Capacitive reactance - Page 1
Forum section: electronics / beginners

author: SimplyElectronics
author_rank: Contributor
timestamp: August 31, 2016, 06:44:32 pm
content: Hi guys,I'm okay with capacitive reactance in terms of theory, and understanding how to calculate the reactance of a cap in an AC circuit.So I'm okay with the Xc = 1/2?FC and can correctly calculate the reactance of the cap using this equation.There are a few elements of this that I am struggling to understand though.1. How do I calculate the required cap value if I already have a reactance value in mind? i.e. what cap value should I choose if my AC is 240V 50Hz and I want to get 20mA from it.2. When you calculate reactance, you get the result in ohms, which means I can the take AC voltage and divide it by the reactance value to give me the output current. But, what if I want to power an LED and I know it requires 3V but I don't know it's current draw.If the capacitor is dropping voltage, how do I understand what voltage the cap is dropping? i.e. I want to drop 247 volts across a cap to give me a remaining 3v for my LED.Ultimately, I want to be able to understand how to choose a capacitor when I know what voltage or current I need to step-down to.I am very sorry for my confusing post. I am just very enthusiastic about being able to understand this.Thank You to anyone who has the time to help me

author: IanB
author_rank: Super Contributor
timestamp: August 31, 2016, 06:57:01 pm
content: Quote from: SimplyElectronics on August 31, 2016, 06:44:32 pmYou just rearrange the equation. For example, to calculate the reactance you have:$$X_c={1 \over 2\pi FC}$$Therefore to calculate \$C\$ you would rearrange the formula to:$$C={1 \over 2\pi FX_c}$$QuoteBut this is not ever the case. An LED is always calculated according to the current (e.g. 10 mA) and not the voltage.QuoteDo not worry about the voltage drop (of the LED) in this case. It is so tiny it doesn't matter.QuoteJust plug the current of 10 mA into the formula above and that will give you a capacitor value. But remember that the mains is dangerous and you would need to use proper mains rated capacitors. You should also have a fuse in the circuit in case of any failures.

author: SimplyElectronics
author_rank: Contributor
timestamp: August 31, 2016, 07:03:51 pm
content: Thank You for your swift and helpful replyI will give the equation a try.Would full AC voltage effectively be supplied at the LED?

author: IanB
author_rank: Super Contributor
timestamp: August 31, 2016, 07:12:32 pm
content: No, it's the other way around (I edited my post above to clarify).There is minimal voltage across the LED. The full mains voltage appears across the capacitor, which is why it must have a suitable voltage rating.

author: SimplyElectronics
author_rank: Contributor
timestamp: August 31, 2016, 07:20:28 pm
content: Quote from: IanB on August 31, 2016, 07:12:32 pmSo this means that if I am supplying 20mA to the LED, and at 20mA the led has a 3V drop, this would mean that the cap would be dropping 237V across it? (Just to clarify in my own head)

author: IanB
author_rank: Super Contributor
timestamp: August 31, 2016, 08:22:11 pm
content: Quote from: SimplyElectronics on August 31, 2016, 07:20:28 pmYes. In fact, such capacitors are often called "dropper capacitors" for this reason.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 31, 2016, 09:01:32 pm
content: Not quite: the capacitor would be dropping very nearly 240V. The voltages add vectorially -- well, assuming the LED acts as a resistor, which really isn't the case (and by the way, you need to use two LEDs back-to-back, or a full-wave rectifier, otherwise the LED will be destroyed). When the voltages are at 90 degree phase to each other (as is reasonably the case here),\$ V_{in} = \sqrt{V_{cap}^2 + V_{load}^2} \$Another catch is that, although a capacitor will do a fine job under steady-state AC conditions, the real world is not steady. Most important of all: when you close a switch, the voltage very quickly (very quickly indeed, even for electronics: within fractions of a nanosecond, sometimes!) rises, which means using the other (more general) capacitor equation,\$ I = C \frac{dV}{dt} \$the dV is, potentially, as high as the peak line voltage (~340V), and dt is potentially nanoseconds. Clearly, the current could be monstrously high!The solution is to add a current-limiting resistor in series with the capacitor, so that not all the voltage is dropped by the capacitor alone. This reduces efficiency (the resistor dissipates power, whereas the capacitor draws reactive current only), but limits the peak current to safe levels.Tim

author: IanB
author_rank: Super Contributor
timestamp: August 31, 2016, 09:06:02 pm
content: To the OP: Have a look at some of Big Clive's videos on YouTube, especially the ones where he takes LED lights apart. He shows lots of examples of circuits using capacitor droppers. You will quickly see how such circuits should be arranged.https://www.youtube.com/user/bigclivedotcom

author: IanB
author_rank: Super Contributor
timestamp: September 02, 2016, 03:26:10 am
content: Oh, look; see this video here:https://youtu.be/IYiItZRUsQUA really nice explanation.