divider:

rectified 6V 3W hub dynamo
4700uF capacitor in parallel with
10ohm resistor and a 480mA white led in parallel with
240ohm resistor and a 20mA red led

cant post an image

Anyone? Tried posting on bikeforums and other forums, can´t get an answer from anyone that knows what they are talking about.

define "work".

What's the goal of the design?

I don't see the purpose of the 10 ohm resistor. If you want to divert current to a red LED, you don't need to add a resistor in series with the white LED to make it happen.

If you want a single white LED (I'll assume with a Vf of 3v at 0.5A) and want to divert 20mA to the red LED (I'll assume a Vf of 2.0v at 20mA), then the resistor in series with the red LED should be equal to (3v - 2v)/0.02A, or 50 ohms.

It's not a great way to design a light, since you're only getting 1.5 watts of power out of the dynamo. There are ways to get more power from the dynamo, but if these are the LEDs that you want, this is how to calculate the value for the resistor in series with the red LED.

tgtbt said:

can´t get an answer from anyone that knows what they are talking about.

Click to expand...

Are you sure about that?

Steve K said:

If you want a single white LED (I'll assume with a Vf of 3v at 0.5A) and want to divert 20mA to the red LED (I'll assume a Vf of 2.0v at 20mA), then the resistor in series with the red LED should be equal to (3v - 2v)/0.02A, or 50 ohms.

Click to expand...

Thats what i did but with different values (6V-2V)/0,02A=200ohms. Where did the rest of V go in a parallel circuit?

tgtbt said:

Thats what i did but with different values (6V-2V)/0,02A=200ohms. Where did the rest of V go in a parallel circuit?

Click to expand...

As I said in the other thread the dynamo is a 500mA current source. So if you put one LED in series you get ~3V, if you put 4 LEDs in series you get ~12V.

znomit said:

Are you sure about that?

Click to expand...

No. Im gonna test everything once i get my dynamo. I posted a picture of parallel leds, someone suggested leds in series and the thread derailed.

tgtbt said:

Thats what i did but with different values (6V-2V)/0,02A=200ohms. Where did the rest of V go in a parallel circuit?

Click to expand...

The generator is not a 6v, 500mA source. It can be modeled as a voltage source in series with an inductance and a resistance. The voltage is proportional to the dynamo's speed. Due the complex and changing nature of the dynamo, the simplest way to make a bike light work is to effectively short out the dynamo. The normal headlight has a resistance of 12 ohms, which is pretty close to a short circuit for the dynamo.

For this reason, it is usually safe to think of the dynamo as a 500mA current source (with some limitations). If you just wire up a single white LED to the rectified dynamo output, then you'll get 500mA through the LED, which will produce 3v. That's why I used 3v in my calculations.

Thanks again. Looks like i learn fast but have a limit to how much i can learn. This is how i brained out my circuit from what i think i understand so far:

1. Source voltage is what comes out of the dynamo and equal to all branches of a parallel circuit, around 6v.

2. Formula for resistors in series with leds is (source voltage minus voltage drop) divided by current=resistance, in this case (6v-3v)/0.48A=6.25ohm and (6v-2v)/0,02A=200ohm.

3. Formula for dividing current is (total current times total resistance) divided by resistance in r1=current through r1, in this case (0.5A*6.0606ohm)/200ohm=0,01515A and (0.5A*6.0606ohm)/6.25ohm=0,4848A. Close enough.

4. Leds get warmer when passing current, they pass even more current when they get warmer, more current gets them even warmer... in this case no more than 500mA, i get that, but

5. leds in parallel will draw more or less of what they are supposed to because of the heat-current thingy and disturb the current balance between them unless controlled with resistors, one in series with each led.

tgtbt said:

1. Source voltage is what comes out of the dynamo and equal to all branches of a parallel circuit, around 6v.

Click to expand...

NO, this is the bit you're getting hung up on, forget the 6V!

honestly, for anything moderately complicated, a person needs to understand circuit analysis such as is taught in EE101. There is probably something on the web, I suppose, but the concepts aren't going to be picked up on a web forum. Also, you need to be able to handle multi-variable linear math. Not calculus, but at least be good at solving simultaneous multiple equations.

I think Znomit and I are so used to working with this stuff that we know how to approximate a lot of it and come up with quick solutions. I don't know how to give that experience to someone else.

In this case, the white LED's Vf of 3v will take the 0.5A from the dynamo and effectively act as a voltage regulator to produce 3v (plus or minus a bit).