Subscribe to EDN
RSS
Reprints/License
Print
Email

Positive regulator makes dual negative-output converter

Edited by Bill Travis

Keith Szolusha, Linear Technology Corp, Milpitas, CA -- EDN, June 26, 2003

Some systems, such as optical networks, require more than one negative voltage. A common procedure is to boost the main negative supply of –5V to –10V and then reduce it with a linear regulator to –9V. The –5V itself comes from a positive supply, typically 5 or 12V. Independently creating each of the two negative voltages requires the use of two switching-regulator ICs. However, a simpler approach uses only one step-down switching regulator and creates both a negative output with low ripple current and a second output with twice the voltage. The circuit in Figure 1 is a positive-to-dual-negative dc/dc converter that converts 5V to both –5V and –10V, using a positive buck regulator. This configuration is not a usual application circuit of a positive buck-regulator IC, which you would typically find in single-output step-down applications. However, by rearranging the connections to ground and VOUT, the dc/dc converter becomes a common positive-to-negative converter.

The circuit features an additional negative-output stage with twice the voltage of the first output by using a second inductor, a catch diode, and an output capacitor. The circuit includes a coupling capacitor, CCOUP, for energy transfer to the second winding and offers regulation of the secondary output without feedback. The use of two inductors, as opposed to one transformer, can provide a flexible circuit layout. This circuit measures less than 3 mm high and delivers high load current (Figure 2). You can replace the two inductors with a single 1-to-1 transformer, but even at the high 1.25-MHz switching frequencies, it may be difficult to find a less-than-5-mm-high transformer that can do the job. The LT1765 buck regulator provides high current capabilities even with small inductors and all-ceramic coupling, input, and output capacitors, thanks to both its 1.25-MHz switching frequency and its 3A onboard power switch.

Figure 2 shows the output capability for the circuit in Figure 1. The load current on one output determines the maximum available load current on the other. At maximum output current, the peak switch current (the sum of the peak inductor currents) is equal to 3A. Forcing the output current higher than the maximum value can cause the output voltage on both lines to collapse and lose regulation. The single feedback signal directly monitors VOUT, but the low-impedance ceramic coupling capacitor with an extremely high rms current rating keeps VOUT2 well-regulated. As load conditions change on both VOUT and VOUT2, VOUT2's regulation becomes slightly compromised over different load conditions. VOUT2 can become unregulated if the load current on VOUT2 becomes extremely small. A preload of 5 mA on VOUT2 is necessary to maintain regulation if the load current falls below 5 mA. Zero load current on VOUT does not cause the converter to lose regulation. Cross-regulation typically stays within 1% of the typical output voltage, an excellent rating for a single-feedback, dual-output converter. Figure 3 shows typical efficiency curves for various values of the load current on VOUT2.

RSS
Reprints/License
Print
Email
Talkback
Canon Resource Center

Featured Company


Most Recent Resources

Advertisement
Related Content

No related content found.

  • 0 rated items found.
Advertisement

KNOWLEDGE CENTER

Datasheets.com Parts Search

185 million searchable parts
(please enter a part number or hit search to begin)
Engineering Careers
Jobs sponsored by
Advertisement
About EDN   |   Site Map   |   Contact Us   |   Subscription   |   RSS
© 2012 UBM Electronics. All rights reserved.
Use of this Web site is subject to its Terms of Use | Privacy Policy

Please visit these other UBM Canon sites

UBM Canon | Design News | Test & Measurement World | Packaging Digest | EDN | Qmed | Pharmalive | Appliance Magazine | Plastics Today | Powder Bulk Solids | Canon Trade Shows