Electronic Design

Power Supply's Output Tracks Negative Input By One-Half

In some fiber-optic communication applications, the transimpedance amplifier needs a negative bias voltage that tracks the negative input rail by one-half for the data signal's common-mode input range. The circuit in Figure 1 uses an MIC2196 controller to accomplish that task. The controller is configured in a boost configuration to generate a −2.5-V output from a −5-V input rail. The one-half ratio is programmed by the resistive divider formed by R8 and R9. Moreover, the ratio can easily be changed to program a different output voltage by the following equation:

VOUT = VIN × \[R8/(R8+R9)\]

For VOUT = 1/2 VIN, use R8 = 10 k Ω and R9 = 10 k Ω. For VOUT = 1/4 VIN, use R8 = 10 k Ω and R9 = 30 k Ω.

The MIC2196's internal error amplifier is disabled by the resistive divider formed by R2 and R6. This divider is tied to the internal 3-V VDD regulator pin to generate a bias on the FB pin of 0.7 V above the chip GND, which is at −5 V. The FB pin is the inverting pin of the controller's internal transconductance error amplifier. The noninverting pin of the error amplifier is tied internally to a reference that is 1.24 V above the chip GND.

Because the FB pin is lower in amplitude than the internal noninverting pin, the internal transconductance error amplifier is forced to source more current into the COMP capacitor, C3, thus raising the output voltage. The MIC7300 is an external error amplifier that sinks enough current from the COMP pin to produce the right duty cycle to generate −2.5 V at the output when the input supply is at −5 V.

Figure 2 shows the circuit's efficiency, which is greater than 85% for load currents up to 2 A. The circuit can easily be modified to handle a different load current by changing the sense-resistor, R4, value.

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