Capacitance or Inductance

[I'mClueless]

Hello,

On the DC side of bridge rectifier which should be first capacitance or inductance?

Yep,
I'mClueless

 

[ljcox]

Normally only a capacitor is used to filter the "DC".

The size of the cap depends upon the current drawn, the source & diode resistances and the magnitude of the ripple that you can live with.

 

[I'mClueless]

ljcox,

Why then do all of the online Capacitance calculators factor in Capacitance, Frequency and Inductance?

Yep,
I'mClueless

 

[Hero999]

Sometimes inductors are used but only on really high current loads (e.g. a welder) from memory I think the inductor is usually connected before the capacitor.

EDIT:

As a general rule use the following formula:

C = (10×I)/Ripple

Where I is the current in mA.
Ripple is the maximum acceptable ripple current.
C is the capacitance in µA

This basic formula only works for 50/60Hz mains frequencies and oversizes capacitors slightly which isn't a problem.

 

[ljcox]

ljcox,

Why then do all of the online Capacitance calculators factor in Capacitance, Frequency and Inductance?

Yep,
I'mClueless

I don't know as I've never looked at the site.

I use the tables that I used as a student.

I agree with Hero999.

There would be some situations where you do need inductance - but not under normal circumstances.

A welder is a possibility as you may want to reduce the transient current from the mains each time the arc is struck. A series inductance would do this.

 

[I'mClueless]

Hero999, ljcox,

Thank you both for your input. The application is on the DC side of a welder. The line current has the potential of 270 Amps / 100+ Volts with up to 600Hz voltage ripple.

Yep,
I'mClueless

 

[bychon]

Online calculators provide the OPTION to use inductors. 4 instance, If they didn't include that, and you needed an inductor, how would you make the calculator work?

as for ripple, √2 C Er F = I

The square root of 2 times capacitance times peak to peak ripple voltage times frequency equals current. This is THE dead accurate formula.

Adding inductors will require a different formula, which I don't know.

 

[ljcox]

I've never designed a welder so I'd be guessing.

 

[crutschow]

Adding a inductor to a rectifier filter will average the current and reduce the peak current pulses from the transformer, thus reducing the IR losses. An inductor will also reduce the ripple. But for cost and weight reasons they are seldom used with smaller power supplies.

If they are used, they are placed in front of the capacitor filter.

 

[Hero999]

An inductor will also cause a large voltage drop but this isn't a problem for a welder.

As it's DC what about using a biased choke to avoid core saturation?

The idea is you stick a powerful magnet to one side of the core to provide a DC field opposite to that created by the current. This should enable you to reduce the core size by half. Just make sure the magnet is powerful enough, otherwise you might repolarise it, maybe a neodymium hard drive magnet is the answer.

 

[Nigel Goodwin]

Inductors (chokes) were used back in the very old valve days to reduce ripple, decent size electrolytics weren't available bcak then, so they had to use an inductor as well.

Using an inductor as well does make a much better filter, but the higer costs involved mean it's cheaper to use modern large capacitors.

 

[Gary B]

It’s been a long time since I messed around in this area but, the choice of capacitive input filter of inductive input filter is the choice of the power supply engineer. The only practical difference is the output of a capacitive input filter is equal to the peak input voltage from the rectifier while the voltage of an inductive input filter will be the RMS equivalent.

 

[Hero999]

The only practical difference is the output of a capacitive input filter is equal to the peak input voltage from the rectifier while the voltage of an inductive input filter will be the RMS equivalent.

That's only true when fully loaded, off load the voltage is equal to the peak, just like a capacitive filter.

 

[Nigel Goodwin]

That's only true when fully loaded, off load the voltage is equal to the peak, just like a capacitive filter.

This is all pretty irrelevant for smoothing purposes, the LC network is a low-pass filter to remove the mains hum and give a smooth DC. It's normally a PI filter, capacitor, inductor, capacitor - and as I said before, the inductor was used because large high voltage capacitors weren't available.

 

[crutschow]

The only practical difference is the output of a capacitive input filter is equal to the peak input voltage from the rectifier while the voltage of an inductive input filter will be the RMS equivalent.

It's a nit, but the output of a loaded filter inductor is the Average of the AC waveform, not the RMS value.

 

[Gary B]

Crutschow, sorry but I must disagree. The average of any symmetric AC wave is going to be zero. The RMS value is defined as the DC equivalent. That is why I stated it that way so all waveforms would be covered including non-sine wave sources like the output of an inverter.

The inductive input filter is really a current filter rather than a voltage filter. All filters have some built in loading in the form of a safety resister to discharge the filter capacitors when power is shut off. In the case of the inductive input filter, it will also define the minimum current in the choke.

 

[Hero999]

He's talking about the waveform after the rectifier, not before.

 

[crutschow]

Crutschow, sorry but I must disagree. The average of any symmetric AC wave is going to be zero. The RMS value is defined as the DC equivalent. That is why I stated it that way so all waveforms would be covered including non-sine wave sources like the output of an inverter.

Of course, I was referring to the output of the rectifier. And the inductor does generate the average of the input voltage. This is true for any waveform including the output of the switch in a switching regulator. The average of a full-wave rectified sine-wave is .626Vp (2/Pi).