2 DC/DC Converter Basics – P_IN vs P_OUT

It is important to understand the relationship between power in, power out, and efficiency when selecting a converter. For an ideal physical system, the power in is equal to the power out. For the system overview depicted in Figure 1, this can be expressed as:

V_Batt × I_Batt = V_CC × I_CC

For example, if the battery voltage V_Batt = 1.3 V, the MSP430 current I_CC = 10 µA, and the MSP430 voltage V_CC = 3.3 V, then, ideally, from the equation above, solving for I_Batt yields:

I_Batt = (V_CC × I_CC) / V_Batt = 25 µA

This means that in order for an ideal dc\dc converter to supply 10 µA at 3.3 V to the MSP430, the battery would need to supply 25 µA at 1.3 V to the dc/dc converter. Ideal implies 100% efficiency but, of course, no real system is 100% efficient. Sources of inefficiency include power dissipation in the form of heat generation, switching losses, and quiescent current of the dc/dc converter itself. The graph in Figure 2 (taken from the TPS60313 datasheet) shows that when V_IN = 1.3 V and V_OUT = 3.3 V, the device is approximately 75% efficient when supplying an output current of 10 µA.

If we reconsider the equation P_IN = P_OUT taking into account the efficiency of the dc/dc converter:

V_Batt × I_Batt = (1 / eff) × V_CC × I_CC

Then solving for I_Batt:

I_Batt = (1 / eff) × ((V_CC × I_CC) / V_Batt) = 34 µA

Using the same values for V_Batt, I_CC, and V_CC, I_Batt is found to be 34 µA. This means that to provide 10 µA at 3.3 V to the MSP430, the battery must supply a total of 34 µA at 1.3 V to the dc\dc converter.