• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Subscribers 63 subscribers
  • Views 635 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

UCC28C45: Lower start up current & operating current than UC2845

Howard Zou
Genius 12365 points
Part Number: UCC28C45
Other Parts Discussed in Thread: UC2845,

Hi,

In the datasheet, it says that "The reduced start-up current of this device minimizes steady state power dissipation in the startup resistor, and the low operating current maximizes efficiency while running, increasing the total circuit efficiency"

1. Both start up current & operating current refer to the current input to VDD which will go through Rstart, right?

2. Start up current is the current input to VDD during the start up phase of the device and won't last long, so it won't generate much heat, right? Then lower start up current won't reduce the heat much because it's not the major source of the heat, right?

3. For operating current, I believe it's related to the drive current which will apply to external MOSFET. So how can it be lower? I mean if the external MOSFET need 1A drive current, no matter we use UCC28C45 or UC2845, the operating current should be almost the same, right? So it won't affect the efficiency greatly, maybe only a little?

  • 0
    TI__Mastermind 34465 points

    Hello Howard,

    (1) Startup current always goes through Rstart . Operating current will only go through Rstart if a bootstrap winding is not used in the circuit.
    A bootstrap winding is shown in fig 20 of the UCC28C45 datasheet.

    (2) Startup current will generate significant power loss in Rstart if the input voltage is high. For example a 240Vac rectified source has a peak voltage of 338Vdc.
    UCC28C45 requires only 50uA of startup current and UC2845 requires 500uA of startup current.

    (3) If you operate the UCC2845 form a VDD=15V then (neglecting MOSFET drive current) the internal power dissipation is 15*2.3 = 34.5mW and if you use the UC2845 the power loss would be 15*11 = 165mW. If you can save 130mW of power this is a significant amount of power in an 8 pin ic.
    The drive losses associated with the MOSFET will add to this power loss as you say. If these drive losses become significant then it is best to use an external gate driver

    Regards

    John

  • 0 in reply to John Griffin1
    TI__Genius 12365 points
    In datasheet it says "Low Operating Current of 2.3 mA at 52 kHz".
    I wonder what's the situation for 2.3mA. For different drive current, I guess this value should change.
  • 0 in reply to Howard Zou
    TI__Mastermind 34465 points
    The test conditions were with a 52kHz oscillator and VDD=15V.
    If you increase the switching frequency then you can expect the operating current to increase also simply because there is more switching cycles per second.
    The power loss due to driving the FETs is equal to 0.5*C*F*V^^2
    where C is the input capacitance of each FET.
    F is the switching frequency
    V is the value of VDD
    You should add this number to the 15*2.3 mW in the previous reply if you want to account for drive losses
    Regards
    John