# Designing a circuit using an LT1073 DC-DC Converter

I'm currently designing a circuit to convert a 24V (±30%) to supply a 3,3V IC used to monitor currents. I need to use the LT1073 which has the following diagram.

The operations of the component and its applications are well-explained in the datasheet. According to the circuit example, I should place a 100uH inductor and a Schottky-Diode at the switch. As far as my understanding goes, the Schottky-Diode is needed to block the discharge of current from going to the ground to allow the current to flow into the load. Is this correct?

This is how the circuit is supposed to be connected

According to the example, I need to put an inductor of 100uH. But what if this isn't possible and I need to pick a smaller inductor? What do I need to consider?

The datasheet mentions:

To operate as an efficient energy transfer element, the inductor must fulfill three require-ments. First, the inductance must be low enough for the inductor to store adequate energy under the worst-case condition of minimum input voltage and switch ON time. The inductance must also be high enough so that maxi-mum current ratings of the LT1073 and inductor are not exceeded at the other worst-case condition of maximum input voltage and ON time. Additionally, the inductor core must be able to store the required flux, i.e., it must not saturate.

And then the datasheet goes on with some formulas that I can't really understand.

(1) Pl = (Vout + Vd - Vin)  * (IOut), where Vd is the diode drop(0.5V for IN5818 Schottky)

(2) PL = 1/2 * L * iPeak^2 *fOsc


These formulas raised a few questions:

1. Where does IOut flows?
2. How can I get fOsc?

I would really appreciate it if someone could give some tips or answers to my questions. Thanks in advance!

• Can you show the entire schematic, and the expected waveforms? May 15, 2019 at 9:20
• I added the schematic... I'm expecting a DC 3,3V to be generated out of the switch SW2. By connecting the right R2-to-R1 resistance ratio, according to the formula given, Vout = 212mV * (R2/R1 +1)(see formula in datasheet) , I should be able to get the voltage that I need. May 15, 2019 at 10:01
• 24 volt at 30% higher is 31.2 volt and the front-page of the data sheet states that the chip works up to 30 volts so, I'd be concerned about using it in your application. May 15, 2019 at 11:57
• The absolute maximum supply voltage says 36 V in Step-Down mode. I think it's going to be fine.. May 16, 2019 at 13:25