Electronic – Voltage drop pulse transformer

buckpwmtransformer

I'm trying to design a buck converter controlled by a PWM signal. The PWM signal and the converter is isolated with a pulse transformer.

My issue is that the pulse transformer behaves differently than expected.

This is the schematics for the pulse transformer part

We use the B82804A0354A110 and the IXDN602SI as driver.

Before we added the pulse transformer to the circuit we measured the output of the IXDN602SI:

This looks good and the pulse are 12V just as specified by the driver.

Now we add the Pulse transformer to the circuit and measured the output of the IXDN602SI again, which is the input to the pulse transformer primary winding:

We also measured the voltage Vgate:

So, the transformer seems to be working as the input is the same as the output (its a 1:1:1 transformer).

The question I have is, why is there a voltage drop? Why is the output voltage of the transformer not 12V pulses?

The pulse frequency is about 1.6 MHz.

Hope you can help.

UPDATE:

As per the answer by @andy aka, I just wanted to show the entire buck converter and explain the reason for my question. The reason I measured the the transformer on its own, is because when I add it to my circuit then the output of the line driver is just 0V and I'm having a hard time understanding why.

This is my schematics and as you can see we do have a gate driver on the secondary side of the pulse transformer.

But my suspection is that there is something wrong with the gate driver, since the entire PWM circuit is driven to 0V. Initially I expected the transformer to be faulty and therefore tested it on its own.

Best Answer

You cannot simply pass DC through a transformer without repercussions. The primary winding is basically an inductor and any DC level is dragged to ground so, the DC level coming out of your driver is facing a big hurdle but coping enough. However, the voltage it is now producing is seriously reduced and, if the pulse transformer had a lower primary DC resistance it would be even more compromised.

The only option you have is to AC couple the driver to the pulse transformer primary but this then exposes a problem on the secondary; if the pulse width is 50% then the output will be equally positive and negative but the positive peak will be 50% of the former amplitude without the pulse transformer. If the duty cycle is a lot lower than 50% then the positive peaks will be closer to the former amplitude but, if the duty is high the positive peaks will be not enough to activate the mosfet.

This means you have to do two things: -

AC couple the primary as previously explained
Restore the DC on the secondary using diodes like this: -

D1 and C2 act to keep the voltage at D1's cathode not less than about 1 volt negative and the other components provide an active clamp to discharge the MOSFET's gate capacitance when the output does fall below 0 volts.

It's never clear-cut with an isolated MOSFET driver.

Related Solutions

Buck not acting as designed

(Voltage drop across inductor) = 17.37V Vload {Voltage at the load) = 0V

Inductors are capable of being destroyed and having 17.37 volts across it totally indicates that it has failed open circuit. This means you will see no voltage on the output and it will likely affect how your bootstrapping operates.

Transformer works on PWM at 50Hz

I looked at the code reference and, as the comments say, it's going to generate a kind of sawtooth - not a sinewave.

void loop() {
    for(up=-1; up <= 255;up++) {
        analogWrite(out,up);
        if(up==255) {
            up=-1;
        }
    }
    delay(20);
}

What you are expecting is the blue sine-wave. What that Arduino code will give is a short sawtooth (ramping up to 255 and back to -1) followed by 20ms of nothing. It won't even give 50 Hz correctly.

If you did get a sine wave out it would give you a 2.5 V DC output with a 5 V peak-to-peak sinewave on top of it. The DC will bias the transformer which expects the input voltage to be symetrical about 0 V (and not 2.5 V).
5 V p-p = 1.7 V RMS so even if you got rid of your DC you still have a very low voltage to work with.
(You think) you have \$5V \times 1A = 5 W\$ to work with but \$5 V_{p-p} = \frac {5}{2 \cdot \sqrt 2} = 1.7 V_{rms}\$ and \$1 A \cdot 1.7 V_{rms} \to 1.7 W\$. Be aware that even if you got this to work efficiently and stepped up the voltage to 120 V the absolute maximum output would still be 1.7 W which is about 14 mA at 120 V.

I suspect that you hope to use your Arduino to power something that would normally run at 50 Hz. I think the answer is "no".

What you are trying to make is an inverter. Have a look on the web for inverter designs. You will find some very cheap designs but they will have terrible non-sine waveforms and poor voltage regulation. Anything better will require specialist and more expensive components.

Edit after Arduino code update.

void loop() {
    for(up=0; up <= 255;up++) {
        analogWrite(out,up);
        delay(20);
    }
    for(down=255; down>=0;down--) {
        analogWrite(out,down);
        delay(20);
    }
}

Your updated code will give an output like this.

The updated code result.

Now your code now steps up by 1 step and waits 20 ms before doing the next step. It will take \$255 \times 20ms = 5.1s\$ to get to 255 and another 5.1 s to come back down to zero. That's one full cycle in 10.2 s or < 0.1 Hz.

No it won't work in a transformer.
As stated, even if it worked and was 100% efficient it could only give 14 mA. You're looking for 44 mA so there is no chance of it working.
It's still not AC. It's a rising and falling DC. The current never changes direction.

It still has a DC offset in it.

Best Answer

Related Solutions

Buck not acting as designed

Transformer works on PWM at 50Hz

Related Topic