# low pass filter in spwm h bridge

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?

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

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

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 H bridge is incapable of delivering 1000 watts efficiently

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

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.

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?