Calculating Capacitor Value for Transformerless Power Supply Design

[seemasdawange]

I designed one transformerless power supply and did lots of calculations for that. can anyone answer me how to calculate value of capacitor using as a smoothing capacitor parallel to zener.and also tell me if the value of capacitor is larger than required then what changes may occur in the actual circuit.?

 

[Mark veniaminavich]

As I know the higher UF of the capacitor the smoother your power supply will run at the expense of starting time, which means that your power supply will need a slightly longer ( one to two seconds ) time to start working properly, also the input and the output matter of the whole system matter.Always safety first

 

[seemasdawange]

As I know the higher UF of the capacitor the smoother your power supply will run at the expense of starting time, which means that your power supply will need a slightly longer ( one to two seconds ) time to start working properly, also the input and the output matter of the whole system matter.Always safety first

If i use the higher UF capacitor then reactance of the capacitor becomes low and causes higher current to flow through it. Isn't that make any major change in the circuit? in that case is our circuit operate as per our requirement?

 

[NascentOxygen]

I designed one transformerless power supply and did lots of calculations for that. can anyone answer me how to calculate value of capacitor using as a smoothing capacitor parallel to zener.

The smoothing capacitance is determined by the maximum load current and the tolerable ripple voltage. There is a formula for this.

and also tell me if the value of capacitor is larger than required then what changes may occur in the actual circuit.?

A larger capacitance may cause a larger inrush current at switch-on. This can put a strain on the rectifiers, and the fuse—and also the on/off switch.

 

[seemasdawange]

The smoothing capacitance is determined by the maximum load current and the tolerable ripple voltage. There is a formula for this.

A larger capacitance may cause a larger inrush current at switch-on. This can put a strain on the rectifiers, and the fuse—and also the on/off switch.

Thanks dats what I'm expecting.

 

[berkeman]

I designed one transformerless power supply and did lots of calculations for that. can anyone answer me how to calculate value of capacitor using as a smoothing capacitor parallel to zener.and also tell me if the value of capacitor is larger than required then what changes may occur in the actual circuit.?

Can you post a copy of your schematic? That would help us to answer any questions that you have. There are lots of ways of making a transformerless power supply. Use the UPLOAD button to attach a copy of your schematic to a post. Thanks.

 

[seemasdawange]

Can you post a copy of your schematic? That would help us to answer any questions that you have. There are lots of ways of making a transformerless power supply. Use the UPLOAD button to attach a copy of your schematic to a post. Thanks.

Here is the schematic.. further circuit is not here. input supply is 230V AC 50Hz as per our country specifications.
Simulation is done in ISIS Proteus 8.0

 

[NascentOxygen]

That series 2kΩ resistor means you can disregard the caution I wrote earlier about capacitor inrush currents at switch-on.

 

[seemasdawange]

That series 2kΩ resistor means you can disregard the caution I wrote earlier about capacitor inrush currents at switch-on.

okay. so what should i do? remove it from series? actually this resister also has power loss in form of heat

 

[NascentOxygen]

Well, what was your reason for placing it there?

 

[seemasdawange]

Well, what was your reason for placing it there?

To limit the current. if voltage goes too high in that case it may protect capacitor by limiting the current

 

[berkeman]

Here is the schematic..

Yikes. I see several problems here. Where did you get this circuit? Can you link to the original source? Overall, this circuit makes no sense, IMO.

For example, you have a 50Hz AC Mains source trying to charge a 20uF capacitor through a 470nF capacitor. Do you see any issue with this (and there is the series resistor problem adding to this problem)...?

 

[NascentOxygen]

Oh, I see—you included it because I mentioned surge current? But you can't make it an arbitrary value—the current through that path becomes the current for the DC side of your power supply. Because you haven't shown what circuits this will be supplying, I can't guess how much DC current it needs to supply. Have you estimated the DC current your loads will draw?

 

[seemasdawange]

Yikes. I see several problems here. Where did you get this circuit? Can you link to the original source? Overall, this circuit makes no sense, IMO.

For example, you have a 50Hz AC Mains source trying to charge a 20uF capacitor through a 470nF capacitor. Do you see any issue with this (and there is the series resistor problem adding to this problem)...?

No I tried to design it by myself. i read from some source to add series resistance there http://www.designercircuits.com/DesignNote1a.pdf check this pdf u may get the answer.
thankx for reply

 

[seemasdawange]

Oh, I see—you included it because I mentioned surge current? But you can't make it an arbitrary value—the current through that path becomes the current for the DC side of your power supply. Because you haven't shown what circuits this will be supplying, I can't guess how much DC current it needs to supply. Have you estimated the DC current your loads will draw?

yes of course. my load is a relay of 12 Volts and have power ratings of 0.36Watts. that means load current is 30mAmps

 

[seemasdawange]

No I tried to design it by myself. i read from some source to add series resistance there http://www.designercircuits.com/DesignNote1a.pdf check this pdf u may get the answer.
thankx for reply

check this image. and look at the output waveform.

Is it ok? or should i make some changes here also?

 

[NascentOxygen]

Before you go further, you do know that modules are available to do all this, more efficiently and probably much more cheaply than you can build it yourself. To wit, that AC capacitor needs mains voltage ratings (600V at least, but I would use 1000V because I know failure will be expensive), so that cap alone will be costly.
Google search, or check out: http://www.digikey.com.au/product-detail/en/cui-inc/PBK-3-12/102-3108-ND/4332645

 

[seemasdawange]

Before you go further, you do know that modules are available to do all this, more efficiently and probably much more cheaply than you can build it yourself. To wit, that AC capacitor needs mains voltage ratings (600V at least, but I would use 1000V because I know failure will be expensive), so that cap alone will be costly.
Google search, or check out: http://www.digikey.com.au/product-detail/en/cui-inc/PBK-3-12/102-3108-ND/4332645

Hmm, u r ryt. can u suggest any module den?

 

[seemasdawange]

Yikes. I see several problems here. Where did you get this circuit? Can you link to the original source? Overall, this circuit makes no sense, IMO.

For example, you have a 50Hz AC Mains source trying to charge a 20uF capacitor through a 470nF capacitor. Do you see any issue with this (and there is the series resistor problem adding to this problem)...?

You may not understood previous diagram as there was transistor and battery of 12Volt shown. the simplified diagram is here. you can check this. is that ok? or still have some problems?

 

[NascentOxygen]

I can't recommend a particular one. Does the one I linked to have suitable ratings? Or search on google, amazon, ebay, or mouser.

 

[seemasdawange]

I can't recommend a particular one. Does the one I linked to have suitable ratings? Or search on google, amazon, ebay, or mouser.

Ok Thanx

 

[berkeman]

You may not understood previous diagram as there was transistor and battery of 12Volt shown. the simplified diagram is here. you can check this. is that ok? or still have some problems?
View attachment 106342

No, it still has errors and makes no sense. Why are you insisting on putting a capacitor in series with the AC Mains input? I can think of no reason in this circuit to do that. And you appear now to have deleted the input storage capacitor. Why?

 

[davenn]

No, it still has errors and makes no sense. Why are you insisting in putting a capacitor in series with the AC Mains input? I can think of no reason in this circuit to do that. And you appear now to have deleted the input storage capacitor. Why?

indeed ...
It would seem that the OP shouldn't be playing with mains voltages as they appear to lack the knowledge to do it correctly and probably not in a safe way

 

[NascentOxygen]

Why are you insisting on putting a capacitor in series with the AC Mains input?

The capacitor is there as a low-heat (i.e., low energy-loss) voltage-dropping impedance. This is standard design of this type of low-power transformerless power supply. A voltage-dropping resistor can be used instead, but this is substituting one problem with another: dealing with the heat that resistor produces.

The design itself is a compromise, and is not suitable for applications where a user could come in contact with any part of the power supply or the devices it powers. Opto-isolation is desirable in achieving this.

 

[berkeman]

The capacitor is there as a low-heat (i.e., low energy-loss) voltage-dropping impedance. This is standard design of this type of low-power transformerless power supply. A voltage-dropping resistor can be used instead, but this is substituting one problem with another: dealing with the heat that resistor produces.

The design itself is a compromise, and is not suitable for applications where a user could come in contact with any part of the power supply or the devices it powers. Opto-isolation is desirable in achieving this.

Thanks NO. At my work, we actually publish a reference design for low-power non-isolated applications that uses capacitor isolation, so I probably should have recognized that. Still, the design seems to have some issues. I wonder if the OP understands what they are trying to do. It would be good to see some specifications for the design from them...

 

[nsaspook]

The capacitor is there as a low-heat (i.e., low energy-loss) voltage-dropping impedance. This is standard design of this type of low-power transformerless power supply. A voltage-dropping resistor can be used instead, but this is substituting one problem with another: dealing with the heat that resistor produces.

Correct, here's a capacitor app note for that type of application: http://www.vishay.com/docs/28153/anaccaps.pdf

I personally think his transformerless power supply with the battery is extremely dangerous.