Abstract:
The invention concerns battery back-up for electronic equipment. A sensor detects a drop in power supply voltage and, in response, connects the back-up battery to the equipment, via a Field-Effect Transistor (FET). The FET causes a lower voltage drop between the battery and the equipment, as compared with a commonly used alternative, namely, a diode.

Description:
The invention concerns circuitry which switches a battery into operation when the power supply of an electronic device fails. 
     BACKGROUND OF THE INVENTION 
     Power supplies for electronic equipment can fail. A common approach to protecting against such failures is to provide a battery back-up. One type of battery back-up is shown in FIG. 1. 
     In this back-up system, a diode is connected from a back-up battery to the equipment. The diode polarity is such that, when the power supply voltage, V POWER , is normal, the diode is reverse-biased (because the battery voltage, V BATTERY , is below V POWER ), and thus the diode acts as an open circuit. However, when V POWER  drops below V BATTERY  by a sufficient amount, such as 0.7 volts, the diode becomes conducting, and the battery now supplies power to the equipment. 
     While this approach works well in many situations, it does have disadvantages. For example, a voltage drop exists across the diode, such as the 0.7 volts (for a silicon device) discussed above. Consequently, if it is desired that the back-up voltage which is actually delivered to the equipment remain equal to V POWER , the battery must deliver a voltage which is high enough to compensate for the voltage drop across the diode. In this example, V BATTERY  must exceed V POWER  by 0.7 volts. Batteries having such fractional ratings can be expensive and hard-to-find. 
     Further, it is not even clear that such an arrangement is workable. That is, if V BATTERY  exceeds V POWER  by 0.7 volts, the diode will be conducting even when the power supply is functional. To inhibit the conduction of the diode, the differential between V BATTERY  and V POWER  must be reduced or eliminated. In practice, many designers reduce the differential by setting V BATTERY  equal to V POWER . However, the reduction also reduces the back-up voltage received by the equipment: when the power supply fails, the voltage delivered to the equipment will equal V BATTERY  minus the diode drop. If V BATTERY  equals 9 volts, then the voltage delivered to the equipment upon power supply failure will be about 8.3 volts. This lowered voltage can be undesirable in some cases. 
     OBJECTS OF THE INVENTION 
     It is an object of the invention to provide an improved battery back-up. 
    
    
     BRIEF DESCRIPTION OF THE DRAWINGS 
     FIG. 1 illustrates a prior-art battery back-up system. 
     FIG. 2 illustrates one form of the invention. 
     FIGS. 3 and 4 are simplifications of FIG. 2. 
    
    
     SUMMARY OF THE INVENTION 
     In one form of the invention, a detector detects when power supply voltage drops below an allowable threshold. The detector turns ON an FET in response, which connects a battery to the electronic equipment. 
     DETAILED DESCRIPTION OF THE INVENTION 
     FIG. 2 illustrates one form of the invention. It may be simplest to begin the explanation with the flip-flop 15. The flip-flop is in one of two opposite states: 
     STATE 1: S is HIGH and R is LOW, wherein Q is HIGH and Q is LOW. 
     STATE 2: S is LOW and R is HIGH, wherein Q is LOW and Q is HIGH. 
     The flip-flop controls which of FETs M1 or M2 is conducting, as the following example will show. 
     EXAMPLE 
     Normal Operation: Power Supply, not Battery, Supplies Power 
     During normal operation, the flip-flop 15 is in State 2. The situation is shown in FIG. 3. Q is LOW and Q is HIGH. Q turns OFF M2 (indicated by an open circuit in FIG. 3) and Q turns M1 ON (indicated by a resistor R). 
     M1 and M2 are p-channel, enhancement-mode MOSFETs. In the ON state, caused by a LOW voltage applied to the gate (with respect to the source), they exhibit a low channel resistance: they act as low-valued resistors, as indicated by resistor R in FIG. 3. In the OFF state, caused by a HIGH gate voltage, they exhibit a high channel resistance: they act as open circuits. 
     Consequently, since Q (now LOW) feeds the gate of M1, M1 is ON. M1 supplies the current required by the equipment, by way of the terminal Vdd. In particular, M1 passes about 5 milliamps of current, as indicated in FIG. 3, with a voltage drop (V DROP ) of about 200-300 millivolts. Conversely, M2 is OFF (i.e., an open circuit) at this time, because Q feeds a HIGH signal to its gate. 
     Because of the voltage drop across M1, Vdd is near 4.7 or 4.8 volts (the voltage will be assumed to be 4.75 volts). This voltage has slightly different effects on Q1 and Q2. Q1 is held in the inactive mode: the base-emitter junction of Q1 is reverse-biased (the emitter is at 3 volts, and the base is at 4.75 volts, which is the potential of Vdd). In contrast, Q2 is held slightly in conduction: its base-emitter junction is slightly forward-biased (the emitter is at 5 volts, and the base is at 4.75 volts.) 
     These states of Q1 and Q2 cause both V1 and V2 to be at logic LOW states. Thus, V1 and V2 are not presently controlling the state of the flip-flop. Instead, the flip-flop&#39;s state is determined by history: the last HIGH signal appearing at R or S determines the present state of the flip-flop. It will now be shown that, because of events occurring during power-up of the system, the present state will be State 2. 
     Power-Up Induces State 2 
     The equipment 18 in FIG. 3, which receives its power from the lead Vdd, inevitably has some capacitance, indicated by capacitor C. (If the natural capacitance is, for some reason insufficient, a 20 pF capacitor can be added.) This capacitance will tend to hold Vdd at a lower voltage than V POWER  during power up: Vdd will lag V POWER . The power-up sequence can be explained by reference to FIG. 4, in which the FETs have been omitted. 
     There are two possible cases. Case 1: the flip-flop is in State 1, in which M2 is ON, applying V BATTERY  to lead Vdd. Case 2: the flip-flop is in STATE 2, in which M1 is ON, applying V POWER  to lead Vdd. Case 1 will be considered first. 
     Case 1: M2 is ON at Start of Power-up 
     As V POWER  climbs from zero volts to 5 volts, the bases B1 and B2 of both transistors Q1 and Q2 are also rising as V C  rises. V C  is rising because of current fed through M2. 
     Assume, for explanation, that V POWER  stops climbing at one diode drop above V BATTERY  (i.e., stops at about 3.7 volts, as indicated in the plot 25). Since V C  is rising, at some instant, V C  will reach V BATTERY , minus the 200 or 300 millivolt drop across M2, and thus reach a voltage of 2.7 or 2.8 volts. At this instant, Q2 passes very little current, because the emitter-base voltage is only 200 or 300 millivolts, and thus V2 is almost zero. 
     However, in contrast, Q1 has a base-emitter voltage of about (3.7-Vdd). Since Vdd is about 2.7 or 2.8 volts, the base-emitter voltage of Q1 is very large. Q1 is thus heavily driven into conduction, and V1 is consequently very large, applying a HIGH signal to the R-input of the flip-flop, driving the flip-flop into State 2. 
     Now, M1 is forced ON, connecting the Vdd-line to V POWER , V POWER  supplies current, as required in normal operation. 
     Case 2: M1 is ON at the Start of Power-up 
     In this case, the flip flop starts up in State 2, and no events occur to change its status. M1 continually connects V POWER  to the Vdd lead. 
     Therefore, no matter what state the flip-flop starts in, during power-up, the capacitance C in FIGS. 3 and 4 causes the flip-flop to reside in State 2 at the end of the power-up transition. 
     It was assumed in Cases 1 and 2 that V POWER  stopped and held at 3.6 volts. This stoppage was postulated for purposes of explanation. In fact, V POWER  climbs directly to 5 volts, and does not stop. Nevertheless, the explanation given above still applies: If Q2 was initially ON, Q2 becomes shut off at some instant during power-up, and Q1 becomes highly conducting, thereby pulling V2 LOW and V1 HIGH, thus driving the flip-flop into State 2. 
     This discussion will now consider the transition to battery back-up status. 
     Power Supply Fails 
     In FIG. 2, if V POWER  falls to a single diode drop beneath V BATTERY , Vdd is pulled below V E1 , turning on Q1. V1 is pulled HIGH, pulling the S input HIGH, driving the Flip-flop into State 1. Now, M2 is driven ON, pulling Vdd to about 2.75 volts (i.e., 3 volts minus the IR drop across the FET, which is the same as that across M1.) 
     Significant Features 
     Several important features of the invention are the following. 
     One. The back-up voltage applied to the equipment is equal to V BATTERY  minus the IR (I is current, R is resistance) drop across M2. This drop will be about 200 to 300 millivolts, for a current of 5 milliamps. This drop is less than one-half the drop occurring in the diode in the prior-art system of FIG. 1. Thus, not only is a higher back-up voltage applied to the equipment, but the power dissipated in M2 is less than half that dissipated in a comparable silicon diode. (The power dissipated in M2 equals current×voltage drop. For equal currents, the voltage drop in M2 is less than half that in the diode in FIG. 1.) 
     Two. There is a momentary drop in voltage on line Vdd, at the instant when V POWER  drops below V BATTERY , and before M2 is switched ON. However, this drop is considered insignificant, as it lasts for only about 2 micro-seconds. That is, the voltage received by the equipment, namely, the voltage on Vdd, momentarily drops to a first value, at which time the flip-flop switches states, and drives M2 into conduction, thereby pulling Vdd to a higher voltage. 
     Three. During normal operation, when the power supply is functioning properly, the circuit of Figure is consuming an insignificant amount of current. That is, currents 30 and 31 in FIG. 2 total less than about 1 micro-amp. The primary reasons for the low current consumption are (1) Q2 is OFF, (2) Q1 is barely conducting (V BE  is about 0.2 volts), and (3) the remaining components, including the flip-flop, are FET devices, which are all OFF. 
     Numerous substitutions and modifications can be undertaken without departing from the true spirit and scope of the invention. What is desired to be secured by Letters Patent is the invention as defined in the following claims.