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i am making bidirectional buck-boost converter. The schismatic of bidirectional buck boost converter is attached here.

Buck Mode = when PWM is applied to Q1 , it work as buck converter. when q1 is OFF D2 will conduct. Q1 and D2 work for Buck converter. power flow is forward.

Boost Mode= when PWM is applied to Q2, it work as boost converter. so Q2 and D1 work for Boost converter. power flow will be opposite. capacitor and load will be connected to V1 voltage side.

i am generating 5v PWM fron function generator and giving it to gate of MOSFET through Gate Driver IR2110. i am facing some problems that are listed below

1) during Boost Mode , when i only use low side MOSFET Q2 and D1 , i achieved 96% efficiency without connecting Q1 high side mosfet. whenever i connect Q1, its efficency goes down. Q1 was heated up. can you tell me what is problem. why Q1 was heated up? why it conducts. i applies its gate to gound. there are some pulses that turns on the Q1 mosfet. it should never turn on during Boost mode.

2) Similarly the case with Q2 , during buck mode. during boost mode , the gate of Q2 is grounded.


A problem that very probably you are having is in the switches' commutation. You need to let a little "dead time" (i.e. both switches off) between the turning off of a switch and the turning on of the other, this is because real transistors dont shut off instantaneously, so during a little time it can happen that one switch is already on, but the other has not turned off yet, so you get a momentary short circuit on every commutation cycle, which can increase switching looses, which explain the temperature rise.

Microcontrollers with pwm generaly have this option to create a dead time. If you want to keep using the signal generator you need to generate the "dead time" by other means (a 555 and some RCs comes to my mind).

"Buck" topology is a different one from "Buck-Boost" topology. I believe you mean "Bidirectional Buck Converter", please edit your question.

  • \$\begingroup\$ thanks @berto .. is there is nay losses of inducter during the dead time. ? how will be the behavior of inductor ? \$\endgroup\$ – Misal313 Mar 29 '16 at 18:31
  • \$\begingroup\$ As you know, the inductor current can not be changed instantaneously, so this current will go by the anti-parallel diodes while the transistor is off. If my answer helped please mark as answered or vote ;). \$\endgroup\$ – berto Mar 29 '16 at 18:34
  • \$\begingroup\$ when the duty cycle is 50% , the efficiency of boost is 95 % . but i increase the duty 70 % , there is a significant increase in input current and decrease in output power. why this happen ? \$\endgroup\$ – Misal313 Mar 29 '16 at 18:35
  • \$\begingroup\$ what if i remove the parallel diode ? what will happen to the inductor current during dead time ? \$\endgroup\$ – Misal313 Mar 29 '16 at 18:37
  • 1
    \$\begingroup\$ @Misal313 Please only mark a question as answered after a few hours, other people might be writing answers as well. The reason why connecting the low side MOSFET causes the MOSFET to heat up is because your diode has such a high forward voltage that the inductor current flows trough the MOSFET body diode instead. Removing the parallel diode would not do anything as long as the low side MOSFET in parallel with it is still connected. \$\endgroup\$ – jms Mar 29 '16 at 18:39

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