![]() Because the loading (R 3,R 4 output divider) was accounted for in the first-stage gain, the second-stage gain input quantity is the Q 2 base voltage, v B2 = v o1. Then the second-stage gain is calculated from the collector of Q 1 which is the output of the first stage. First way: the gain of the first stage is calculated including the loading of the R 3,R 4 resistor divider. The total voltage gain can be calculated in either of two ways. ![]() Because the input resistance of the second stage (resistors R 3 and R 4) forms a voltage divider with the output resistance (R C1) of the first stage, the total gain is not simply the product of the gain for the individual (separated) stages. Two cascaded common emitter stages are shown in figure 10.1.2. The complication in calculating the gain of cascaded stages comes from the non-ideal coupling between stages due to loading. We generally refer all noise to the input of the signal chain, taking out the effects of the gain stages.ġ0.1 Cascade of two single transistor stagesįigure 10.1.2 DC coupled Common Emitter stages Even if AC-coupled, noise from preceding stages gets amplified by each downstream amplifier stage, making for nothing but a noise source after a while. There are practical reasons why you just can't continue cascading stages “forever…” If DC-coupled, real-world offsets can be impossible to trim out. By the time you reached that point, other adverse effects would have caused much more trouble, for example, the fact that noise from each successive stage is added to the noise coming into that stage and is further amplified on down the cascade of amplifiers. ![]() In fact, you would have to go to a cascade of 100 stages with these specifications before you even lost 1% of the expected ideal gain ( i.e. To confirm this assertion, assume a low performance op amp with R out = 100Ω and R in = 1MΩ, what is the gain with two stages of gain A 1 and A 2 in series? (assume R L = 1 MΩ) That effect, if any, is modeled in the R out.įor most integrated circuit amplifiers where R in is in the MΩ to GΩ range, and R out is in the 50 to 100 Ω range, the gains are pretty close to being the simple product of the gain stages. The above equations assume that the individual amplifier gains, A do not change with output loading. As a matter of fact, we really only need R out to go to 0 to have the resistor dividers to go to 1. If you try to design gain stages using Ib you will run into alot of trouble since you would have to measure every transistor's Hfe in your design.As we would expect, the overall equation reduces to the ideal case of A V = A 1*A 2 for two ideal stages when we let the R out go to 0 and the R in go to infinity. And we have a expression for the Ic current as a function of only the voltage divider bias and the emitter resistor since we can regard the BE junction drop as 0.7 for most applications. See, we never used Hfe in these calculations. Ie = Ve / Re = (Vbase-0.7V) / Re = (Voltage_divider_bias - 0.7V) / ReĪnd for Hfe >= 100 we see that Ie is within 1% or less from being as large as Ic. Since the Emiter current is approximately the Ic current we see that we can set the Ic current by simply choosing the correct value of the emitter resistor as long as we know how we set the voltage divider bias Which means that the emitter voltage will be Vbase-0.7V. The BE junction of a conducting transistor is around 0.7V over most of its normal operating range. The base voltage is decided by the voltage divider is it not ?. But because the Ib current is so small it hardly affects the voltage divider at all. Ib will set itself automaticly up by drawing the current it needs from the voltage divider to maintain Ic at the value set by the emitter resistor. The use of voltage divider bias enables you to do the opposite.
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