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Design Ideas |
Figure 1 shows a negative low-dropout (LDO)
regulator capable of providing output voltages from –2 to –16V, with
load currents greater than –3A. LDO regulators, which can maintain output
regulation with less than 1V of I/O span, are popular because of their increased
efficiency compared with standard linear regulators (which require 2 to 3V
headroom). You have many positive LDO regulators to choose from, but the choice
of negative devices is very limited. If you need load currents greater than –1A,
your choices become vanishingly small. Using an LM3411 controller IC, it's
possible to build a versatile, high-current negative LDO regulator that provides
good electrical performance and uses inexpensive components.
The LM3411 has an internal 1.22V reference between the IN and COMP pins. When the voltage across R5 reaches 1.22V, the LM3411 sources current from its OUT pin to provide control of the negative output voltage at –VOUT. The pass transistor Q3 and its driver Q2 sink –3A of load current at the regulated output. The bias current to turn on Q2 flows through R3. Controlling the voltage at the R3-Q2 node provides the output-voltage regulation.
When the regulated output voltage reaches the design set value, the voltage across R5 becomes 1.22V. This value causes IC1 to source current from its OUT pin, forcing the voltage at the base of Q2 to become more positive. This action reduces the base current available for Q2, thus reducing the output load current. This negative-feedback action locks the output voltage at the design value. Capacitor C1 is necessary for compensation, and R4 is required for proper start-up.
For component selection, use the following guidelines: You can increase C2 and C3 without limit. Their voltage ratings must be adequate to accommodate –VOUT and –VIN. R2 sets the maximum base-drive current available for Q3. For –3A output current, you should provide 50 mA of base drive to Q3. You can use the following expression to determine R2 (round off to the nearest standard value): R2=20(VIN–1.5). You must also consider the power dissipation in R2, using P=(VIN–1.5)2/R2.
The following guidelines provide a quick selection of the wattage ratings for R2: If VIN<6V, use a 1/4W resistor; for 6V<VIN<11V, use a 1/2W resistor; and for 11V<VIN<18V, use a 1W resistor. You should select R3 so that a maximum of 10 mA will flow through it, using the expression R3=100(VIN–1.5). The value of R6 determines the regulated output –VOUT. To select R6, use the expression R6=82(VOUT–1.22). For improved accuracy, use 1%-tolerance resistors for R5 and R6. You must provide a heat sink for Q3 that allows the transistor to safely dissipate the maximum power it can encounter in the application: P=(VIN–VOUT)·ILOAD.
For the purposes of obtaining test data, you can use a design that targets a –15V application, assuming an input voltage of approximately –16V. The calculated component values are R2=300[ohms], 1W; R3=1.5 k[ohms], 1/4W; and R6=1.1 k[ohms], 1/4W. The heat sink is a Thermalloy 7020B. The voltage rating for C2 and C3 is 25V. The tests for regulation and dropout voltage are performed at 20°C ambient.
Regulation test: VIN=–16V. As the load current varies from 0 to –3A, the change in –VOUT is 7 mV. The regulation, defined as the change in –VOUT divided by –VOUT, is 0.05%.
Dropout-voltage test: With –3A load current, –VIN is reduced until –VOUT drops 10 mV below its nominal value. The dropout voltage, defined as the difference between –VIN and –VOUT at this point, is 15.08–14.76=0.32V. (DI #1946)
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