domingo, 14 de febrero de 2010

Op Amps as Regulators:






If you need a high quality linear regulator, an op amp can save a lot of effort. In the following demo, you can see that there is a simple zener shunt regulator connected to the positive input of the op amp. This becomes the reference voltage. If the zener is a 6.2 volt device, the reference will be 6.2 volts. Actually the reference voltage will likely be a little more or less than 6.2 volts (due to tolerances and the actual current flowing through the diode). If the voltage is precisely 6.2 volts on the positive input, the output of the regulator (the emitter of the current boost NPN bipolar transistor) will be precisely 6.2 volts. The feedback line from the emitter to the negative input of the op amp allows the op amp to monitor the output and compensate for changing load current. If the load resistor decreases in resistance, the output current increases (because we have a regulated voltage source). Without the feedback, the output from the regulator would likely drop a little. In most cases that would be fine. In some circuits, however, the change in voltage would be unacceptable.













When you push the button in the following demo, the resistance will decrease. You will notice that the regulator output current through the resistor increases in proportion to the fall in resistance. You will also notice that the output voltage is rock solid. If you look carefully, you can see that the output voltage from the op amp increases slightly to increase the current through the base of the transistor (which is needed to maintain the proper output voltage).





























This next section will examine the current flow in the regulator components. In the following diagram, the first thing we set up is the zener shunt regulator. We must select a resistor that will allow enough current to flow through the zener to allow it to operate but the resistor must also limit the current flow through the zener. If too much current flows through the zener, the power dissipation will be too great and it will cause the zener to fail. Since the op amp will not have any appreciable current flow into its positive input, there will be (essentially) no branch current to worry about. The resistor can be chosen solely on the desired current flow through the zener. If we use a diode like a 1N5234 1/2 watt zener diode, we know that it has a breakdown voltage of 6.2 volts and a maximum continuous power dissipation of 1/2 watt. If we use Ohm's Law, we can calculate the maximum allowable current with the following formula:






P = IE or I=P/E




I = 0.5/6.2





I = 0.080 amps (Maximum allowable current)

















Since we don't want to use the diode at its limit, we will chose a value of half of the maximum allowable current flow. This means that we want 40 milliamps of current. To calculate the value of the resistor, we need to know the voltage across the resistor. If we're using a supply voltage of 12 volts and have a zener voltage of 6.2 volts, we know that the voltage across the resistor will be 12 minus 6.2 or 5.8 volts. We can again use Ohm's Law for the calculations.






V = IR or R=V/I (This is designated as Rzener in the following drawing)




R = 5.8/0.040





R = 145 ohms (150 ohms would be close enough)

















Now that we have the zener properly biased, we can look at the rest of the regulator. On the bipolar transistor page we mentioned the transistor's 'beta'. It tells you the DC current gain of the transistor. For this example, let's use a 2N3055 transistor. It is very commonly found in regulated power supplies. This transistor's beta is given as 40-70. For the calculations, we'll use the median value of 55. This means that for every 1 amp of emitter current, we will need 1/55 of an amp of current from the op amp's output. In this regulator, the maximum output current of the op amp will be the limiting factor. Most op amps can not be relied on to deliver more than about 15ma of current. For this example, let's say that we have a load resistance of 10 ohms. If the regulated voltage is 6.2 volts and the load is 10 amps:






I = V/R




I = 6.2/10





I = .62 amps

















This means that the current through the transistor and Rload is 0.62 amps. If the transistor has a beta of 55, the current flow through the transistor's base and Rbasewill be 0.62/55 or 11ma. If Rbase is 100 ohms and the current flow through it is 11ma, the voltage drop would be 1.1 volts. If you remember that the transistor has a 0.7 volt drop from the base to the emitter, you can calculate the voltage that the op amp will have to produce at its output.






Vop amp = Vout+Vbe+VRbase




Vop amp = 6.2+0.7+1.1





Vop amp = 8 volts

















Fuente:
Amplificador Operacional como regulador
Gerald Soto, EES.
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