This way, the AC input signal will be perfectly “centered” between the amplifier’s high and low signal limit levels.Ĭlass A: The amplifier output is a faithful reproduction of the input.Ĭlass B operation is what we had the first time an AC signal was applied to the common-emitter amplifier with no DC bias voltage. To achieve this, sufficient DC bias voltage is usually set at the level necessary to drive the transistor exactly halfway between cutoff and saturation. ![]() Class A operation can only be obtained when the transistor spends its entire time in the active mode, never reaching either cutoff or saturation. Although I didn’t introduce this concept back in the common-emitter section, this is what we were hoping to attain in our simulations. Amplifier class operation is categorized with alphabetical letters: A, B, C, and AB.įor Class A operation, the entire input waveform is faithfully reproduced. Because it is possible to operate an amplifier in modes other than full-wave reproduction and specific applications require different ranges of reproduction, it is useful to describe the degree to which an amplifier reproduces the input waveform by designating it according to class. In fact, some applications may necessitate this very kind of amplification. This addition was called a bias voltage.Ī half-wave output is not problematic for some applications. The solution to this problem was to add a small bias voltage to the amplifier input so that the transistor stayed in active mode throughout the entire wave cycle. Since our purpose at that time was to reproduce the entire waveshape, this constituted a problem. In the common-emitter section of this chapter, we saw a SPICE analysis where the output waveform resembled a half-wave rectified shape: only half of the input waveform was reproduced, with the other half being completely cut off. The enclosed drawing shows the construction of the stabilization line.\) The actual DC operating point exist where both curves cross each other - the exponential BJT characteristic and the linear R2 characteristic Ic=f(Vb-Vbe), which has a negative slope. In this case, both graphs can be plotted in one single diagram - and it becomes clear that Ic does depend only very little on the temperature-dependent Ic=f(Vbe) characteristic. This is nothing else than to plot the ohmic law for the resistor R2 - expressed not by the emitter voltage Ve but by Ve=Vb-Vbe. It is very easy to visualize this effect using the method of "stabilizing line" in the Ic=f(Vbe) characteristic. As a consequence of the negative feedback, the collector current does not depend too much from the base-emitter voltage, which normally is chosen (for calculating purposes) to be app. This is the typical design approach because - at the same time - the emitter resistor R2 provides current controlled voltage feedback (voltage Ve) which can work satisfactorily only for a "stiff" base voltage Vb. Without these constraints, it would be even better to chose a factor between both currents of 20 or 50 instead of only 9). (Due to some other constraints like power consumption and input resistance the voltage divider must not be much lower in resistance. ![]() That means: As independent as possible on the base current which is connected with large tolerances (due to B variations). In general, the voltage divider is chosen with the aim to produce a base voltage Vb "as stiff as possible". For this example, we will assume 9 times the base current for a total of. As far as I understand your problem, the key point is in the sentence (quote): " In order to provide a stiff base voltage, resistor R4 should have a current of about 5 to 10 times the base current.
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