Hi,

I need to power some new latching solenoids, which will consume about 0.7A at 12V.

My power source is a 4.2V 8000mAh Li-Poly battery.

I need to design a step-up DC-DC converter, and I plan to design it such that the output of the circuit is 12V at 1A. Therefore, the input current at 4V would need to be about 3A.

Does this mean that the switching current for my requirements would need to be at least 3A?

I am planning to use the LTC3124, which I believe would be adequate unless there is an alternative option.

I also tried the following solution, which seems to work well, but I can't find the manufacturer of the IC. It appears to be a fabless Chinese company. I was surprised that it was a simple SOT-23 device able to provide 1A at 12V.

Youmile 10PCS SX1308 DC-DC Step-Up Boost Converter 2-24V to 2-28V 2A Adjustable Power Supply Module Voltage Regulator with 2PCS 24AWG Wire : Amazon.co.uk: Business, Industry & Science

Regards

DJ

 

Your initial calculations "seem" to assume that you can make the process 100% efficient. This will not be possible. I would, for planning purposes, start with an assumption that you can achieve an efficiency of 80%. On that basis I would calculate that 12 watts of output power will require:

\( 12 \text{ watts}/80\text{%}\;=\;15\text{ watts, input power} \)

As your batteries discharge you will increase the current demand from the batteries causing them to discharge faster. At 4V the batteries will be drawing:

\( 15 \text{ watts}/4\text{ Volts}\;=\; 3.75 \text{ amperes} \)

You may need to rethink your choice of chips based on this new information. You should pay particular attention to inductor selection with respect to DC resistance and saturation current as well as your capacitor selection with respect to ESR (Equivalent Series Resistance).

 

Your initial calculations "seem" to assume that you can make the process 100% efficient. This will not be possible. I would, for planning purposes, start with an assumption that you can achieve an efficiency of 80%. On that basis I would calculate that 12 watts of output power will require:

\( 12 \text{ watts}/80\text{%}\;=\;15\text{ watts, input power} \)

As your batteries discharge you will increase the current demand from the batteries causing them to discharge faster. At 4V the batteries will be drawing:

\( 15 \text{ watts}/4\text{ Volts}\;=\; 3.75 \text{ amperes} \)

You may need to rethink your choice of chips based on this new information. You should pay particular attention to inductor selection with respect to DC resistance and saturation current as well as your capacitor selection with respect to ESR (Equivalent Series Resistance).

Yes , my calculation was therotical. I think I will use LTC3124.