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LM2766: Can it be used with no diodes?

Stefano Infante
Expert 2310 points
Part Number: LM2766
Other Parts Discussed in Thread: OPA2170, , TLV2402

Hello!

I am using LM2766 with 3.3V input and 6.6V output. It only powers a dual amplifier, OPA2170 (about 250 uA total current).

LM2766 has an enable signal, which should be used to shutdown OPA2170 in stand-by mode, but the diode between V+ and VOUT makes VOUT always greater than about 3.1V, and OPA2170 does not shut down. 250 uA is three times the entire system current in stand-by, and of course this is not acceptable.

This leads me to two possible options:

1) remove the diode

2) replace OPA2170 with a pin-to-pin compatible device with lower current consumption (such as TLV2402), so that the small current consumption does not have a relevant effect on the overall stand-by current.

I would prefer the first choice, as it would let my system behave as expected, but is it ok?

Thanks in advance and regards

Stefano Infante

  • 0
    TI__Guru 61255 points

    Hello Stefano,

    The diode is necessary for startup, see here from the datasheet page 9:

    The Schottky diode D1 is only needed to protect the device from turning on its own parasitic diode and potentially latching up. During start-up, D1 also quickly charges up the output capacitor to VIN minus the diode drop thereby decreasing the start-up time. Therefore, the Schottky diode D1 must have enough current carrying capability to charge the output capacitor at start-up, as well as a low forward voltage to prevent the internal parasitic diode from turning on. A Schottky diode like 1N5817 can be used for most applications. If the input voltage ramp is less than 10 V/ms, a smaller Schottky diode like MBR0520LT1 can be used to reduce the circuit size

    So I do not recommend option 1 above and recommend using option 2.

  • 0 in reply to Brigitte
    Expert 2310 points

    Hi Brigitte,

    yes, I had read that data sheet section, but I hoped the very small current I use could give me a chance of removing the diode.

    I will go for option 2.

    Thanks and regards,

    Stefano