Capacitor question

experiMental

I have a circuit as follows:

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LED light (25 mA)

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[capacitor]

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[ GENERATOR ]

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*this is a representation of a circuit, all components connected in parallel

So I have a capacitor, rated at 0.1uF and a dynamo-based generator. For the circuit to work, should the capacitor be charged fully in a fraction of a second, or can it take any amount of time to charge it up, even as much as 30 seconds? Given the latter charging time, would the circuit still work?

DublinDilbert

experiMental wrote: »

would the circuit still work?

Have no idea, what's the circuit supposed to do?

What sort of generator are you using?

What's limiting the current flowing in the LED?

spideog7

If you presume the generator is a consistent power source then the capacitor will charge at a rate determined by the time constant RC, since you don't have any resistance in series with the cap then technically the time constant is very low.

But you don't have a resistor in series with your LED either so in short no it's not going to work. The capacitor doesn't do anything in the circuit and your LED doesn't do anything (except maybe burn).

Red Alert

You will need a resistor, about 5k ohms in series with the LED usually. Why is the capacitor there?

Gurgle

experiMental wrote: »

a dynamo-based generator

Need more info.

Probably AC from the dynamo, voltage ?

You'll want a resistor in series with the LED, how big depends on the LED's current rating, LED's voltage drop and your dynamo's max voltage.

Whats the intention of the capacitor?
If its for smoothing, then you'll want to rectify the voltage first. Maybe the dynamo has a rectifier built in?

Back to 'need more info'

backboiler

experiMental wrote: »

I have a circuit as follows:

|

---

LED light (25 mA)

---

|
|
|

---

[capacitor]

---

|
| |
|

---

[ GENERATOR ]

---

|


*this is a representation of a circuit, all components connected in parallel

So I have a capacitor, rated at 0.1uF and a dynamo-based generator. For the circuit to work, should the capacitor be charged fully in a fraction of a second, or can it take any amount of time to charge it up, even as much as 30 seconds? Given the latter charging time, would the circuit still work?

Assuming the generator's DC (dynamo usually means that but not strictly) and forward biasing the LED, what happens is that, at the point when the voltage across the capacitor reaches the forward voltage of the LED, practically all the current available from the generator diverts through the LED. The charging time depends on the internal resistance of the generator and the ESR of the cap at the frequency of the supplied waveform, assuming the resistance of the LED to be as good as infinite until it's biased on. The calculations aren't that complicated but seeing as you'll probably have a half-wave rectified output from the dynamo with a frequency and peak value dependent on the speed it's turning it's impossible to make an attempt but it's going to be a fraction of a second with likely values.

Since the generator will almost certainly be able to provide more than 25 mA (or whatever If(max) of the LED is) you'll see smoke from the LED fairly quickly. As someone above said you need a current-limiting resistor in series with the LED, value impossible to say. If you want it to work in the real world you'll need a fatter capacitor, a thousand times your value above at least.