Obtain higher voltage from a buck regulator

Several semiconductor vendors' current-mode buck controllers have input-voltage ranges of 30 to 36V but have output-voltage ranges from the reference voltage to approximately 6V. This output-voltage constraint arises from the common-mode-voltage limitation of the current-sense amplifier. In real-world applications, the power-supply designer must be able to generate high output voltage for printers, servers, routers, networking, and test equipment. Using a conventional buck regulator to generate higher voltages is a challenge. The circuit in Figure 1 meets the challenge by using an external op amp, a small-signal pnp transistor, and a low-output-voltage buck regulator to deliver 20V output voltage from a 27V input supply for load currents as high as 2.5A. You can easily program the circuit to provide higher load currents by merely lowering the sense resistor, R2. IC1, the controller in Figure 1 is the MIC2182, and IC2, the operational amplifier, is the MIC6211. Resistors R3 and R6 program the output voltage as follows: VOUT = 20V = VFB(1+R3/R6).

You can generate 20V output using a standard, current-mode buck regulator.

This curve shows the efficiency of the regulator in Figure 1

CSH (Pin 8) and VOUT (Pin 9) of the buck controller normally connect across the sense resistor, R2, for output voltages as high as 6V. The controller asserts current limit when it senses approximately 100 mV across these two pins. In the case of VOUT=20V, the operational amplifier and Q3 level-shift the voltage drop across R2 from 20V down to 5V, which is within the input common-mode range of the internal current-sense amplifier of the buck regulator. You can readily understand the circuit operation by assuming a voltage drop of 40mV across R2 for a load current of 1A. The current through R8 is 400µA ((20.04Ð20)/100), which is also the current through R9 and R4 (via Q3). This current generates a voltage drop of 40mV across R4 (400 µA3100½). The controller's VOUT pin connects to the internal 5V regulator (VDD) and R4; the other side of R4 connects to the CSH pin. The voltage on the VOUT pin is 5V, and the voltage on the CSH pin is 5.04V. The voltage difference between these two pins is exactly the voltage drop across R2. The simple circuit in Figure 1 allows you to achieve greater than 95% efficiency for load currents as high as 2.5A with VOUT=20V, using a conventional low-output-voltage, current-mode buck regulator. Figure 2 shows the efficiency of the regulator.