Diode full bridge Rectifier; inductor to reduce current ripple (on AC or DC side)

Hi

Basically, increasing the capacitor value reduces the output DC voltage ripple.
But than input AC current (rms and peak) and current through DC capacitors increases. To reduce this inductor is use.

From simulation point of view there is no difference whether I put same value inductor on AC or DC side (equal rms current \[\to\] equal power dissipation).
However, are there any practical limitations or (dis)advantages when putting inductor on AC or DC side?

Hi,

typically a rectifier followed by a capacitor is used to supply a load with electrical energy.
This load usually specifies
* the allowed input voltage range ( maybe ripple, too)
* and the input current.

These are the target values to design your circuit.

Each load has other requirements, therefore we can´t recommend a universal solution.

--> for a detailed answer we need the specifications.

************

From simulation point of view there is no difference whether I put same value inductor on AC or DC side (equal rms current equal power dissipation).

Click to expand...

Not true.
With an inductor on the input side you will reduce the input ripple current, but lower the (loaded) capacitor voltage.
With an inductor at the output side you may influence the voltage to the load in a way that the load will refuse to work. (depends on load). But you won´t decrease input ripple current significantely.

Please tell what "power dissipation" you talk about. (load, capacitor, inductor, rectifier...)

****
You talk about a simulation:

I assume you used a constant load current or constant load resitance... this often is not realistic.
--> Just to see the difference: use a varying load_current switching form 50mA to 100mA with 1kHz and measure the voltage across the load. Then you will see how much it differs when inductor is on the one or other side.

****
An inductor on the DC side sometimes is not useful to reduce the ripple current.

Klaus

KlausST said:

typically a rectifier followed by a capacitor is used to supply a load with electrical energy.
This load usually specifies
* the allowed input voltage range ( maybe ripple, too)
* and the input current.

These are the target values to design your circuit.

Each load has other requirements, therefore we can´t recommend a universal solution.

Click to expand...

Rectifier is only the first stage.
Main AC outlet (230VAC) rectified to 200VDC, than step up converter to 700VDC and inverter for 3×400VAC, 3kW). So, for the first stage I need constant DC voltage.

But If I add capacitor to reduce ripple, input current increases (for this case from 40 to 200Arms!!).

KlausST said:

With an inductor on the input side you will reduce the input ripple current, but lower the (loaded) capacitor voltage.

Click to expand...

This is not a problem since step-up converter will in the next stage increase DC voltage.

KlausST said:

Please tell what "power dissipation" you talk about. (load, capacitor, inductor, rectifier...)

Click to expand...

I² through inductor is the same.

KlausST said:

I assume you used a constant load current or constant load resitance... this often is not realistic.

Click to expand...

I used ideal resistor as a load.

Hi,

there are so many issues in your post.. I find it hard where to start from.

Main AC outlet (230VAC) rectified to 200VDC

Click to expand...

230VAC RMS rectified without capacitor will give 230V RMS.
...but with capacitor and no laod you will see 325V DC
...with capacitor and with load it will be somewhere inbetween, but usually more close to 325V.

How do you come to 200V DC?

******

Rectifier is only the first stage.
Main AC outlet (230VAC) rectified to 200VDC, than step up converter to 700VDC and inverter for 3×400VAC, 3kW). So, for the first stage I need constant DC voltage.

Click to expand...

and

This is not a problem since step-up converter will in the next stage increase DC voltage.

Click to expand...

One time you say you need constant voltage, next time you say this is no problem.... --> I don´t understand.

******

But If I add capacitor to reduce ripple, input current increases (for this case from 40 to 200Arms!!).

Click to expand...

No way.
Maybe the peak current goes from 40A to 200A, but not RMS. (At least not with same output conditions)

******

I² through inductor is the same.

Click to expand...

Please provide more information.
(If ripple current increases, the RMS current will increase also, so is is not the same and I^2 is not the same...At least with same output conditions)

******

I used ideal resistor as a load.

Click to expand...

I recommend to use realistic simulations...otherwise a simulation is not useful.

Klaus

KlausST said:

How do you come to 200V DC?

Click to expand...

Sorry. My mistake. I used wrong parameters.
It is:
without inductor: Vout = 318 V, Iin = 41 A (peak 225 A) (η = 35 %)
with inducotor: Vout = 277, In = 18 A (peak 38 A) (η = 62 %)

KlausST said:

One time you say you need constant voltage, next time you say this is no problem.... --> I don´t understand.

Click to expand...

Constant, as in, low ripple DC value voltage.
Ideally, I would like to get as high voltage as I can get, but then I have high current (exceeds the current rating of the circuit breaker!) and efficiency is low.

So basically, my question is: how to achieve high-efficient passive AC/DC conversion?
And, is increasing an inductor value (and subsequently reducing the output voltage) the only way?

What I gather that you are attempting to do, is to create a 400 volt, 3 phase converter from a single phase 230 volt outlet....correct?
Then in your simulation, you are finding out that the peak input currents are excessive, and want to tame them with an inductor input....correct?

You are not going to achieve this with a passive filter input. You require an active power factor correction circuit. The advantage to this approach is that you step up the voltage and correct the power factor in a single stage.
Although because of the high power involved, it will have to be an interleaved PFC circuit.

Hi,

without inductor: Vout = 318 V, Iin = 41 A (peak 225 A) (η = 35 %)
with inducotor: Vout = 277, In = 18 A (peak 38 A) (η = 62 %)

Click to expand...

How did you calculate η?
It is not efficiency.

Efficiency is P_out / P_In.
With or without inductor efficiency should be be well above 95%.

*****

(exceeds the current rating of the circuit breaker!)

Click to expand...

Usually thermal circuit breakers react on RMS_current.
Magnetic circuit breakers will react on I^2*t.

You should not use high value inductors, it makes the output voltage weak.
With medium value inductors you decrease current peaks, decrease EMI...

For high efficiency look for low DC_resistance inductors. Let´s say you want disspated power to be less than 5W with a 20A RMS load, then look for 10W / (20A)^2 = 25 mOhms DC resistance.
(230V x 20A = 4600W. So the loss in inductance is only about 0.2%)

Klaus

schmitt trigger said:

What I gather that you are attempting to do, is to create a 400 volt, 3 phase converter from a single phase 230 volt outlet....correct?

Click to expand...

Yes. But in multiple stages.

schmitt trigger said:

Then in your simulation, you are finding out that the peak input currents are excessive, and want to tame them with an inductor input....correct?

Click to expand...

Yes... Stuck on the 1st stage :smile:

schmitt trigger said:

You are not going to achieve this with a passive filter input.

Click to expand...

Actually, I am more worried about how the step-up converter (2nd stage) and DC/AC inverter (3rd stage) will turn out. For now I only need DC voltage; less active components - better. And, since it is going to be supplied from LV AC outlet, good efficiency would be nice!

- - - Updated - - -

KlausST said:

How did you calculate η?
It is not efficiency.

Efficiency is P_out / P_In.
With or without inductor efficiency should be be well above 95%.

Click to expand...

without inductor:
Vin = 230 Vrms, Iin = 41 Arms \[\to\] Pin = 9430 W
Vout = 318 V, Rout = 30 Ω \[\to\] Pout = Vout²/Rout = 3371 W
π = 35 %

with:
Vin = 230 Vrms, Iin = 17 A \[\to\] Pin = 3910 W
Vout = 277 V, Rout = 30 Ω \[\to\] Pout = Vout²/Rout = 2558 W
π = 65 %

Hi,

two issues:

You have a 1 Ohms resistor in series.. it surely dissipates power.

DC power calculation:
U_average x I_average is a good assumption. Even better: integral (I x U)

AC power calculateion:
U_RMS x I_RMS x cos(phi).
--> You forgot cos(phi)

Imagine: If you have 9430W input and 3371W output then the dissipated power is about 6000W. This is more than heating power of my kitchen oven with stove.
I hope you see it makes no sense this way.

Klaus

Added:
U_RMS x I_RMS x cos(phi). ..

another thaught: Maybe this is only true for sinusoidal waveforms.
For pulsed / arbitrary waveforms you may need to use: integral (I x U)

With an inductor on the input side you will reduce the input ripple current, but lower the (loaded) capacitor voltage.
With an inductor at the output side you may influence the voltage to the load in a way that the load will refuse to work. (depends on load). But you won´t decrease input ripple current significantly.

Click to expand...

Can't confirm the statement. A classical rectifier circuit which was very popular before the age of switched mode power supplies uses a relative large inductor between bridge rectifier (either full bridge or three-phase bridge) and output capacitor. With sufficient large inductor value, the input current is a square wave, not a sine, but the rms value is much lower than without an inductor. Output voltage is load independent above a certain load current. A series inductor on the AC side reduces the peak current but above a certain value, it also creates an unwanted load dependent voltage drop. It's rarely extended above 5 or 6 X percent.

Hi,

I agree in terms..

You talk about: AC --> rectifier --> inductance --> capacitor
But here the situation is: AC --> inductance --> rectifier --> capacitor

In the first case the inductance see´s DC current and thus is low impedance for DC currents, but attenuates ripple current. (I agree it tends to be square wave form)
In the second case there is no DC current through the inductance. Not that low impedance DC path. The current will not become square wave shaped.

Klaus