@#12 @Alec_t
OK, so let me summarize what I think you guys are suggesting and make sure I am on the same page with you.

Have some sort of power supply higher than the dynamo, say 36V that powers the LM3914 as well as the partial strips of Christmas lights. The dynamo will go through a diode and cap in series to act as a peak detector. That output of the peak detect feeds the LM3914 which is setup for bargraph mode (instead of dot mode). I would then take every other output and tie them to some sort of PNP like the BS250 and switch the lights on/off that way. Sound like I am on the same page?

 

You don't have to power the chip with more voltage than the dynamo can produce. You can use anywhere in the 10v to 20v range because that's plenty for the chip to drive a mosfet, and then reduce the dynamo voltage that you are sensing with 2 resistors.

The dynamo makes 36 volts? Use 2 resistors to reduce that to... I almost said, "10 volts" but there is a calculation to set the LM3914 chip up for whatever range you need. See the datasheet.

I used an LM3915 for the audio system in a church and that puppy could max out on 5 watts, 50 watts, or 200 watts, depending on how you decide the input resistors. The main problem is that I did that in 1985 and I forgot the instructions in only 30 years.

You don't have to use, "every other output". Just set the range of operation to fire however many levels of display you want.

Gee, that's really complicated!
Gee that's really versatile!
Take your pick. The only part where you really have to follow instructions is the range setting resistors. Bring a pencil, I never got all the math right on the first try.

 

@#12
OK, OK, looks like I should really sit down a draw up a schematic at some point so I stop making silly mistakes .

Let me try again. The dynamo says it outputs 6V, 12V, 24V, so I don't really know its max, I sort of chose 36V as an upper limit, I will have to test it to see what the upper limit really is, but it seemed like a reasonable value for doing some napkin math. So let's say the dynamo runs through a voltage divider. I need to pick a voltage that works well for the LED Christmas light segments, but let's say 12V for now. So the voltage divider is a 3:1 divider. This reduced voltage goes through the peak detect circuitry and into the LM3914 signal input. The LM3914 will be running from the 12V supply. Some sort of PNP FET on the output of the 3914 to drive the LED lights which will be running off of 12V supply as well. I was thinking of using every other output since I will probably only have ~5 segments of Christmas lights to light up, so I wouldn't need all 10 outputs. So if I scaled the 3914 to full range, I would take every other one. That sound reasonable? I have a digital background, so I do projects like these to expend my knowledge of FETs and other sorts of things going on in the analog world...

BTW, I looked up the 3915 to read up on it and it appears to be EOL know. Wonder why the 3914 survived but the 3915 didn't (probably demand vs price points).

 

a schematic

Might look something like this :-

 

I wouldn't need all 10 outputs. So if I scaled the 3914 to full range, I would take every other one.

I recommend you use the bottom range of operation. That way, if the input voltage gets too high, it won't damage the chip by being out of the allowable input range. You can add some small LEDs to each output on the control board so you can see what's happening as you adjust the input potentiometer. It will give you a feel for how much to turn the knob.

 

Might look something like this :-
View attachment 110550

Thank you for the schematic, that is perfect!!! Any recommendation for the M parts? The BS250? Something else?

 

that's a 1/4 amp transistor used as a hard switch. it should work.

 

OK, parts on order, I'll keep you guys posted. I couldn't find a BS250 on Tayda, so I went with IRF9540N Power MOSFET instead.

 

Just a little update. The dynamo Should be here shortly, but I managed to gather some specs on it.

I was told that at the following RPMs, it would produce the following voltages:
70 RPM - 12V
90 RPM - 19V
150 RPM - 24V

Then a little more research showed: "The average cadence on an exercise bike usually ranges between 50 and 110 rpms, depending on the rider, the resistance on the bike and the goal of the training. Beginners and occasional riders usually ride somewhere between 50 and 60 rpm. More advanced riders, especially those riding bikes with weighted flywheels as opposed to motorized exercise bikes, ride closer to 80 or 100 rpm when simulating level terrain."

So It looks like I should be under 24V pretty consistently with a normal rider on it.

 

You need another piece of data, the same 3 points under load(s). For instance attach the light bulb that came with the dynamo, and measure your voltage-versus-rpm again. With no load, you have measured the EMF, which is useful, but you need to know about the dynamo's internal impedance.

Think of it as like the internal resistance of a battery. A 12V car battery has a far lower resistance than 8 AAs in series. With no load, you can not tell them apart by measuring only the voltage which will be 12V in both cases.

 

You need another piece of data, the same 3 points under load(s). For instance attach the light bulb that came with the dynamo, and measure your voltage-versus-rpm again. With no load, you have measured the EMF, which is useful, but you need to know about the dynamo's internal impedance.

Think of it as like the internal resistance of a battery. A 12V car battery has a far lower resistance than 8 AAs in series. With no load, you can not tell them apart by measuring only the voltage which will be 12V in both cases.

This is a good point. But I suspect that if I go with the FET approach outlined above, the load on the dynamo will be practically nil, no?

 

The load is the lights, which are so far undefined. How you switch them (MOSFET, etc.) should be mostly irrelevant.

 

The load is the lights, which are so far undefined. How you switch them (MOSFET, etc.) should be mostly irrelevant.

But that is sort of my point, the load is not the lights in this case. The lights will be powered by some DC voltage (say 24V that is controlled via a FET being switched by the LM3914. In this case, the only load on the dynamo would be the current through the voltage divider and the current sunk into the LM3914's inputs (unless I am missing something, which is TOTALLY possible).

 

But that is sort of my point, the load is not the lights in this case. The lights will be powered by some DC voltage (say 24V that is controlled via a FET being switched by the LM3914. In this case, the only load on the dynamo would be the current through the voltage divider and the current sunk into the LM3914's inputs (unless I am missing something, which is TOTALLY possible).

You can use a separate power supply for the lights, but the dynamo should have some load to keep the output voltage from fluctuating so wildly. If you don't want the load visible, use a resistor. somewhere around 240 ohms, 5 watts is my first guess.
like a pair of 100 ohm, 2 watt resistors in series. Nothing precise, just a load of some sort.

 

You can use a separate power supply for the lights, but the dynamo should have some load to keep the output voltage from fluctuating so wildly. If you don't want the load visible, use a resistor. somewhere around 240 ohms, 5 watts is my first guess.
like a pair of 100 ohm, 2 watt resistors in series. Nothing precise, just a load of some sort.

Hmmm, OK, makes perfect sense. I was just envisioning the only way to get this to work was to use the approach @Alec_t was nice enough to spell out for me (in which case I would need a constant load based on your comment, and that makes sense).

 

referring to post #24
put a load resistor on the left side of the dynamo.
C1 and R1 are going to make a peak detector that takes a couple of seconds to fade to zero. I just think the peaks shouldn't be so awfully wild as an unloaded dynamo tends to do.

 

But that is sort of my point, the load is not the lights in this case.

My bad, I confused this with another thread.

 

referring to post #24
put a load resistor on the left side of the dynamo.
C1 and R1 are going to make a peak detector that takes a couple of seconds to fade to zero. I just think the peaks shouldn't be so awfully wild as an unloaded dynamo tends to do.

OK, thanks. I am still waiting on the dynamo to arrive, so I really have had no idea what to expect (in terms to how the output voltage varies). Thank you for the advice!

My bad, I confused this with another thread.

No worries, I appreciate the attempt!