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The integration of power management solutions for today's portable application processors is growing. The current consumption of total power, standby, and deep sleep can affect battery size, bill of materials (BOM) cost, and product awareness.

When designing portable devices—such as smart phones or PDAs—system designers must consider many power supply variables. As they consume more and more power, smart phones require highly integrated power management solutions to achieve the longest battery life design goal in the smallest possible PCB area.

Today's application processors need to provide different power domains for the core, I/O, memory, and external devices. For example, the LP3971 is a power management unit (PMU) designed to meet all of these requirements, utilizing three high-efficiency buck converters and six low dropout (LDO) regulators.

Application processors require multiple supply voltages that can be optimized through core power management and system architecture. The LP3971 features an I2C-controlled output voltage, factory-configurable power-up sequencing, and default output voltage to meet a wide range of system requirements.

This article focuses on how to use a device such as the LP3971 in combination with a buck converter and LDO function to power a low voltage microprocessor in a PDA/smartphone application.

When designing a system, the architecture must be balanced, including cost, PCB area, component size, talk time, standby time, battery capacity, and schedule. The microprocessor RAM requires a 1.5V supply with a maximum current of 400mA. Let's start with the simplest, lowest cost solution—the LDO that connects directly to the lithium battery—see Figure 1 below.

The battery voltage will drop from 4.2V to 3.2V, at which level the system goes into deep sleep until the battery is charged or replaced. Figure 2 shows a typical lithium battery discharge cycle.

For the configuration shown in Figure 1, the efficiency of the LDO 5 would be:

LDO percentage efficiency = [(Vout * Iout) / Vin * (Iout + Iq)] * 100

For this and other examples in this article, Iq is neglected (40 mA) because it is very small relative to Iout (400 mA).

The efficiency equation then becomes:

Percent efficiency = [(Vout) / (Vin)] * 100.

For Vin = 4.2V and Vout = 1.5V, the efficiency of LDO is 1.5/4.2 = 36%. Total power Pt = 4.2 * 0.400 = 1.70W.

All power that is not delivered to the output load will be dissipated as heat inside the LDO. The power dissipated is:

Dissipation (Pd) = (Vin - Vout) * Iout = (4.2 - 1.5) * 0.400 = 1.1W, dissipated as heat.