![]() We have to take the diode voltageĭrop into consideration because the value at the base is 0.7V higher than at the output. For a 12V ceiling and 0.7V floor, the middle value would be (12 - 0.7V)/2 +0.7V= 6.35V. If the base voltage is biased incorrectly or well off the midway, the transistor can have significant clipping This gives a maximum peak-to-peak excursion for the signal and gives it the most room to operate Therefore, we want to bias the base voltage so that it falls in between Why this matters in regards to the voltage that we supply to the base is that whatever voltageĬomes into the circuit has a floor of 0.7V and a ceiling of 12V. The transistor can swing from 0.7V up to 12V. Therefore, with a power supply of 12V and a ground 0.7V, Therefore, 0.7V is the floor of the circuit, and not ground or 0V. The transistor cannot swing fully to ground is because the transistor will not turn on unless the diode is switched on and that happensĪt about 0.7V, the barrier voltage for the internal diode. On the other end of the spectrum, the transistor can swing down to near ground, about 0.7V. Transistor can swing up to 12V, which is V CC. That can be present at the output of the transistor is dependent on the DC voltage range that the circuit has. What value do we want across the base? To answer this, we look at the complete picture. Next, now knowing V CC, we bias the base voltage of the circuit, which we means we select the voltage that we want to fall across theīase of the transistor. ![]() For our circuit, we will choose a reasonable voltage value of 12 for V CC. This is typically 40-60V on most transistors. The transistor can handle on the collector. You can really choose any value from above V B, the base voltage, to the maximum value that V CC, applied to the collector of the transistor. So the first thing is we decide how much voltage we're going to supply this circuit. We will now get into the details of why the various components in the circuit are needed and how to choose the values for them. The voltage amplifier circuit that we will build with a BJT NPN transistor and a few resistors and capacitors Find as close of a value to the calculated capacitor values as possible and use those. The same approximations should be used for the capacitor values. You can simply connect a 1KΩ resistor instead of 1.1KΩ. For all basic purposes, this is not necessary. ![]() So you can either approximate it down to 1KΩ or put a 1KΩ resistor in series with a 100Ω resistor. ![]() This resistor value is close enough and is suitable forĪ 1.1KΩ resistor also you will not find. For example, below we calculate one of the resistors toīe 565Ω. For this circuit, we choose the 2N3904 transistor.įor a lot of the values of the components we actually compute, you won't be able to find those precise values. The transistor that is chosen can really be any BJT NPN transistor. This circuit works as a great voltage amplifier when a more precise and sophisticated op ap IC isn't available or a We will also show the maximum gain that a transistor can get based on the values we choose without clipping. We will show all the components that are necessary to build this circuit as well as how to choose the values of these components. Therefore, in this case, a voltage amplifier is built with all simple, discrete components. ![]() However, with a transistor and the correct biasing, we can produce the same voltage amplification effect of an op amp Voltage amplifiers, many times, are built with op amp circuits. Input 1V into the circuit, we can get 10V as output if we set the circuit for a gain of 10. In this project, we will show how to build a voltage amplifier with a transistor.Ī voltage amplifier circuit is a circuit that amplifies the input voltage to a higher voltage. How to Build a Voltage Amplifier Circuit with a Transistor ![]()
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