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Current rating Problem in battery charging in sine wave inverter

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Mithun_K_Das

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Today I was trying to add the battery charging mechanism in my sine wave inverter. As per the theory, I kept off the upper two MOSFETs and only PWM-ed lower two MOSFETs together at 7KHz with duty cycle 4-20% for testing. What I found that, the ampere rating in the winding of the transformer is 16Amp but on the other hand battery is charging at 2.5Amp rating.

I don't think it is normal. Isn't it a power loss in transformer? What is the suggestion?
 

does the 16A in transformer is the rating specification or a measured one?

if i take it tobe 16A measured (max), then at 4-20 % duty cycle , it may average to 2 to 3A (not calculated but only approximation.)
 

Capture.JPG

It will help you to understand. The transform is giving 16Amp but I'm getting only 2.6A in battery side. rest of amps are dissipated among MOSFETs (?).
 

PWM-ed lower two MOSFETs together at 7KHz with duty cycle 4-20% for testing
"Together" means what it says? Switching-on left and right low side simultaneously? 4% duty cycle might be already too much, depending on transformer leakage inductance.

How did you measure currents? Did you look at current waveforms?
 

I followed some good reference designs. According to them it should work like this. But no one there measured as I measured. That's why I was asking if I'm wrong or its ok what I'm getting.
 

The circuit is working as boost converter, respectively input will be higher than output current. If you measure RMS or even peak (you still didn't tell), there will be an additional form factor. Additionally you get MOSFET switching losses.

If you can't measure current waveforms, you would want to check which parts are heating up.
 

I'm going to make a question: How and with what type of instrument did you measure the current?

Some measurement equipment will only measure sinewaves or pure DC.....if your waveform is chopped, there may be a measurement error.
 

The circuit is working as boost converter, respectively input will be higher than output current. If you measure RMS or even peak (you still didn't tell), there will be an additional form factor. Additionally you get MOSFET switching losses.

If you can't measure current waveforms, you would want to check which parts are heating up.


OK, so its normal in sine wave inverter. I was confused if I'm having any problem. I'll try again soon with that circuit.

- - - Updated - - -

I'm going to make a question: How and with what type of instrument did you measure the current?

Some measurement equipment will only measure sinewaves or pure DC.....if your waveform is chopped, there may be a measurement error.

I used a clip on meter to measure the current.
 

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