Converting Negative Input Voltages To Positive Output Voltages
Some applications require a positive voltage to be generated from a negative input voltage. This can be done either by using a transformer based topology, such as a flyback converter, or with a negative-input buck-boost converter. Such a buck-boost converter can be implemented using a low-side controller, such as National Semiconductor's LM5022.
Fig. 1 shows a simplified schematic of the negative buck-boost converter operating in continuous conduction mode (CCM). In this mode of operation, the MOSFET Q1 is on for a time interval, DT, of the switching period, T, while the diode, D1, is on for an interval of (1-D)T. The positive terminal of the input voltage, which is also the negative terminal of the output voltage, is connected to one end of the inductor. The negative terminal of the input voltage is connected to the source of the switching FET Q1.
When Q1 is on, output-diode D1 is reverse biased and the input voltage is applied across inductor L1. The inductor current ramps up at a rate of:
When Q1 goes off, D1 becomes forward biased and the voltage across the inductor becomes equal to -VOUT so that the inductor current is discharged at a rate of:
Under steady-state conditions, the positive and negative volt-seconds across the inductor are equal, so that:
From here we easily get the familiar input-out voltage conversion equation for a buck-boost converter:
Fig. 2 shows a low-cost, high-efficiency negative-input buck-boost converter based on the LM5022. It converts ?12 V to 5 V with a maximum output current of 2.5 A. The ground pin of LM5022 (IC1) is connected to the negative-input terminal, or system common.
Because its reference, or ground pin is negative with respect to the output voltage, the LM5022 must monitor and regulate output voltage through a level-shifting circuit. This level-shifting is implemented by PNP transistor Q1. The output voltage is sensed across resistor R7 and the VBE of Q2. In Q2, the resistor establishes an emitter current of:
This translates to a nearly equal collector current of IC=αIE flowing through R8. Hence the voltage across R8 is
(See equation 6)
The LM5022 will try to maintain by regulating output voltage at a constant value.
The VBE of Q2 has a negative temperature coefficient of about -2.2mV/°C, which will cause the output voltage to vary with temperature. To minimize this variation, Q2 needs to be placed away from the main power-dissipating and heat-producing components on the PCB.
For a buck-boost converter, average inductor current is:
(See equation 7)
In this design the switching frequency is: fs = 500 KHz, and the duty ratio, D, is 0.3 (for a 12-V input). At the full load current of 2.5A, the average inductor current is IL = 3.6A. Thus the peak to peak inductor ripple current is:
Therefore, the inductor should be rated for at least 3.9 A.
The buck-boost converter topology has high ripple currents in both the input and output capacitors. The rms values of input and output ripple currents are about the same, which is approximately equal to:
In this design, low-ESR, high-ripple-current ceramic capacitors are selected to filter ripple currents. Large capacitor values are needed at the output to meet stability and load-transient response requirements.
The main power-dissipating component in this converter is the output diode d1. This needs to be a Schottky diode with a low forward-voltage drop. The average diode current is equal to the load current, ILOAD. The diode conduction loss is:
where VD1 is the diode's forward voltage drop.
The diode used here has VD1=0.5V, so that PD1=1.3 W at full load. This is 10% of the output power. With other losses added, the full load efficiency of this converter is about 83%.