![]() This resistance may, or may not, influence the result. Whenever we study the input and output resistance of a network, it is advised by many to always take into consideration the source resistance. I introduced the source load in calculations for 2 reasons.ġ. As with saw in a previous article, it has a unity gain, making it an ideal circuit to isolate different stages when designing electronic circuits. With a small output resistance and a large input resistance, the common-collector amplifier is mostly used as a buffer. If R E is 1 kΩ, and β+1 is 200 Ω, one can easily calculate that R source+r ∏ can be neglected in the r input equation. How small? R source can be around 10 kΩ, while r ∏, is around 1 kΩ. In many texts, R source+r ∏ is neglected, because it is a lot smaller than (β+1) R E. This equation shows that the common-collector amplifier has a large input resistance, due to the product (β+1) R E. So the small signal input resistance of the common-collector amplifier is As such, from v test, R source, r ∏, R E loop, v test can be written as in the following equation: Inspecting the circuit in Figure 3, we can see that i test = i b.Īlso, we notice that R E receives two currents, i test and β i test. For this task, let’s replace the transistor with its small-signal equivalent as in Figure 3.Īccording to Ohm’s Law, r input is calculated by dividing the test source voltage at its current. Its value can be easily derived if we know the test source and the current it sources into the circuit under test. The input resistance is usually calculated with a test source connected at the amplifier input. The output resistance depends mainly on the source resistance R source, the transistor input resistance r ∏, and it is small, since these two resistor values are divided by a large number, β+1. Which is the equation shown in most articles or textbooks. If we neglect (R source+r ∏)/R E, r out becomes This shows that R E influence is small in the calculation of the common collector output resistance or, at most, it decreases r th, which is exactly what we need for a circuit output. ![]() Judging by the usual resistor values of R source, r ∏ and R E, the ratio (R source+r ∏)/R E is small as compared to β+1. Therefore, r out can be written as in the following equation. If we short-circuit this source, the circuit output resistance as seen from R E is R E in parallel with r th. In Figure 2, v oc is an independent source. The calculation of the common-collector output resistance reduces to the calculation of the output resistance of this circuit, because the two circuits are equivalent. In the same article, I showed that the common collector amplifier is equivalent to a Thevenin source that feeds R E (see Figure 2). In a previous article, Derive the Transfer Function of the Common Collector Amplifier with Thevenin’s Theorem, I used Thevenin’s Theorem to demonstrate, step by step, how to derive the small-signal transfer function of this amplifier. Perhaps 95% or 98% of the original signal is all you can expect? But nothing greater than 100%.The output resistance of this amplifier is the resistance seen by the next stage, as looking to the emitter resistor R E, as in Figure 1. So the emitter will just be about \$700\:\text\$ needs to source a lot more to whatever is connected there. The still more simplified circuit becomes:įrom this, you should be able to easily see that the emitter will be about a diode drop below the base voltage - no matter what the base voltage is. In your circuit, though, you directly drive the base of \$Q_1\$ with a voltage source. Simulate this circuit – Schematic created using CircuitLab (The base pair is terribly sized, for example.) But let's take it slowly. You have lots of problems in your understanding with this circuit.
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