To implement a step-up converter with a current output, designers often simply connect the load in place of the top resistor in a resistive-divider feedback network. The bottom resistor then serves as a current-sense resistor. Though simple, this approach is inefficient. Low efficiency results from the relatively high sense voltages – usually, 1.25 V but as high as 2.5 V for some ICs. A switch-mode dc/dc converter configured as a 20-mA current source minimizes the efficiency loss by lowering the sense voltage to 200 mV (Figure 1). Advantages of this circuit include the factor-of-six gain in efficiency; minimal board area; and readily available, low-cost components. Applications include battery charging, LED drive, and general-purpose current sources.
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| Figure 1. | The use of a current mirror in the feedback loop boosts efficiency and stability in this current-output step-up converter. |
Resistors R1 and R2 form a voltage divider that derives 200 mV from the IC’s reference output. This sense voltage connects to one emitter of the current mirror comprising Q1 and Q2. Both collectors connect to the output voltage via 200-kΩ resistors. The collector of Q2 also connects to the IC’s feedback pin, and Q2’s emitter connects to the low-side current- sense resistor, R5. The feedback network appears to the IC’s control loop as a common-base amplifier. Selecting a 2N3904 for Q2 yields sufficient emitter-to-collector gain for the purpose: approximately 80 V/V. Moreover, the network’s large bandwidth (characteristic of common-base configurations) prevents instability in the IC’s control loop.
