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"Chapter 1 - The Op Amp's Place in the World" - HTL Wien 10

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Current Divider Rule<br />

2-4<br />

put voltage. Remember that <strong>the</strong> voltage divider rule always assumes that <strong>the</strong> output resistor<br />

is not loaded; <strong>the</strong> equation is not valid when <strong>the</strong> output resistor is loaded by a parallel<br />

component. Fortunately, most circuits follow<strong>in</strong>g a voltage divider are <strong>in</strong>put circuits, and<br />

<strong>in</strong>put circuits are usually high resistance circuits. When a fixed load is <strong>in</strong> parallel with <strong>the</strong><br />

output resistor, <strong>the</strong> equivalent parallel value comprised of <strong>the</strong> output resistor and load<strong>in</strong>g<br />

resistor can be used <strong>in</strong> <strong>the</strong> voltage divider calculations with no error. Many people ignore<br />

<strong>the</strong> load resistor if it is ten times greater than <strong>the</strong> output resistor value, but this calculation<br />

can lead to a <strong>10</strong>% error.<br />

2.4 Current Divider Rule<br />

When <strong>the</strong> output of a circuit is not loaded, <strong>the</strong> current divider rule can be used to calculate<br />

<strong>the</strong> current flow <strong>in</strong> <strong>the</strong> output branch circuit (R 2). <strong>The</strong> currents I 1 and I 2 <strong>in</strong> Figure 2–6 are<br />

assumed to be flow<strong>in</strong>g <strong>in</strong> <strong>the</strong> branch circuits. Equation 2–9 is written with <strong>the</strong> aid of Kirchoff’s<br />

current law. <strong>The</strong> circuit voltage is written <strong>in</strong> Equation 2–<strong>10</strong> with <strong>the</strong> aid of Ohm’s law.<br />

Comb<strong>in</strong><strong>in</strong>g Equations 2–9 and 2–<strong>10</strong> yields Equation 2–11.<br />

Figure 2–6. Current Divider Rule<br />

I I 1 I 2<br />

I1 I2<br />

I R1 R2<br />

V I 1 R 1 I 2 R 2<br />

I I 1 I 2 I 2<br />

R2 R1 V<br />

I2 I2R1 R2 R1 Rearrang<strong>in</strong>g <strong>the</strong> terms <strong>in</strong> Equation 2–11 yields Equation 2–12.<br />

I2 I R1 R1 R2 (2–9)<br />

(2–<strong>10</strong>)<br />

(2–11)<br />

(2–12)<br />

<strong>The</strong> total circuit current divides <strong>in</strong>to two parts, and <strong>the</strong> resistance (R 1) divided by <strong>the</strong> total<br />

resistance determ<strong>in</strong>es how much current flows through R 2. An easy method of remember<strong>in</strong>g<br />

<strong>the</strong> current divider rule is to remember <strong>the</strong> voltage divider rule. <strong>The</strong>n modify <strong>the</strong> voltage<br />

divider rule such that <strong>the</strong> opposite resistor is divided by <strong>the</strong> total resistance, and <strong>the</strong> fraction<br />

is multiplied by <strong>the</strong> <strong>in</strong>put current to get <strong>the</strong> branch current.

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