SN6501: Question on the SW current during the dead time

Hans Zhang

Part Number: SN6501
Other Parts Discussed in Thread: SN6507

Tool/software:

Hi TI,

I am using the SN6501 IC. The following are the testing waveforms of VSW1, VSW2, and ISW1/2 under no load condition and approximately 4Vin. The testing currents should be the magnetizing currents.

My question is that why there are several mA currents holding during the dead time. There seems a big capacitor between the SW1/2 and GND. What is the value of this capacitor? Or is there any other explanations on this behaviour? In my opinion, the magnetizing current during the dead time should be maintained at a low value.

Looking forward for your reply.

Thanks,

Hans

over 1 year ago

0 Andrew J over 1 year ago

TI__Mastermind 31815 points

Hello Hans,

Thank you for reaching out. The 20mA of current that you are seeing during the dead time is expected and it's the result of the switch's body capacitance.

Best,
Andrew

0 Hans Zhang over 1 year ago in reply to Andrew J

Prodigy 70 points

Hi Andrew,

Thanks for your timely reply. As far as I am concerned, the body capacitance of an internal MOSFET can be just several pF. According to ic = C*dv/dt, the body capacitance should be larger from the waveforms. Could you give me some principle explanations?

Best regards,

Hans Zhang

0 Andrew J over 1 year ago in reply to Hans Zhang

TI__Mastermind 31815 points

Hello Hans,

The internal MOSFET is designed to handle the SN6507's normal application. Unfortunately, I cannot provide further design details.

Best,
Andrew

0 Hans Zhang over 1 year ago in reply to Andrew J

Prodigy 70 points

Hi Andrew,

Thank you for your reply. Let me add one more question.

The current during the dead time is the following currnet of the magnetizing current or the charging/discharging of the Cds capacitor? I just think the capacitor is large.

Thanks,

Hans Zhang

0 Andrew J over 1 year ago in reply to Hans Zhang

TI__Mastermind 31815 points

Hello Hans,

The peak current is because primary inductance of the transformer and the on-time duration. After that the current drops during the dead time. The rate of drop will depend on the body capacitance and that is what we see happening here.

Best,
Andrew

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SN6501: Question on the SW current during the dead time