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Low Energy Base Drive Using Regenerative Turn Off for Switching Regulators

IP.com Disclosure Number: IPCOM000038953D
Original Publication Date: 1987-Mar-01
Included in the Prior Art Database: 2005-Feb-01
Document File: 3 page(s) / 51K

Publishing Venue

IBM

Related People

Hitchcock, LJ: AUTHOR

Abstract

A known base drive for a proportional drive Dual Switch (DS) experiences circuit limitations, a chief one of which is that the transformer saturates, causing excessive current to flow. A reconfiguration of the circuit solves these problems in the following manner. The known base drive is shown in Fig. 1. When transistor Q1 turns off, magnetizing current is forced to flow in windings N2 and N3. The DS turns on and proportional drive begins. As Q1 turns off, Q2 turns on and begins charging capacitor C1 toward VB . Q1 then turns on, and the voltage on C1 is applied to N1 (neglecting the drop in D2 and Q1). The DS begins to turn off, and reverse turn-off current is supplied by C1. When turn off is complete, C1 will still have a voltage present. C1 must continue to discharge, and its energy is transferred to N1 as Im .

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Low Energy Base Drive Using Regenerative Turn Off for Switching Regulators

A known base drive for a proportional drive Dual Switch (DS) experiences circuit limitations, a chief one of which is that the transformer saturates, causing excessive current to flow. A reconfiguration of the circuit solves these problems in the following manner. The known base drive is shown in Fig. 1. When transistor Q1 turns off, magnetizing current is forced to flow in windings N2 and N3. The DS turns on and proportional drive begins. As Q1 turns off, Q2 turns on and begins charging capacitor C1 toward VB . Q1 then turns on, and the voltage on C1 is applied to N1 (neglecting the drop in D2 and Q1). The DS begins to turn off, and reverse turn-off current is supplied by C1. When turn off is complete, C1 will still have a voltage present. C1 must continue to discharge, and its energy is transferred to N1 as Im . After C1 discharges, D1 conducts to try to keep Im from falling until Q1 turns off again, starting the next cycle. If C1 and R2 are sized to give the proper turn-off drive for the DS at all line and load conditions, T1 is likely to saturate while

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discharging C1. This will cause excessive current to flow in D1 and Q1 even after C1 is discharged. There is more DS turn-on drive at low line. At high line C1 doesn't charge to as high a voltage while the DS is on and it discharges to a lower voltage when the DS turns off, leaving less energy to increase Im . Also, at high line there is more time for Im to decrease while D1 conducts due to longer off...