this is the design for the circuit I'm just not really sure how I would go about finding the capacitance or the resistance. I've looked around for the answer but I can't really find anything. if you could explain it as simple as possible that would be great too cause I'm new to electronic design.

 

You need to have an expression for the ripple voltage. A large C along with a large R will be good for minimizing ripple. The smaller R is in parallel with the load on the supply, the bigger C will need to be. the product of R and C will give you the discharge time for no load. This needs to be much larger than 1/2 cycle of the line voltage.

 

The resistance is not part of the rectifier circuit, it is the load resistance, so what load do you want to power?

The capacitance then depends upon the load and how much ripple the load can tolerate.

 

Choosing the right component depends on the results you want. For instance, if you're building a simple amplifier, a small capacitor may let a lot of AC hum leak through into your audio. A larger capacitor will control more of it. However, if you go to extremes you may get rid of all the AC hum but now you're building with a huge cap that is not needed, and the inrush of current when turned on can damage your rectifiers. There's more to it than I know, so I only offer this as just a little bit of an explanation to hopefully shed some understanding on matters. With ANY rectifier and capacitor there will ALWAYS be ripple. The Load will pull so much current and the capacitor will be there to fill in the gaps. Stay below the ripple and you'll have a clean signal.

 

The load resistor is there only to calculate how much ripple voltage will be present.
You don't choose the load resistor. This will be determined by what you intend to power with your power supply.

To calculate the ripple voltage, have a look at this:
http://www.zen22142.zen.co.uk/Design/dcpsu.htm

For 50Hz AC LINE frequency,

Ripple Voltage = 10 x I / C

where I is current in mA
C is capacitance in μF

 

Hello there
If all that seems too confusing just try different values of smoothing capacitor and load resistance (impedance) in your circuit to see the effects on the output waveform. That's the fun part.
This is the point where the fun begins . A rule of thumb, we are looking to have a ripple voltage of less than 100mV peak to peak.
That's it, simple right?
Have fun!

 

BTW, in your schematic, the ground symbol should go in the junction of the D2 and D3 anodes.

 

The load resistor is there only to calculate how much ripple voltage will be present.
You don't choose the load resistor. This will be determined by what you intend to power with your power supply.

To calculate the ripple voltage, have a look at this:
http://www.zen22142.zen.co.uk/Design/dcpsu.htm

For 50Hz AC LINE frequency,

Ripple Voltage = 10 x I / C

where I is current in mA
C is capacitance in μF

this website actually helped a lot thank you, once more thing though. I changed the input to be the same as the Uk mains with 240rms and 50Hz what types of diodes should I be looking for that? also how would i find what type of power rating the load would need to have so it doesn't get destroyed

 

how would i find what type of power rating the load would need to have so it doesn't get destroyed

The load only takes the power it needs so you don't have to worry about the supply having "too much power".
So you need to know the load you are driving and the power it requires (either its rating on the nameplate or from calculations), which determines the minimum power the supply must generate.

 

The load only takes the power it needs so you don't have to worry about the supply having "too much power".
So you need to know the load you are driving and the power it requires (either its rating on the nameplate or from calculations), which determines the minimum power the supply must generate.

this is my current circuit and when I connect a multimeter to the load it shows 2.5kA are being drawn which makes me think something is wrong and if it's not wrong how would I find the diodes to withstand the current?

 

1) How do you connect a multimeter to a load to measure current?
Thank goodness this is only a simulation. In real life you will have let out the magic smoke and blow your ammeter.

2) As this is a simulation, connecting 240Vrms 50Hz will do no harm on paper. In real life you can kill yourself. So think twice before you do this for real. We will always recommend that you use an isolation transformer between you and AC MAINS.

 

when I connect a multimeter to the load it shows 2.5kA are being drawn which makes me think something is wrong

You likely incorrectly connected the ammeter in parallel with the load instead of correctly in series.

 

You likely incorrectly connected the ammeter in parallel with the load instead of correctly in series.

yeahh i was just being an idiot one more question. the diodes in the circuit i showed have a forward voltage drop of 1.1 V but for a rectifier with an input of 5Vrms and 100kHz that seems like a big drop you wouldn't happen to know a different type of diode you could recommend