1
\$\begingroup\$

I am working on a power electronic application. I am using push-pull converter topology.

enter image description here

I am using IRF630 power MOSFET to do switching. I also use primary side center tapped transformer Np/Ns = 1/13 ratio. I am using IR2113 MOSFET driver. This is my schematic. I am getting HIN and LIN from STM32F407VG.

enter image description here

These are the inputs which I am getting from STM32F407VG.

enter image description here

There is also dead time to prevent same time turn off/on for MOSFETs.

In this application my purpose was getting 230 V sine wave. This signal is what I got. By the way oscilloscope probe was in 10x. So I am assuming that signal was 316 V.

enter image description here

What could be the mistake?

\$\endgroup\$
7
  • 3
    \$\begingroup\$ What is the problem? Why would you expect sine wave on the secondary side? \$\endgroup\$
    – winny
    Jun 11, 2021 at 12:47
  • \$\begingroup\$ I thought i would be getting sine wave with oscillation between +E and -E(in the first schematic) just like dc to ac inverting. \$\endgroup\$
    – Bowman
    Jun 11, 2021 at 13:02
  • \$\begingroup\$ Did you analyze the impedance ratio and mutual coupling load effects? 15^2 \$\endgroup\$ Jun 11, 2021 at 13:04
  • \$\begingroup\$ @TonyStewartEE75 No, i did not. I am using a transformer which was used in a high frequency application. \$\endgroup\$
    – Bowman
    Jun 11, 2021 at 13:09
  • 1
    \$\begingroup\$ Your circuit will give you an approximate square wave, not a sine wave. \$\endgroup\$ Jun 11, 2021 at 13:52

3 Answers 3

2
\$\begingroup\$

IR2113 is a bootstrapped high and low side (half H-bridge) driver. Your MOSFETs are both low-side. Consider using low-side drivers instead. These are basically inexpensive high current buffers, to push enough current into the gate to switch quickly.

But it may still work.

Here's the internal schematic of the driver:

enter image description here

The top driver is normally bootstrapped, but it will probably work if it is used as a low side driver. However, to turn the top FET on, it uses the bootstrap cap (between VB and VS, C1 on your schematic) as power supply to push current into the gate. For this to work, pin VS must be connected to the FET source:

enter image description here

...otherwise the current has nowhere to go. But on your schematic, VS is not connected to the FET source, so it can't work.

If this driver can work with both channels low-side, you could replace D1 with a short, since the diode is only useful if you need a bootstrap supply for a high-side drive, which is not the case. So shorting the diode will power the driver directly. Then, connect pin VS to the source of the FET Q2, ie GND.

If this doesn't work, then you need a pair of low-side drivers.

\$\endgroup\$
2
  • \$\begingroup\$ LO and HO are connected to both MOSFETs gate pin. I checked their voltage value. Both are 17.20 V . Is not it enough ? \$\endgroup\$
    – Bowman
    Jun 11, 2021 at 13:25
  • \$\begingroup\$ Oh, sorry, I mixed up the 5V VCC and the 15V drive voltage. But looking at the schematic again, I spotted the problem. Will edit. \$\endgroup\$
    – bobflux
    Jun 11, 2021 at 14:13
0
\$\begingroup\$

IR2113 is used for a push-pull topology as an half bridge.

If you use digital level MOSFET, you can drive directly your switches. IRLZ44N (55v-47A) may be suitable or better STP40NF10L (100v-40A).

The output will be a DC voltage in your initial schematic.

Be careful with the connection to the earth.

\$\endgroup\$
2
  • \$\begingroup\$ Should i change MOSFET ? To be honest i don't have experience about digital level MOSFETs. Can you give me advice to use which one ? \$\endgroup\$
    – Bowman
    Jun 11, 2021 at 13:04
  • \$\begingroup\$ Search "logic level MOSFET, with a standing VDS of minimum the double of your 15V supply (50 V or 100V, price the lower) , current ad-hoc as you will. IRLZ44N may be suitable. \$\endgroup\$
    – user288518
    Jun 11, 2021 at 13:10
0
\$\begingroup\$

can you please explain me why my circuit won't give sine wave ?

Your circuit is typical of circuits where MOSFETs are used as switches. In such circuits, the mosfet is either ON or OFF. When the upper mosfet is ON, current flows through the center tap of the transformer, through the upper turns, through the mosfet, then to ground. When the lower mosfet is ON, current flows through the center tap of the transformer, through the lower turns, through the mosfet, then to ground. The voltage across the respective transformer turns will be the better part of the supply voltage. The emf induced in the secondary is proportional to the back emf induced in the primary, which is close to the voltage applied to the primary. So, the output voltage at the secondary will be approximately a square wave.

A common way to get a sine wave power output is to use pulse-width modulation, where the pulses are at a much higher frequency than the desired sine wave, and the pulse widths vary according to the desired sine wave frequency. Then the high frequency component is filtered out leaving the sine wave.

\$\endgroup\$
2
  • \$\begingroup\$ Thank you for making it clear for me. I am looking for a topology to get sine wave from secondary of transformer. I am currently searching microcontroller based dc to ac converters. What would you suggest me to do to get sine wave from transformer's secondary side? \$\endgroup\$
    – Bowman
    Jun 15, 2021 at 14:44
  • \$\begingroup\$ Added explanation of pulse width modulation technique of making power sine waves. \$\endgroup\$ Jun 15, 2021 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.