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I'm trying to obtain a sinusoidal signal (\$f_{sin}=50Hz\$) by using a ferrite transformer. I am going to apply a PWM signal at the primary side (\$100kHz\le f_{PWM}\le300kHz\$). I'm going to use one (or two?) H-bridges for driving the transformer. What kind of circuit topology should I use for applying this PWM?

I have three choices in my mind, but each one has its own flaw from my point of view.


Choice 1

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At this first scheme, PWM signal is applied in one respective direction for obtaining each positive and negative cycle of the sine wave. My concern is the DC bias on the windings that may saturate the transformer core or make it work near saturation point.


Choice 2

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The primary voltage swings between the positive and negative values. My concern is, lots of energy will swing (be wasted?) between the secondary windings and the LC filter during the zero-crossings of the sinusoidal wave since the duty cycle of the PWM will be around 0.5.


Choice 3

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Similar to the first one. I have to use two H-bridges and more window area of the transformer because of the extra primary windings.


Which of these topologies are problematic? Are there any other topologies commonly used? When to use each one?

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2 Answers 2

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There's no real difference between "choice 1" and "choice 3", except for the drive mechanism.

There is no real "DC bias", since it only lasts for a half-period of the output waveform. This is no worse than driving the transformer with an actual sinewave.

Note that the transformer needs to be constructed so that it can handle the power level at the output frequency, not the PWM frequency. This is one reason that UPSs in general have hefty power transformers inside them.

"Choice 2" is generally only used when waveform distortion is the primary concern, as in class-D audio power amplifiers. Otherwise, the other two configurations tend to be more efficient.

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  • \$\begingroup\$ I didn't understand the reason behind what you said in your third paragraph. The transformer will be transferring the high frequency PWM pulses, why does it have to be designed for the low frequency sine wave? Can you please give me a link for reading more about this? \$\endgroup\$ Apr 28, 2014 at 11:46
  • \$\begingroup\$ The transformer is handling a superposition of a number of signals. Yes, the PWM signal on the primary is one of them, but so is the output sine waveform on the secondary. One way to think of it is that the voltage waveform is the PWM signal created by the H-bridge, but the current waveform is the output sinewave. In other words, even when the PWM waveform is a 0 volts, there will still be a hefty current flowing through the transformer and the H-bridge (you can't just open-circuit it). The transformer needs to be able to handle all of this without saturating. \$\endgroup\$
    – Dave Tweed
    Apr 28, 2014 at 12:26
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Dave answered your question nicely, but I want to point out a additional issue with choice 3.

Choice 3 is a perfectly viable alternative, but the voltages are usually flipped around. The main advantage of driving a center-tapped primary is that both switches can be on the low side, which generally makes them easier to drive. For example, they can be NPN bipolar transistors or N-channel FETs with the emitter (bipolar) or source (FET) grounded. The center tap is then connected directly to the power supply.

The drawback is that the side of the primary that is not being driven will act like a secondary. Another way to think of this is that the primary winding works like a auto-transformer. The result is that the undriven end of the primary will go to twice the supply voltage. That means your low side switches have to be able to withstand twice the voltage than if a regular primary were driven with something like a H-bridge.

This higher voltage stress on the switches is one reason you see center-tapped primaries more when the input voltage is low. For example, if the power is coming from a 12 V battery, then the switches only need to handle around 30 V. There are many nice FETs to choose from with such a low voltage rating. However, if the power supply is rectified AC, that would be 170 V at least here in the US. The switches would have to withstand 340 V at least, so probably 400 V devices in practise. The available transistors will be fewer, have less desirable characteristics, and be more expensive. If the power supply is rectified line voltage anywhere in the world, then the voltage ratings of the switches becomes really inconvenient.

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