Check my math???

[Andy.wpg]

This is similar to a problem I was looking at last year. I wanted to cut Candy Cane strings so I could control them individually.

There are 17 bulbs in each cane (x3 canes is about 50ct)

voltage per bulb=2.4V (total volts dropped by lights = 17x2.4=40.8 volts)
amps = 0.1 amps
Resistance = 729 ohms (theoretical resistance of missing lights)

Given the information above, I would be looking to dissipate 79.2 volts total.
Given c=1/(2*Pi*f*R)
1/(2*3.14*60Hz*729ohms)= 0.0000033492201829 farads
I should be able to drop the voltage by using a 3.35 microfarad non polarized capacitor?

Tony

Looks right to me. You will, of course, have to use the next standard value. Make sure the capacitor you use has the voltage rating with a good safety margin.

[CaptKirk]

Make sure the capacitor you use has the voltage rating with a good safety margin.

Which around here means you want something like a 1.21 Jigawatt capacitor...

[Andy.wpg]

You cannot measure it without a load. The load would then become the meter. It would also be different, depending on the meter.

Ooops, yeah, I forgot about that......

[Andy.wpg]

but I have done this to run LED's and things before and it works. Since they are incans, the will only draw the current they need.

I reread this and I should clarify it. When using incans, they will only draw the current they need - NOT SO WITH LED'S!!! You still need to limit the current for LED's.

[Andy.wpg]

Wow, that would be very cool if that works. I was thinking (trying to remember back to my electronics theory classes in college) if the reactance needed the load to operate? Can you actualy measure it without a load?

Heh Heh, I just got the joke (intended or not?). We were talking about 'heatless voltage drop' and you said cool

[CaptKirk]

Not but I will take credit... It was a subconscious nod to the heatless voltage drop...

[Zeph]

I will be interested to hear how well this works.

I'm trying to recall if the capacitor will be accomplishing this by phase shifting the current vs voltage sine waves by creating a reactive vs purely resistive load, and if so what side effects that will have, beyond the less than unity power factor (watts vs volt-amps).

Is the ESR of the cap an issue?

I wouldn't be holding the cap when it first powers up, just in case there's a discrepancy between ideal theory and practical components.

[Andy.wpg]

I will be interested to hear how well this works.

I'm trying to recall if the capacitor will be accomplishing this by phase shifting the current vs voltage sine waves by creating a reactive vs purely resistive load, and if so what side effects that will have, beyond the less than unity power factor (watts vs volt-amps).

I was kinda wondering what the effect will be when PWM is used to to dim the channel - it may not work. I have never used it in this situation before.

I wouldn't be holding the cap when it first powers up, just in case there's a discrepancy between ideal theory and practical components.

Of course, the "proper" way to drop voltage is to use a transformer. And you are right, stand back when this thing first powers up!

[ErnieHorning]

I’ve never used just a capacitor only for this. I’ve used this method for low current where I only needed a few milliamps. In this case, I also add a low ohm resistor which lowers in in-rush current to the capacitor. This also has the benefit of lowering the potential of a catastrophic explosion if the capacitor should short.

I suppose there’s also a way to calculate a particular resistor so that as bulbs burn out, the current is progressively reduced because neither the resistor nor capacitor can supply the additional current. This would eliminate the ‘Nuclear Meltdown’ effect when too many bulbs burnout.

[CaptKirk]

So the resistor acts as a current limiter until it starts acting like a fuse?