I have a problem with the filtering stage of an spwm H bridge inverter,i have spwm signal 25khz and fundamental of 50hz.i make a low pass filter with a 1,5mh inductor in series an capacitor 2 μf parallel.All its perfect the signal is ok the voltage 220 volt ok..when i connect a load in the output the voltage drops dramaticaly,and the efficiency for 1000w inverter is very bad..! (i thing current can`t pass from the inductor) what is the wrong?

Thank you for your answer's.it's an 12v-220v inverter .i have build a 340v dc-dc pwm stage with feedback for dc-dc bus,this stage has 4 of 250w hf feritte transformers.I connect the positive from the dc bus to the bridge for the ac stage.All wires are copper,4 wires of 1,5mm all striped together from the bus to the bridge,all negative conections are together.I connect the load normaly in Vout of bridge after the cap,as above.Earlier i run this iverter with square wave signal 50hz in the bridge(for no need filter)with a very bad and old battery(only 50Ah!)and works fine in 500 watts(i dont try bigger load,and the reason is the small battery size).Voltage was stable,the load was lamp's,and two 250 watt home motor's and all work's fine.After that i decide to make the signal spwm in the bridge for sinewave.Now after the filter choke and capacitor i have problem even at very small load such a lamp of 25w! my filter is exacly like the first picture above and my load connected to the V out.My filterS choke is a inductor when i find in an other machine,not too big but not too small,just like a ring with your finger,i measure it with lc meter and is 1,5mh, with yellow color feritte ring,with 2 of 1,5mm copper together,and a non polarize cap of 2 μf.My expected efficiency is at least 70% in full load,for that type of inverter.It has very small no load current and almost zero losses in first small loads.(i meter that with ampmeters in output an input,with small loads and with square wave signal earlier. Now even a small 25w lamp can't bright fast and full the problem is to the filter,but what is the most possible senario for this fenomeno,thank you.

NOTE,the size of core in the choke is like a ring of my finger.

  • 2
    \$\begingroup\$ Schematic or it didn’t happen. \$\endgroup\$
    – winny
    Commented Aug 3, 2018 at 22:18
  • \$\begingroup\$ The inductor needs to be small enough to pass the Di/dt that the load needs \$\endgroup\$
    – user16222
    Commented Aug 4, 2018 at 8:32

2 Answers 2


Your inductor value and capacitor value seem OK (resonant at 2.9 kHz) and loading with 48.4 ohms (1000 watts on 220 VAC) should not be a problem.

But the devil is in the detail: -

  • Your inductor has too much series resistance
  • Your inductor is saturating
  • Your H bridge is incapable of delivering 1000 watts efficiently
  • Your H bridge bus voltage is drooping under load

Here's a bode plot so you can see it should be OK: -

enter image description here

enter image description here

The red curve tells you that the frequency response is flat in the 50 Hz area and that you get "as-expected" roll off of the PWM artefacts above 3 kHz.

Picture source.

With perfect components there should be very little attenuation at 50 Hz


What does this output filter need todo?

  1. it needs to attenuate the switching frequency
  2. it needs to pass the load current.

From the information you have provided:

  1. Inverter is SPWM
  2. Switching frequency is 25kHz
  3. fundamental is 50Hz
  4. Output voltage is 220AC
  5. Output "power" is 1kW

You have realised this out of 1.5mH:2uF

as AndyAKA has stated, conceptually it should be fine

\$f_o = \frac{1}{2\pi\sqrt{LC}}\$ = 2.905kHz.

and at 50Hz there should be no attenuation.

However... this filter can be realised out of an infinite combination of L and C to produce such a cutoff frequency and herein lies the problem. This L could be too big; inductors are referred to as chokes for a reason as they choke the change in current.

You mention SPWM and this has a DClink utilisation of around 50% but without a statement about your DClink voltage or how you are deriving it, not a lot more can be derived (basically your output voltage could be lower thus your load current could be higher adding to the problem).

You state an output power of 1kW... for now I am assuming you are driving into a purely resistive load and thus VA = W (in practice you will need to consider DPF and PF in the efficiency).

At 1kW and 220V, the load current is 4.5Arms

The maximum currrent at maximum frequency (50Hz) is thus:

\$i(2\pi50\times t) = \sqrt{2}\times4.5\times sin(2\pi 50 \times t) \$

The maximum rate of change of current is thus:

\$2\sqrt{2}\pi 50 \times 4.5= 0.001999A/\mu s\$

If we assume the filter capacitor will tend to short the load at 25kHz and thus zero voltage occurs at the maximum di/dt

\$ \hat{L} = \frac{\check{V}}{2\sqrt{2}\hat{f}\times\check{I}} \$

\$ = \frac{220}{0.001999A/\mu s} = 110mH \$.

This is the MAXIMUM usable inductor upto 25kHz, for such a load change. you wouldn't want this size due to practicality but also there is about -1.7dB of attenuation IF the filter was set at 3kHz as you have.

so from a current throughput you should also be fine.

So questions...

  1. What have you exactly built?
  2. What exactly is your load (resistive?)
  3. what sort of cable length are you connecting to?
  4. Have you done something silly and done EARTH referenced loading?
  5. What efficiency are you measuring
  6. How are you measuring and calculating the efficiency
  7. What is your expecting efficiency

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