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Supply current

Yusong Di
Prodigy 30 points
Other Parts Discussed in Thread: LM3488, LM2587

Hi, all

I have a question about the definition of supply current. What is supply current? On DC-DC converter side, Is that same as input current?

Moreover, I bought a DC-DC Converter, which is boost converter LM2587, and SEPIC LM3488, They all can achieve 13V 2A output from the datasheet, so output power can be about 20Watts, consider the cost power on DC/DC, the input power must be larger than 20W, Right??

BUT, I did many experiment on that, all these chips CANNOT stand that large input power. They Burn, of course with 100ohms output resistor connected.

So confused, then I found datasheet about the supply current, so, ask for help!!!

Thanks!

God bless all~

  • 0
    TI__Guru 74225 points

    First, the supply current listed on the data sheet is usually the current required to run the chip and not the total input current.

    As an example, the LM2587 has a supply current of 11mA typical on the data sheet.  This is the current that is required to

    power the chip, but is not related to the input current or power that the chip can supply.  Also, it may increase with load.

    Second, you are correct about the input power.  It will always be larger than the output power by some small amount depending on the

    efficiency of the system.  The input power is calculated as follows: (input power) = (output power)/efficiency.

    For example with an efficiency of 80%=0.8 and an output power of 20W, the input power is 25W.

    If the input voltage is 5V, the input current will be about 5A.

    Third, a boost converter is rated differently from a buck.  On a buck data sheet the current rating on the front page refers to the

    maximum output current, and therefore power, that the device can supply.

    A boost converter is rated by its switch current.  The switch current is related to the input current and not the output.

    As an example, suppose you are boosting from 5V input to a 13V and 2A output.  And also assume that the efficiency is about 80%.

    The input current is about: 13V*2A/(0.8*5V) = 6.5A.  The switch current rating on the data sheet must be greater than this.

    So, a part like the LM2587 will not work.  Unless the input voltage is larger or the efficiency can be made grater.

    Frank D