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LM22676-ADJ Trying to generate a -10V Rail

Martyn Turner
Prodigy 30 points
Other Parts Discussed in Thread: LM22676

Hi,

I am trying to generate a -10V rail @ 100mA load current using the LM22676 but currently i cannot get an output. I am using the SO PowerPAD package.

I have attached my schematic.

As a note, i have exactly the same components creating a +10V rail also on the PCB and this is working fine.

I am unsure what i have done wrong and would be extremely grateful for any assistance.2043.-10V Rail.docx

  • 0
    TI__Genius 16025 points

    Hi Martyn,

    What is your input voltage? Remember Vdevice(max) = Vin + |Vout|.

    Also, make sure the device thermal/exposed pad or DAP is connected to -10V as well.

    Ensure your output capacitors are rated to at least 20V higher than the positive rail design.

    If none of these work, please provide your layout.

    Thanks,

    Anston 

  • 0 in reply to Anston Lobo
    Prodigy 30 points

    Hi Anston,

    Many thanks for the reply.

    My input voltage is 11.1 VDC from a lithium-ion battery pack.

    Could you explain further why the exposed pad must be connected to -10V and why the output caps need to be 20V higher in rating to that of the +10V rail.

    Apologies for the naivety but i will need something concrete to convince my engineers that they have made an error.

    I have also attached my layout and schematic.

    Many Thanks

    Martyn

    4812.-10V Rail.docx

     

  • 0 in reply to Martyn Turner
    TI__Genius 16025 points

    Hi Martyn,

    Why to isolate the exposed pad?

    The exposed pad is connected internally to IC GND. Since all IC GND pins are connected to -Vout, having the exposed pad connected to system GND is simply shorting the output.

    On closer inspection of your layout, resistor R83 has a pad connected to the exposed pad of the IC U7. While R83's pad should make contact to system GND, U7's exposed pad SHOULD NOT be connected to system GND.

    If you look at the GND plane, there are 3 via's that connect the IC exposed pad to the GND plane. This will cause the IC to shutdown and not supply any current, since the output is short circuited.

    If your engineers are not convinced, as a simple test, Isolate the IC's exposed pad by placing tape over it and then solder the IC connections. The part should work. Do not place any load on the negative output, since there is no exposed pad to dissipate the heat.

    Capacitors

    In my earlier post I accidentally mentioned 'Output' capacitors. This increase in 20V over the positive rail should only apply to C64 in your schematic. The reason, In a positive rail scenario, this capacitor sees only 11.1V which is from VIN. However in a negative rail scenario, this capacitor sees VIN + |VOUT|, which is 11.1V + |-10|V = 21.1V. 

    Almost all ceramic caps derate to half their rated capacitance at half their rated voltage and it only gets worse as you approach the rated voltage. As an example, 10uF, 30V ceramic is only 5uF at 15V and hardly 1uF at 25V. As it approaches 30V, there is almost no significant capacitance left. This applies only to ceramic.

    To buffer your IC against high frequency noise due to operation, it is vital this capacitor be a 0.5uF or 0.1uF rated to 50V.

    I hope this will suffice.

    Thanks,

    Anston

  • 0 in reply to Anston Lobo
    Prodigy 30 points

    Anston,

    Thanks for the clear explanation, i will pass this onto my engineers and see how we get on.

    Many Thanks

    Martyn