LT1945 Design for 100mA on -15V & +15V

I am designing a power supply using LT1945. Here is the environment:

Input :12V
Output +15V, -15V
Current: 100mA, -100mA

Please find the Schematics attached for reference.

Issue: I have designed the circuit taking cue from the datasheet, (all the application circuits shown are of lower currents viz, 10mA, 20mA), however, while simulating it, the negative rail is stable only for currents lower than 40mA, even though the datasheet mentions it can source currents upto 350mA. At 100mA, the voltage drops to ~10V or so, and I am not able to solve the problem nor make it work at 100mA. 

Can anyone help me with an explanation or tell me what I am missing?

  • 0
    •  Analog Employees 
    on Nov 19, 2018 4:39 PM over 1 year ago

    Hi,

    The datasheet says both converters are designed with a 350mA current limit, that however, is referring to the switch current limit, and not to the output current available.

    The schematics provided at the front and last pages of the datasheet are useful for providing an idea of the load current available for the input and output voltage shown on those applications. As a general rule, as the output voltage is increased, the output current available will drop.

    If you really need 100mA load for each output, you need to pick a converter that has at least 500mA switch current capability, guaranteed, for the positive output, and about 700mA for the negative. Please note the LT1945 has a minimum of 250mA of guaranteed switch current, so this part will not support your needs.

    The LT8330 has a 1A switch and can be used for positive or negative applications to provide the power you need, though it will require one part per output.

  • Hi

    I'm using LT1945 in my design. Below is the circuit. My application will need 30V and -5V with </=10mA.

    So Vpos =30V,@10mA Vneg = -5V@10mA  Vin =5V

    How can I calculate current required by 5V input to LT1945.

    Is it using below approach?

    ((5V * 10mA /5V) + (30V * 10mA /5V)) = 70mA.

    Whether this is proper?. Please let me know.

    Regards

    Nidhi P Shetty

  • 0
    •  Analog Employees 
    on Apr 16, 2020 9:49 PM 3 months ago in reply to nidhi.shetty@iwavesystems.com

    Hi Nidhi,

    Your calculation would be true if the converters were 100% efficient. They are not. Assume efficiency is abut 70% and you will get a more realistic value.

    On a different note,

    In a boost converter the output has to be higher than the input, and the negative converter has the same magnitude of the input voltage, 5V. I would suggest you replace D30 with another 10uH inductor. Then, the input voltage can be higher  or lower.

    Or, you can just drop the input voltage level, or raise output level a little.

    Cheers!

    DV

  • Hi

    Thanks for your reply. If I consider converter efficiency of 70% by worst case for 10mA load on both positive and negative output I may need 100mA supply current from input.

    But when I simulated using LTspice I'm observing more current being drawn from supply.

    As my input 5V @200mA input is also shared with  other regulator in actual design I have to decide on worst case current requirement by LT1945 input.

    Also as per my understanding for both positive and negative output there are independent circuit inside. Hence I'm configuring positive output as boost type and negative as inverting, not boost. Still is it necessary to replace D30 with a inductor.

    Please help me out.

    Regards

    Nidhi P Shetty

  • 0
    •  Analog Employees 
    on Apr 17, 2020 2:34 PM 3 months ago in reply to nidhi.shetty@iwavesystems.com

    The inverter you are using has the same limitations as the boost, in terms of the magnitude of the output voltage. It simply cannot be lower than the input. The addition of an inductor does not change the inverting function, but it allows for Vout to be higher or lower than the input.

    In terms of input current needed; I would recommend budgeting for 100mA, for your power level.

    Why your simulation draws 200mA, I simply don't know, but the real circuit should not.

    Best.

    DV