At the beginning of this chapter, we saw how could be used as switches, operating in either their “saturation” or “cutoff” modes. In the last section we saw how transistors behave within their “active” modes, between the far limits of saturation and cutoff.
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Because transistors are able to control current in an analog (infinitely divisible) fashion, they find use as for analog signals. Transistor as a Simple SwitchOne of the simpler transistor amplifier circuits to study previously illustrated the transistor’s switching ability. (Figure )NPN transistor as a simple switch.It is called the common-emitter configuration because (ignoring the power supply battery) both the signal source and the load share the emitter lead as a common connection point shown in Figure.
This is not the only way in which a transistor may be used as an amplifier, as we will see in later sections of this chapter.Common-emitter amplifier: The input and output signals both share a connection to the emitter.Before, a small solar cell current saturated a transistor, illuminating a lamp. Knowing now that transistors are able to “throttle” their collector currents according to the amount of base current supplied by an input signal source, we should see that the brightness of the lamp in this circuit is controllable by the solar cell’s light exposure. When there is just a little light shone on the solar cell, the lamp will glow dimly. The lamp’s brightness will steadily increase as more light falls on the solar cell.Suppose that we were interested in using the solar cell as a light intensity instrument. We want to measure the intensity of incident light with the solar cell by using its output current to drive a meter movement.
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It is possible to directly connect a meter movement to a solar cell (Figure ) for this purpose. In fact, the simplest light-exposure meters for photography work are designed like this.High intensity light directly drives light meter.Although this approach might work for moderate light intensity measurements, it would not work as well for low light intensity measurements. Because the solar cell has to supply the meter movement’s power needs, the system is necessarily limited in its sensitivity. Supposing that our need here is to measure very low-level light intensities, we are pressed to find another solution.
Voltage Series Feedback Amplifier Derivation
Transistor as an AmplifierPerhaps the most direct solution to this measurement problem is to use a transistor (Figure ) to amplify the solar cell’s current so that more meter deflection may be obtained for less incident light.Cell current must be amplified for low intensity light.Current through the meter movement in this circuit will be β times the solar cell current. With a transistor β of 100, this represents a substantial increase in measurement sensitivity. It is prudent to point out that the additional power to move the meter needle comes from the battery on the far right of the circuit, not the solar cell itself. All the solar cell’s current does is control battery current to the meter to provide a greater meter reading than the solar cell could provide unaided.Because the transistor is a current-regulating device, and because meter movement indications are based on the current through the movement coil, meter indication in this circuit should depend only on the current from the solar cell, not on the amount of voltage provided by the battery. This means the accuracy of the circuit will be independent of battery condition, a significant feature! All that is required of the battery is a certain minimum voltage and current output ability to drive the meter full-scale. Voltage Output due to C urrent Through a L oad ResistorAnother way in which the common-emitter configuration may be used is to produce an output voltage derived from the input signal, rather than a specific output current.
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Common Emitter Bjt Amplifier
Let’s replace the meter movement with a plain resistor and measure voltage between collector and emitter in FigureCommon emitter amplifier develops voltage output due to current through load resistor.With the solar cell darkened (no current), the transistor will be in cutoff mode and behave as an open switch between collector and emitter. This will produce maximum voltage drop between collector and emitter for maximum V output, equal to the full voltage of the battery.At full power (maximum light exposure), the solar cell will drive the transistor into saturation mode, making it behave like a closed switch between collector and emitter. The result will be minimum voltage drop between collector and emitter, or almost zero output voltage. In actuality, a saturated transistor can never achieve zero voltage drop between collector and emitter because of the two PN junctions through which collector current must travel. However, this “collector-emitter saturation voltage” will be fairly low, around several tenths of a volt, depending on the specific transistor used.For light exposure levels somewhere between zero and maximum solar cell output, the transistor will be in its active mode, and the output voltage will be somewhere between zero and full battery voltage. An important quality to note here about the common-emitter configuration is that the output voltage is inverted with respect to the input signal. That is, the output voltage decreases as the input signal increases.
Are you constrained in the methods you can use to solve this problem?Most of your calculations that I compared to my own were close to the same as mine. There would be better agreement if you would carry more significant digits in your calculations; I carried 12 digits in mine.One value which I think you have wrong is Rif. The various components of Ri are not all multiplied by the (1 + Av.B) factor. For example, the resistance Rs is not increased by the feedback.Another interesting thing is that the ratio Av/Af is apparently not the same as Ri/Rf.You might have to calculate a complete solution to the entire circuit when the feedback resistor is connected.
I don't think that application of feedback factors to parameters such as Av and Ri will correctly give the corresponding Af and Rf. Click to expand.The problem with the standard formalism (the A and B blocks) is that the gain (A) and feedback (B) blocks are assumed to be unilateral.
That is, there is no transmission of signal in the reverse direction. In the case of the feedback block, this can be seriously in error. The circuit you are analyzing here has signal transmission from the emitter of the first transistor to the output through the 6800 ohm feedback resistor. This causes the actual values of various parameters (Af, Rif, etc.) to be different from what you would get if the feedback resistor could only transmit signals in one direction.
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