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Lecture Notes for Analog Electronics - The Electronic Universe ...

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1.5 <strong>The</strong>venin <strong>The</strong>orem (contd.)<br />

Class <strong>Notes</strong> 2<br />

Recall that the <strong>The</strong>venin <strong>The</strong>orem states that any collection of resistors and EMFs is equivalent<br />

to a circuit of the <strong>for</strong>m shown within the box labelled “Circuit A” in the figure below.<br />

As be<strong>for</strong>e, the load resistor RL is not part of the <strong>The</strong>venin circuit. <strong>The</strong> <strong>The</strong>venin idea,<br />

however, is most useful when one considers two circuits or circuit elements, with the first<br />

circuit’s output providing the input <strong>for</strong> the second circuit. In Fig. 6, the output of the<br />

first circuit (A), consisitng of VTH and RTH, is fed to the second circuit element (B), which<br />

consists simply of a load resistance (RL) to ground. This simple configuration represents, in<br />

a general way, a very broad range of analog electronics.<br />

V TH<br />

Circuit A<br />

R TH<br />

1.5.1 Avoiding Circuit Loading<br />

Vout<br />

Figure 6: Two interacting circuits.<br />

Circuit B<br />

VTH is a voltage source. In the limit that RTH → 0 the output voltage delivered to the load<br />

RL remains at constant voltage. For finite RTH, the output voltage is reduced from VTH by<br />

an amount IRTH, whereIis the current of the complete circuit, which depends upon the<br />

value of the load resistance RL: I = VTH/(RTH + RL).<br />

<strong>The</strong>re<strong>for</strong>e, RTH determines to what extent the output of the first circuit behaves as an<br />

ideal voltage source. An approximately ideal behavior turns out to be quite desirable in most<br />

cases, as Vout can be considered constant, independent of what load is connected. Since our<br />

combined equivalent circuit (A + B) <strong>for</strong>ms a simple voltage divider, we can easily see what<br />

the requirement <strong>for</strong> RTH can be found from the following:<br />

�<br />

�<br />

RL<br />

Vout = VTH<br />

RTH + RL<br />

5<br />

=<br />

VTH<br />

1+(RTH/RL)<br />

R L

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