# Capacitor to drop AC

Status
Not open for further replies.

#### premkumar9

##### Member
Hi,
In the attached circuit LED represents a series parallel combination of LEDs with series resistances which require 12V, 300mA. To drive it from full wave rectified 230V AC (applied between A abd B), can I use a series capacitor to control the current through my LED load? If so, how to select the type and value of the capacitor ?

#### Attachments

• 3.2 KB Views: 650

#### Diver300

##### Well-Known Member
Most Helpful Member
You can't use a capacitor to limit current once it has been rectified.

What you can do is put a capacitor in series with AC to limit the current, and rectify afterwards. It won't isolate from the mains, you need a capacitor rated to the full mains voltage (1.4142 times the rms) and the current depends on the supply voltage, frequency and the capacitor tolerance.

However, 12V and 300 mA is a lot for this sort of power supply. Also, the current though the capacitor will not be reduced with the addition of a small resistor, so you would be better to put all the LEDs in series so that the overall current is less. It is a good idea to put some resistance in series with the capacitor to reduce the in-rush current.

The impedance of the capacitor is 1/ωC where ω is 2πF so it is 314 for 50 Hz

For 300mA the capacitor needs to have an impedance of 240/.3 = 800Ω

So C needs to be about 4 µF. You can get capacitors of that size, rated to 250 V or more. They are called "motor start capacitors".

However, they are larger and more expensive than a small transformer that would supply 300 mA at 12 V.

I wouldn't use a series capacitor as a current limit unless the current is 50 mA or less.

#### Attachments

• 6.6 KB Views: 413

#### Mr RB

##### Well-Known Member
If you put the leds in parallel reversed, you can do away with all the diodes...

And just use a cap and a resistor to drive them.

#### MrAl

##### Well-Known Member
Most Helpful Member
Hello there,

I have to agree that 300ma is a little high for a series cap power supply solution.
It would be better to use a dc wall wart.
Maybe you can find large caps on sale somewhere online, but you have to also
take measures to limit the turn on surge current which is quite high with a
series cap power supply like that.
You can get surplus dc wall warts pretty cheap that will put out 12v or more and
300ma.

#### premkumar9

##### Member
You can't use a capacitor to limit current once it has been rectified.

What you can do is put a capacitor in series with AC to limit the current, and rectify afterwards. It won't isolate from the mains, you need a capacitor rated to the full mains voltage (1.4142 times the rms) and the current depends on the supply voltage, frequency and the capacitor tolerance.

However, 12V and 300 mA is a lot for this sort of power supply. Also, the current though the capacitor will not be reduced with the addition of a small resistor, so you would be better to put all the LEDs in series so that the overall current is less. It is a good idea to put some resistance in series with the capacitor to reduce the in-rush current.

The impedance of the capacitor is 1/ωC where ω is 2πF so it is 314 for 50 Hz

For 300mA the capacitor needs to have an impedance of 240/.3 = 800Ω

So C needs to be about 4 µF. You can get capacitors of that size, rated to 250 V or more. They are called "motor start capacitors".

However, they are larger and more expensive than a small transformer that would supply 300 mA at 12 V.

I wouldn't use a series capacitor as a current limit unless the current is 50 mA or less.
Thank you for the information. Based on this I will try to work out a solution for my requirement and come back.

#### Willbe

##### New Member
You need to put a diode across the LED to protect the LED from reverse breakdown during the negative half cycle of the AC waveform.

#### premkumar9

##### Member
You need to put a diode across the LED to protect the LED from reverse breakdown during the negative half cycle of the AC waveform.
That is not required when we use the full wave rectifier. Right?

#### Willbe

##### New Member
That is not required when we use the full wave rectifier. Right?
Right; I missed that.

#### Mr RB

##### Well-Known Member
Or when you put the leds in parallel reversed.. Which I already said.

#### premkumar9

##### Member
Or when you put the leds in parallel reversed.. Which I already said.
Can you please elaborate a little bit?

#### Sceadwian

##### Banned
If you put the LED's in two separate strings facing the opposite direction they'll act like a rectifier, one string will protect the other from reverse over voltage.

#### Mike_2545

##### Super Moderator
The formula for capacitance reactance is Xc=1/2piFC.
So for 5v, 300mA from 120V 60 Hz AC, you would need a 6.9 mF high voltage cap to do the job.

Or for 12v at 300mA from 230 v 50 Hz Ac you will need a 4.38 mF high voltage cap to do the job.

Last edited:

#### Willbe

##### New Member
The 200v bipolar caps that I used blew out; use 400v caps.

#### MrAl

##### Well-Known Member
Most Helpful Member
Hi,

I have to agree that 7uf is about the right size for the cap for 5v and 300ma,
but the input surge is going to be high without some series resistance like
100 ohms.

They used to make oil filled capacitors for AC with high values like this,
but they used to be around 20 dollars each! Sometimes you can find
higher value AC caps in old microwave ovens though.

I'd go with a wall wart myself.

Last edited:

#### Mike_2545

##### Super Moderator
The OP was asking "can I use a series capacitor to control the current through my LED load? If so, how to select the type and value of the capacitor ?"

If he wanted someone to design a circuit for him, I am sure he would have asked.

Also the voltage of the capacitor should be 1.4 times greater than stated. So 230 Vac requires a 322v or higher to be on the safe side.

Status
Not open for further replies.

Loading