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UCC2804 Note Question

Other Parts Discussed in Thread: UCC2804

Can someone clarify  note 2 with respect to the Vcc voltage on the UCC2804 datasheet page two ?  I am getting hit by reliability because we have 11.74V on the Vcc pin.  They way I read this note is that as long as I am below 30mA the 12V maximum does not apply. 

Note 2 states In normal operation Vcc is powered through a current limiting resistor.  Absolute maximum of 12V applies when Vcc is driven from low impedance source such that Icc does not exceed 30mA.  We are less than 5 mA.

Then the note states:

The resistor should be sized so that the Vcc voltage, under operating conditions is below 12V but above the turn on threshold.

  • Hi Erik

    I can sympathize with your problem but unfortunately both maximums apply.

    As you say it is difficult to get the current up to 30mA and usually the 12V limit is the one most likely to apply.

    You would normally not want to operate close to this limit and how close to the limit is normally determined by your company's derating policy and your reliability department.

    John Griffin

  • John,

    Thanks for the response.  I think reliability is happy because of this bit of information I received from one of the app engineers.

    This is 15V process and VCC abs max rating is 15V. The reason of specifying 12V abs max rating is to keep below the min zener of 12V (see below) unless it is current limited to the 30mA. If customer needs 15V, I suggest to put 100ohm between VCC pin and Supply. The decoupling cap should be on VCC pin after the 100ohm resistor

                                                                                     UCC180X     UCC380X

    Vcc Internal Zener Voltage Icc =10mA           12 13.5 15    12 13.5 15  V


    If I use 15V we should be ok.

    Thanks

  • I think it is not true that both maxima apply as John states in his answer.

    From the block diagram it can be seen that the IC is internally protected by a zener.

    Common known is when a voltage larger than the zener voltage is applied, the zener will conduct. If the current is not limited, the zener will blow up.

    Note 2 simply says you should not blow up this zener. To prevent this you should either

    1. stay below the minimum specified zener voltage when you apply a low impedance source, i.e. a source that is hardly or not current limited
    2. use a current limiting resistor suchs that the current will not exceed 30mA

     

    Because you have 11.74V on the Vcc pin, option 1 is satisfied. Note that the zener will not conduct. If you can prove this voltage will not increase to 12V or higher, then your application is reliable.

    If the voltage can fluctuate and can exceed 12V, you should prove the current cannot exceed 30mA. If so, your application is reliable.

     

    About the second part in note 2: it simply tells you how big the resistor should be if you go for option 2.

    In general in rest IC's draw a different current than in operation mode. This current flows through the current limiting resistor causing a voltage drop. The specification of the UCCx80x gives two currents: a Start-up Current and an Operating Supply Current. Note that the operating current is the current the IC uses itself. To drive the external mosfets, also current is needed. That's why the current in your application is about 5mA.

    Note 2 says you should choose the current limiting resistor such that the voltage drop will cause a voltage on the Vcc pin which is lower than 12V but higher than the turn off threshold voltage for all currents, so at 0.1 mA Start-up Current  and 5mA operating current.

  • Another thing about reliability:

     

    When using option 2, the zener will clamp the voltage to the zener voltage, let's say 12V. If your source is say 15V, then a part of the current entering the Vcc pin is is used by the IC itself, but the rest is dissipated by the zener.

    Example: Source = 15V, zener = 12V, Ic needs 1mA, driver needs 4 mA.

    If you use a current limiting resistor of 100 Ohm, the current entering the Vcc pin is: (15V-12V) / 100 Ohm = 30mA.

     Ic needs 1mA, driver needs 4 mA, so 30-1-4 = 25mA is consumed by the zener. That means 12V * 25mA = 0.3W is dissipated by the zener!

    This dissipation will heat up the IC. And in general applies the hotter an IC gets, the lower its lifetime will be ... which will influence your reliability.