The first step in this experiment is to measure and record the common emitter characteristic curves for a silicon NPN 2N1480 transistor. These are diffused junction transistors with a fairly small base current gain.
So far we have been concerned only with the properties of the transistor itself. In practical amplifier applications, it is desirable to design a circuit whose properties are predictable, ie which don't depend strongly on your accidental choice of a specific transistor. In Fig. 1 we show a very common configuration of base and emitter biasing for the common emitter configuration of amplifier. This circuit is reasonably stable against variations in transistor properties (for b = hfe ranging from 14 to 40, the Q point stays within 20% of VCE @ 6.5 V).
The base-emitter voltage drop for a turned on Si transistor is 0.6 V. Therefore
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The emitter current IE, will be IE = [(VE)/(RE)] from Ohm's law. The base current IB = [(IE)/(b)]. You can load your Thevenin equivalent of the base bias circuit with the current IB and calculate a new VB. You could solve these equations simultaneously to get the exact operating point, but this is seldom necessary. If the design is good (meaning not too sensitive to b, a single iteration will yield a very accurate approximation to the operating point.
From the equation for the collector-emitter circuit:
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Before actually measuring these quantities, it is enlightening to estimate them.
Use the transresistance model, in which the transistor is replaced by an equivalent circuit containing a current generator ic = bib in series with the base emitter resistor rtr (called the transresistance). The equivalent circuit is shown in Fig. 2. By differentiating the basic diode equation (for a p-n junction), it has been shown in class that the dynamic base emitter resistance or ``transresistance'' is inversely proportional to the current IE.
rtr = [(¶V)/( ¶I)] = [(0.026 ohm/A)/(Ie)].
[This is commonly expressed in mixed non-SI units by measuring IE in milliamps IE(mA) and expressing the ratio as; ``rtr = [26/(IE(mA))]''. However, we prefer you to avoid this expression and use the normal equations given above and below with consistent SI units as rtr = [(0.026 ohm/A)/(IE)] which is the same as
rtr = [0.026 volt/(IE)].]
Thus, the transresistance is:
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where we have assumed IC @ IE and the ``2 ohm'' is a rough average value for the ohmic resistance (rb) between the lead and the base emitter junction. Note that an AC signal introduced on the base lead can pass to AC ground through three paths, R1, R2, and through the transistor. So, ri (input impedance) is calculated (with ``||'' meaning ``in parallel'' so you add reciprocals to get the reciprocal of the combination) as:
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