hi, can someone help me to clarify some doubts? I wanted to know the relationship between the capacitance of the filter capacitors after a rectifier bridge and the amount of maximum current that the rectifier diodes can accept. let me explain: I know that after the bridge, if the capacitance of the capacitors is too small there is too high a ripple, if instead it is too large, you can have a peak current at ignition that is too high which could damage the bridge. so given a certain voltage, what capacity to use ... what is an acceptable ripple (1v?)? (however using a downstream regulator)
Things like wiring resistance and fuses and so on make a massive difference to peak currents, as do differences in ESR of the caps.
Most diodes specify surge current ratings and these are magnitudes more than the normal steady state currents.
An acceptable ripple is one that is comfortably (but not excessively) to low (yes low) for the application. Adding lots of capacitance (if genuinely needed to solve some issue) means the basic design is poor in things like ripple rejections.
To much capacitance can also give the mains transformer a hard time as it needs to deliver more current over less time (over each cycle).
Most diodes specify surge current ratings and these are magnitudes more than the normal steady state currents.
An acceptable ripple is one that is comfortably (but not excessively) to low (yes low) for the application. Adding lots of capacitance (if genuinely needed to solve some issue) means the basic design is poor in things like ripple rejections.
To much capacitance can also give the mains transformer a hard time as it needs to deliver more current over less time (over each cycle).
Diodes are chosen on several parameters -voltage-current-speed etc and its on those parameters filter capacitors are chosen .
As I keep saying --solid state diodes are not perfect --valve/tube rectifier present a more perfect rectifier .
Capacitor Smoothing Circuits & Calculations >> Electronics Notes
https://www.eeeguide.com/half-wave-rectifier-with-capacitor-filter/
power supply - Full wave rectifier capacitance calculation - Electrical Engineering Stack Exchange
As I keep saying --solid state diodes are not perfect --valve/tube rectifier present a more perfect rectifier .
Capacitor Smoothing Circuits & Calculations >> Electronics Notes
https://www.eeeguide.com/half-wave-rectifier-with-capacitor-filter/
power supply - Full wave rectifier capacitance calculation - Electrical Engineering Stack Exchange
The current is limited by the series resistance of the transformer primary and secondary windings mainly. For toroidals those are usually very low, meaning very high peak currents.
At switch on the capacitor is effectively just its ESR, the capacitive impedance is not really relevant to the instantaneous inrush current as it will be at zero volts at that instant.
If the primary is switched on at a mains zero-crossing the pulse will be much less as the cap can charge as the voltage builds up, if the primary is energized at the peak of the mains waveform the inrush current will be a maximum.
A soft-start circuit or thermistor can be used to reduce the inrush spike.
At switch on the capacitor is effectively just its ESR, the capacitive impedance is not really relevant to the instantaneous inrush current as it will be at zero volts at that instant.
If the primary is switched on at a mains zero-crossing the pulse will be much less as the cap can charge as the voltage builds up, if the primary is energized at the peak of the mains waveform the inrush current will be a maximum.
A soft-start circuit or thermistor can be used to reduce the inrush spike.
You are mainly designing for the load on the power supply, not the transformer itself. Even if you can supply 10A, the concern is actually the 2A load. There is generally a correlation between uF and ripple ratings on capacitors, however, recently I stated in another thread that I use enough capacitance at where the ripple ratings of the capacitors add up to 2x my supply capacity. The uF, while still important, is not as much a concern as the ripple ratings. If you have a load that can draw 2A, I would design for 4A. In the case of low voltage power supplies, one can easily crank up the uF and ripple usually with not much impact on the cost of doing so. At high voltages, the capacitors get expensive and much larger the bigger the capacitance and there needs to be more attention to the cost to content ratio.
You are designing for your load and desired ripple with capacitance. The rectifier is designed generally with the transformer potentials.
You chose the capacitor related to your load and the allowed ripple.
Rule of thumb: 1A of load discharges a 1000uF capacitor about 1V between charges. So that's 1V ripple.
So if the load is 1A and you allow 0.25V ripple you need 4000uF. Very rough rule of thumb, but gets you in the ballpark.
Jan
The real answer today is to buy diodes *much* bigger than your transformer.
5A transformer, 10A or 20A diodes.
There's several foolish approximations here. But diode=transformer served me well in early days when even good gear came with undersize diodes. Today diode prices are SO cheap that there's no reason to cut it too close.
5A transformer, 10A or 20A diodes.
There's several foolish approximations here. But diode=transformer served me well in early days when even good gear came with undersize diodes. Today diode prices are SO cheap that there's no reason to cut it too close.
thank you all for your contribution! Well, I have to power a pico psu of a mini pc for audio use.
Being a pc the load is variable, from about 1A to 2.2A while loading windows.
The regulator is a 7.5A mic29752 and the transformer has two secondaries each of 5.33A, 15V.
I was thinking of using 10 x 3300uF capacitors to obtain a ripple of 0.66V calculated on 2.2A.
Calculating 2.2A as the rated current (but in reality it is less), 0.6V as ripple and 21V as voltage (15V x 1.414 ca.) I calculated a peak current (Ifsm) of about 28A. the formula I used is: I (fsm) = (3.14x2.2A) / (2x0.6V/ 21V) ^.
^ = square root.
The diodes are 10A schottky vishay mbr1045 with 150A peak current IFsm.
Being a pc the load is variable, from about 1A to 2.2A while loading windows.
The regulator is a 7.5A mic29752 and the transformer has two secondaries each of 5.33A, 15V.
I was thinking of using 10 x 3300uF capacitors to obtain a ripple of 0.66V calculated on 2.2A.
Calculating 2.2A as the rated current (but in reality it is less), 0.6V as ripple and 21V as voltage (15V x 1.414 ca.) I calculated a peak current (Ifsm) of about 28A. the formula I used is: I (fsm) = (3.14x2.2A) / (2x0.6V/ 21V) ^.
^ = square root.
The diodes are 10A schottky vishay mbr1045 with 150A peak current IFsm.
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Are you talking about inrush current?
Average ripple current is about 3 x I(load).
Yes, with 33 000 uF and 2.2 A you'll get about 0.5V ripple amplitude (about 18VDC).
But it seems to me you may have it better and easier if you use 10 000 uF x 0.47R 5W x 10 000 uF CRC-filter.
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