Linear Voltage Regulators

Transcription

1 8/6/ inearoltageegulators(oeriew).doc / 8/6/ inearoltageegulators(oeriew).doc / inear oltage egulators The schematic below shows a pretty darn good design for a linear regulator. t has good regulation, high output power, and acceptable efficiency (for a linear regulator, that is!). = = ma = emember, a zener diode is just a pn junction diode. t has three operating modes: forward biased, reerse biased, and breakdown. With most junction diodes, we try to aoid breakdown. Howeer, breakdown is the typical operating mode of zener diode! ecall that a zener diode in breakdown (but only in breakdown!) has the following characteristics: > Q: Yikes! Can t we start with something simpler? A: ure. The simplest linear regulator is the shunt regulator. We simply add a zener diode to the preious oltage diision regulator: where is the zener breakdown oltage of the diode (a deice parameter!). n other words, a zener diode in breakdown behaes a lot like a oltage source with a alue o =.

2 8/6/ inearoltageegulators(oeriew).doc 3/ 8/6/ inearoltageegulators(oeriew).doc 4/ z _ Meaning that the regulator (if the zener is in breakdown) can be modeled as: The regulated load oltage is =! This shunt regulator proides both good line regulation, and good load regulation. Howeer, it is particularly inefficient! Q: Why is that? > z _ > = ener CD reakdown Model Thus, the load current (and then some) must flow through resistor. Q: Yikes! The load current might be quite high wouldn t this cause quite the power dissipation in resistor? A: Absolutely! Moreoer, if the load current is small, then all of must flow through the zener diode again creating a power dissipation nightmare! Q: s there some solution to this? A: Yes! We can insert a buffer between the zener diode and the load. = = ma A: From KC it is eident that =. Moreoer, since and are both positie, we conclude that >.

3 8/6/ inearoltageegulators(oeriew).doc 5/ 8/6/ inearoltageegulators(oeriew).doc 6/ ecall that the circuit: in i = is known as a oltage follower. t is essentially an amplifier with a oltage gain of one (i.e., A o =., so out = in ). Q: A oltage gain of one! What good is that?? A: The important aspect of a oltage follower is that its input resistance is ery large the input terminals of an opamp draw almost no current. Perhaps an example would help. Consider this oltage diider, with an opencircuit output oltage of 8. olts. in K K out = in 4 ma 4 ma out = 8 Now let s attach a K load to the output of this circuit; note the K 6 ma output oltage drops to 6, and the current increases to 6 ma. K K out = 6 Q: ut thought the output oltage 3 ma 3 ma of this was 8. ; why is it now 6.?? A: 8. was the opencircuit output oltage the output is 8. only when an open circuit is attached to it! As this example shows, attaching a load to the output will cause the output oltage to change. One way to preent this is with a oltage follower: 8. K K 4 ma 4 ma i = in OpAmp supplies not shown 4 ma K out = 8

4 8/6/ inearoltageegulators(oeriew).doc 7/ 8/6/ inearoltageegulators(oeriew).doc 8/ Q: ut the output of the oltage diider is not connected to an opencircuit in this case it s connected to the input of the oltage follower. Why then does the output oltage not change? A: Ah, but the output of the oltage diider is attached to an open circuit! emember, the input resistance of the opamp is ery high essentially an open circuit. This is why the oltage follower works; it senses the oltage on its input, but does not draw any current from the circuit it is sensing it senses directly the opencircuit output oltage, and then places that oltage across the load. Thus, the current through the load does not hae to come from the original circuit the load current instead is proided by the opamp (through its power supply). An application of this then is with our regulator circuit: = ma OpAmp supplies not shown = Note here that the load current no longer passes through resistor, or the zener diode. Thus, only a small amount of current is required (e.g., ma). This will bias the zener into breakdown, and create a reference oltage the oltage follower to sense. A eference oltage Circuit > ) ( z ref = for Q: ay want a different regulated oltage then =, must find a new zener diode? A: You can, but an easier way would be simply to change the gain of the oltage follower. Consider this circuit: in = out ref =

5 8/6/ inearoltageegulators(oeriew).doc 9/ 8/6/ inearoltageegulators(oeriew).doc / ecall that an opamp is a differential amplifier, such that: out d ( ) = A where and are the oltages on the inerting () and noninerting () input terminals, and A d is the differential gain of the opamp. ecall that this differential gain is typically ery, ery 7 large for example, A = or more. d Q: don t understand. After all, a gigantic A d would mean that the output o = Ad ( ) would likewise be ery large thus saturating the amplifier unless of course the inputs were approximately equal ). ( Thus, if the output of the opamp is not saturated, the condition must be true! o, returning to this circuit, we find from oltage diision: From K: ( ) = out = in ecause of irtual short: = in out And of what use is a differential amplifier if the inputs must be approximately equal?? A: Opamps are generally not implemented by themseles! nstead, they typically are but one component of many in a feedback amplifier (see the aboe circuit as an example). n these applications, we will indeed find that but we will also find that this is a desirable condition! The condition is known as a irtual short. f this is not true, the output of the opamp will be saturated. Combining these equations, we find: ( ) ( ) = = in out out in Thus, the gain of this noninerting amplifier is: o Ao = = i Q: Hey wait! thought you said the gain was something really 7 huge, like??

6 8/6/ inearoltageegulators(oeriew).doc / 8/6/ inearoltageegulators(oeriew).doc / A: The differential gain of the opamp (i.e., A d ) is really huge, but do not confuse an opamp with the amplifier. Our amplifier consists of an opamp and two resistors. An opamp is but one element of this feedback amplifier an amplifier whose gain is a mere! Combining these elements, we now hae a regulator whose oltage is adjustable by the proper selection of resistors and : = OpAmp supplies not shown ( ) = Q: Yikes! A oltage regulator will likely need to delier more load current than that; is there some solution to this problem? A: Absolutely! We can add a pass transistor to the circuit in this way the load current will not need to pass through (and thus limited by) an opamp. Collector Typically, this pass transistor is a npn ipolar Junction Transisitor (JT). ecall that a npn JT is constructed such that it has two pn junctions. We call these junctions the Collector ase Junction (CJ) and the Emitterase Junction (EJ). i ase C E Emitter i C CE i E = Note in the circuit aboe that the load current is deliered by the opamp. This can actually be a bit of a problem, as the current at the output of an opamp is typically limited to a rather moderate amount(e.g., 5 ma). Each of these junctions can either be forward biased or reerse biased, and therefore there are four possible operating mode of a npn JT. Each mode has its own name:

7 8/6/ inearoltageegulators(oeriew).doc 3/ 8/6/ inearoltageegulators(oeriew).doc 4/ The reerse actie is typically not used the three fundamental bias states of an JT are thus Cutoff, Actie, and aturation. Each of these modes can be described mathematically (or, at least approximately so). Cutoff MODE EJ CJ Cutoff eerse eerse Actie Forward eerse eerse Actie eerse Forward aturation Forward Forward An npn JT in cutoff exhibits these properties: i = ic = ie = E < C > Actie An npn JT in actie mode exhibits these properties: E 7. i = β i 7. CE > i > C A transistor is somewhat like a ale used to control liquid current. n this analogy we find: This is (sort of) like this Electric current iquid current oltage Pressure Collector ale nput Emitter ale Output ase ale Control Knob Collector Collector Current i C (liquid) aturation An npn JT in saturation exhibits these properties: ase CE ase P CE Pressure ic < β i 7.. E CE Emitter Emitter Cutoff is analogous to the alue being completely closed no current will flow through the alue, regardless of how much pressure ( CE ) is applied.

8 8/6/ inearoltageegulators(oeriew).doc 5/ 8/6/ inearoltageegulators(oeriew).doc 6/ aturation is analogous to the alue being completely open it takes almost no pressure ( CE ) to get a lot of current to flow through the ale. Note from KC that: = C Actie mode is analogous to haing the alue partially open it requires some pressure ( CE ) to get current to flow. Moreoer, this current can be increased by further opening the ale (increasing base current i ) or decreased by further closing the ale (decreasing base current i ). t is the actie mode that we desire for the pass transistor in our regulator circuit. The base of the npn is attached to the opamp output, the emitter to the load, and the collector to the source oltage: Typically we find that << C and <<, so that: C n other words, the load current no longer comes from the opamp. nstead, the load current is deliered directly from the source (e.g., the output of the power supply filter). This means that the load current can potentially be ery big. Howeer, there is still one problem: = = C = β CE = ( ) = This big load current must trael through the npn JT! As a result, we potentially hae a major power dissipation problem in the JT. The power absorbed by the JT is approximately: ( ) P JT CE E This is the major cause of the inefficiency associated with this regulator. f we hae designed our regulator well, then the JT should be the only deice that dissipates much heat. n other words, the efficiency of this regulator is approximately:

9 8/6/ inearoltageegulators(oeriew).doc 7/ 8/6/ inearoltageegulators(oeriew).doc 8/ where P =. P η = P P We can also conclude from conseration of energy that the power deliered by the input source is: JT P PJT P = ( ) = = a result that is hopefully eident when examining the circuit aboe! Q: ut how does the opamp know what this base current should be? After all, it has no knowledge of what the JT parameter β is! True, but what it does know (indirectly) is what the load oltage is. The load oltage is diided by resistors and, this oltage is then sensed as the inerting input oltage. emember, if the load oltage is at its proper leel, then = = the irtual short is enforced! Howeer, if the load oltage drops below its regulated alue (for example, if the load instantaneously decreases), then: Q: o what about the current at the output of the opamp? What is its purpose now? The opamp increases or decreases its output oltage until the base current becomes the correct alue to proide the required load current of β. n other words, the base current controls our ale increasing or decreasing the base current (metaphorically) adjusts the ale control knob one way or the other. < =, causing the opamp output oltage to increase, causing the base current to increase, causing the load current to increase, causing the load oltage to increase, causing to increase until we again arrie at the steady state condition wherein the irtual short is enforced = = ). (

10 8/6/ inearoltageegulators(oeriew).doc 9/ 8/6/ inearoltageegulators(oeriew).doc / = C = β CE = urns told me to keep the pressure at 4.3 psi, but for some reason the pressure has dropped. better open the alue a bit more, until the pressure increases back to 4.3 psi. Analogous to eference oltage Analogous to unregulated supply < < Analogous to Differential OpAmp Analogous to JT Of course, if > =, the opposite set of reactions will occur, causing the regulated oltage to drop to its correct alue. This system demonstrates the beauty of feedback. You do not need to know precisely how much to adjust the knob ; since you hae a way of measuring (sensing) the alue you wish to control. You just keep turning until the quantity returns to the desired alue! Analogous to Feedback oltage Note here that water pressure is analogous to oltage; water flow is analogous to electrical current. Analogous to oad (with change current requirement!) Current