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UCC28180: Power supply from boostrap

Benjamin Loyer
Prodigy 10 points
Part Number: UCC28180
Other Parts Discussed in Thread: TIDA-00779

Hello,

I tried to find an answer on the internet but wasn’t able to. So here I am !

I was trying to find a way to supply my ucc28180 and a convenient way would be to use a bootstrap from the main line. However, it is mentionned in the datasheet (tldr: you can’t):

The TIDA-00779 design requires an external bias supply to power the UCC28180 PFC controller
UCC27531D gate driver, and relay, which is used to shunt the inrush current limiting resistor.
TI recommends powering these devices from a regulated auxiliary supply. These devices are not intended
to be used from a bootstrap bias supply. A bootstrap bias supply is fed from the input high voltage through
a resistor with sufficient capacitance on the VCC pin to hold the voltage on the VCC pin until the current
can be supplied from a bias winding on the boost inductor.
The UCC28180 has a UVLO of 11.5 V and the UCC27531D has a UVLO of 4.5 V, whereas the minimum
voltage required to turn on the relay is 9.6 V (for a 12-V relay), so the bias voltage for board operation
must be ≥ 12 V. The total current required for these devices is approximately 55 mA

(TIDUBE1C)

Could you help me please to understand why ? I suspect a question of power dissipation through the bootstrap components or maybe a question of voltage stability ? 

Thank you for your time.

Benjamin

  • +1
    TI__Mastermind 32500 points

    Hello Benjamin, 

    Thank you for your interest in the UCC28180 PFC controller. 

    Your suspicion of power dissipation is correct.  
    Using a boot-strap network to provide 55mA of bias current represents a (usually) unacceptable continuous power loss.

    Consider that a PFC usually runs from 85Vac to 265Vac, with rectified peaks at 120V to 375V.
    To obtain 55mA at low line, you'd need Rboot < (120V-12V)/0.055A = 1964R.  Power loss here is V^2/R = (108V)^2/1964 = 5.94W.  
    At high line, this same resistance must sustain (375-12)^2/1964 = 67W!  and boost strap current rises to 185mA, so you have to do something with that extra 130mA (like dump it into a zener diode clamp). 

    That is why this is usually not done.  It can be done, but doesn't make economic sense. 

    You can embellish the bootstrap with a shut-off circuit that disconnects the bias from the high-voltage rail to save all that loss (except during start-up).
    But that circuit needs to be very reliable.  If it fails, maybe a fuse or PTC can be used to limit loss and avoid damage.

    Usually, a PFC stage is not used alone, but powers a follow-on DC-DC converter.  This converter also needs bias power and a separate bias supply serves to power both the PFC and the DC-DC stages as well as all ancillary circuitry.  It is usually most efficient and economical to do it this way. 

    Regards,
    Ulrich 

  • 0 in reply to Ulrich Goerke
    Prodigy 10 points

    Hello Ulrich,

    Thank you very much for your complete answer.

    Regards,

    Benjamin