Capacitive Reactance (Output Voltage)

[Suraj143]

Reactance of a 0.22µF/400V capacitor running in the mains frequency 50Hz will be 14.4K.
Xc = 1/[2 X 3. 14 X F X C]

Adding to mains voltage the output current will be
I = 230 V / 14. 4 = 15.9 mA.

That means I can get 15.9mA out from 0.22uF capacitor.

My question is there is no any document what will be the output voltage on the capacitor!! How to calculate that?

 

[panic mode]

you just used 230V in your calculation. This is the voltage, is this what you are expecting?

 

[Suraj143]

Nop.

For example if I need to power up 7805 its maximum input voltage is 35V.So I still don't know whats the output voltage of the power capacitor ( that is still my question).

 

[panic mode]

so you want to use 0.22uF as series impedance to get low voltage out? with no isolation from mains? there must be also resistor in parallel with 0.22uF so when disconnected from mains, potentially high voltage across 0.22uF capacitor will discharge (for example 220k, 1/2W).

at mere 15-16mA, there is no need for 7805, why not use bridge rectifier, filtering capacitor and zener diode to clamp the voltage?
supposedly zener is 15V or so, then you can use SMPS to get 5V and higher current output (~45mA).

 

[Suraj143]

Lol you still do not get what I mean.

My question is whats the output voltage of the capacitor?Nothing else.

All other components I know & I have made lots of transformer less supplies.Of course they all have bridge rectifier,1Meg resistor,zenner & a small cap.

 

[Suraj143]

I know this is a tricky question, lots of people don't know this scenario & internet that is doesn't have many articles regarding output voltage on this capacitor.

 

[panic mode]

what do you mean by "output voltage"?

 

[Suraj143]

What is the voltage that is going to the bridge rectifier? (Voltage that will out from capacitor).

 

[Mithrandir2008]

@Suraj143

The way you are putting the question is confusing.

If you connect a 230V/50Hz supply across a 0.22uF capacitor, the equivalent resistance offered is 14.4Kohm. However, I don't understand why you say you can "get 15.9mA from a 0.22uF capacitor". It is necessarily the current flowing through it. The current you can "get" from a capacitor depends on the resistance you connect across it during discharge.

Back to your original question, the voltage across a practical capacitor can be known if you consider the Equivalent Series Resistance(ESR) of the capacitor. Add the ESR and the impedance to get the actual equivalent impedance. Then calculate the current and remove the drop across ESR to get the output voltage across the "true" capacitance. Note that this value will be not be very different as the ESR becomes significant only at very high frequencies, so for most practical purposes it is ignored(unless you are working for some ripple losses in power converters).

I have no idea if this is what you wanted to ask.

 

[Suraj143]

Please see my post 8 that will clear most of doubts

 

[Mithrandir2008]

By my understanding for this particular case, it is easier to imagine the entire circuit to be consisting of 2 parts. One is the capacitor and the other the equivalent load after shifting it to the AC side(diode bridge and all).

This reduces the circuit to a simple resistance divider network. You can easily figure the voltage drop by applying KVL. I can't recall exactly the shifting procedure, but I am sure it is available in a quite a few network theory textbooks.

Hope that helped.

 

[panic mode]

this cannot be answered until circuit is completely described. btw. this is what simulators (spice) are for.
as you have already estimated, line current would be approx 15mA when bridge is shorted (no output).
when bridge is not shorted, there will be some voltage drop across it (and some DC voltage at the output).
suppose zener is 5V, then at max current of 15mA, there will be voltage drop across R3:
100*0.015=1.5V
there will also be about 0.7V across conducting diodes (two at a time).
hence voltage drop across bridge is about:
5+1.5+2*0.7=8V voltage across C2 is expected to be about 5+1.5=6.5V.

if the zener was 12V then this drop would be about 15V. voltage across C2 is expected to be about 13.5V.


rated voltage of C2 should be higher of course, probably 2x what was estimated.

is this what you are after?

 

[Suraj143]

Hi panic mode nice calculations.You have done without taking any capacitor reactance formulas that is nice.

Your 5V zenner calculation is correct.

But 15V zenner ones I have a doubt.It should be
If the zener was 15V then this drop would be about 18V. voltage across C2 is expected to be about 16.5V.

Am I correct?

 

[panic mode]

correct... as said this is an estimate only. you should simulate it to get better result. for example we didn't reduce line current although impedance of the circuit has increased (voltage across .22uF capacitor is no longer 230V, it is about 10% less - so current will be also about 10% less or in the order of 14.5mA).
since the output current is so small, using linear regulator is not very useful. for example if you need 5V out, you can use 18V or 20V zener and add switching regulator like

https://www.mouser.com/ProductDetai...15-W36-C/?qs=uJpRT2lXVNWg5/3Kq428yerU/ojeO/N1

of the cheaper one as demonstrated by MrRB. switching regulators are more efficient and can convert voltage difference into current boost. that would approximately triple output current without increasing value of 0.22uF capacitor which is largest component in the circuit.

 

[unclejed613]

OH NO... it's the rebirth of the "hot chassis"..... your "ground" is floating at 110Vac from your house ground...

 

[panic mode]

OH NO... it's the rebirth of the "hot chassis"..... your "ground" is floating at 110Vac from your house ground...

floating yes, 110VAC no
(he is using 230VAC)
ground... well, it is floating
do NOT connect any part of circuit to chassis or anything else (must be sealed in insulated enclosure).

 

[MrAl]

What is the voltage that is going to the bridge rectifier? (Voltage that will out from capacitor).

Hi,

With the circuit you have posted, the 'output' from the capacitor will depend on the load. If the zener is clamping then the output will be the zener voltage plus the drop across the 100 ohm resistor plus two diode drops. If the zener is not clamping then the voltage will be the LED voltage plus the drop across the 1k resistor. If you assume 15ma for the current then you can approximate the drops across the resistors with V=I*R.

This is approximate and we could do a better calculation to get more accurate results, but the main idea is that the voltage you see out of the right side of the cap to ground will depend on the LED or zener voltage plus either the 100 ohm or 1k ohm voltage drop because that is what limits the cap voltage.

A couple notes:
1. The fuse should be a much lower value, 500ma max probably.
2. The cap used there should be a cap made for use in an AC circuit.

 

[crosslakeguy]

not sure what either? the output should be eqaul to the input voltage/ and capacitance= duration as understood if you have a 400v cap and charge to 4v you still get 4v and capacitor is charged up1/100th? so at 5% tolerance a 400v cap charged at 4v? the rating is at full voltage not 1% so what was the question, what is the voltage curve as related to charge at rated voltage, compared to lower voltages than rated?

 

[colin55]

I notice that no-one has knocked this request on the head.

When you place a capacitor in front of a bridge rectifier on 240v you are producing what is called a CONSTANT CURRENT GENERATOR or CONSTANT CURRENT CIRCUIT.
In your case the constant current is 15mA. This 15mA will flow through a short circuit, a LED or a 1k resistor.
If the load is removed, the output voltage will be 240 x 1.4 = 336v.
The actual current will depend on the voltage developed across the load.
When the voltage across the load reaches 100v, the current will drop to 10mA. When the voltage reaches 200v, the current will drop to 5mA. When the voltage reaches 330v, the current will drop to 0mA.

If you are just putting a capacitor on the 240v mains, the output voltage will be 336v peak.

 

[MrAl]

Hi,

Good observation Colin. The rule of thumb is roughly constant up to about one-third of the capacitors reactance. So for R from 0 to about 5000 ohms we can assume close to 16ma current. This makes the output voltage easier to calculate.
Output voltage roughly 0.016 times the resistance.
If there is an LED then we can assume the resistance is the voltage of the LED divided by 0.016, so for a 1.6v LED that means it will be about 100 ohms. If there is another 100 ohms in series with it, then the total voltage out of the cap is 3.2 volts plus two diode drops which means about 4.6 volts.
If there is a resistor in the other lead then that drop has to be considered too though. I'll leave that to the reader.