Some of the newest high-performance processors require a 1.5V bus terminator that can source currents as high as 7A. This terminator requires a very fast transient response, because the device must react to high peak-current changes that have extremely fast rise times. The 1.5V terminator circuit in Fig 1 meets these demanding requirements. The circuit is a 7A linear regulator that provides a 1.5V output from a 3.3V input. The primary objectives are low cost, tight regulation, and fast transient response.
The pass element of the regulator, which supplies the load current, is the three-transistor Darlington configuration comprising Q2, Q3, and Q4. The bias current flowing through R1 drives the pass element until Q1 pulls down and removes the drive from the base of Q2. The circuit uses an LM3411 precision controller to regulate the output to 1.5V. This IC sources current at the IC's output when its input reaches the setpoint voltage. You use R7 and R8 to adjust the setpoint voltage to 1.5V. This divider's resistance is low enough to override the LM3411's built-in divider. You can, therefore, use an LM3411 having any voltage option.
When the voltage at the input of the LM3411 reaches 1.5V, the output sources current and turns on Q1, which, in turn, steals the base drive for Q2. The result is a reduction in the current available from the 1.5V regulated output. This negative-feedback loop locks the output at 1.5V. C1 and C2 are ceramic capacitors used for compensation. C4 is necessary for circuit stability, and C3 and C4 affect transient response. If you employ capacitor types other than those in Fig 1, make sure to use high-quality, low-equivalent-series-resistance units.
Performance measurements show that the output voltage changes less than 0.1 mV as the load increases from 0 to 7A and the input voltage varies from 3 to 3.6V. The dropout voltage, or the minimum input-output differential required to maintain a regulated output, is 1.4V. Thus, you need a minimum input voltage of 2.9V to keep the 1.5V output in regulation. The test for transient response uses a 0.2(ohm) power resistor, applied to the output using a mechanical contact, to provide a 0 to 7A load-current step. The change in output voltage is lower than 10 mV with a total recovery time of about 30 µsec.
Heat sinking is important in this application. As with any linear regulator, the power dissipated in the pass transistor is (VIN - VOUT) x ILOAD. You must provide Q4 with adequate heat sinking so the junction temperature never exceeds 150°C. Fig 2 shows the calculated heat-sink-to-ambient thermal resistance. You must provide for this thermal resistance with various values of load current. These values assume a maximum ambient temperature of 50°C, 3.3V input, and a TO-220 pass transistor mounted with thermal grease and a mica insulator.
You can calculate the thermal resistance by using a combination of heat-sink/airflow parameters, which are published in heat-sink data sheets. For example, a design that can supply 6A continuous load current requires a heat sink with thermal resistance of about 5.5°C/W or less. The following heat-sink/airflow combinations satisfy this criterion:
Aavid 5510B with 150-fpm airflow,
Aavid 5298B with no airflow,
Thermalloy 6101B with 100-fpm airflow, and
Thermalloy 7023B with no airflow.
The design trade-off is that smaller, lighter, and less-expensive heat sinks require more airflow to provide the needed value of thermal resistance.
If the regulator is to respond quickly to changes in load current, you must carefully select the input and output capacitors. The output capacitors are the most critical. These capacitors must supply current to the load during the time it takes the regulator loop to sense the output-voltage change and turn on the pass transistor. A good choice for an output capacitor is the Sanyo-Oscon device in Fig 1.
The input capacitors provide an energy reservoir from which the regulator sources current to force the output back up to its nominal value. The Panasonic HFQ type in Fig 1 is a good choice. If you wish to substitute the specified input or output capacitors, you must verify the transient performance to ensure proper regulator operation.
To optimize performance, you must minimize parasitic inductance inherent in connecting traces. For all paths shown as heavy lines in Fig 1, you should use traces that are as wide and short as possible. You should also place components such as to minimize lead length. Resistors R7 and R8 should be close to IC1. Also, you should directly make connections to the input and ground pins of IC1 at the regulator output to avoid poor load regulation that could occur because of voltage drops along traces.
The power transistor you select for Q4 must have high current gain at 7A and should also have a large bandwidth (fT) for the circuit to work as described. The D44H2 represents a good cost/performance trade-off. The current gain of Q4 dictates the power dissipation in its driver, Q3. The driver supplies the base current to Q4. The lower the gain of Q4, the more current Q3 must supply to the base and the higher the power Q3 must dissipate.
The D44H2 has a guaranteed minimum gain of 40 at 4A, but typical gain is higher. Assuming about 30% lower gain at 7A from data-sheet curves, the gain at 7A is more than 28. So, to support a 7A load current, Q3 must supply a 250-mA worst-case base current. Assuming a 3.3V input, Q3 does not dissipate more than about 250 mW, which a 2N3906 in a TO-92 case (whose thermal resistance is about 180°C/W) can easily handle. However, you may have a problem if you use a small surface-mountable device.
If you wish to make substitutions for the cited transistors, pay careful attention to the current-gain/power-dissipation trade-off, as well as to the fT of the transistors. The specified transistors are very fast--a necessary quality to guarantee fast loop response in the regulator.
