DP 3-2 The resistance RL in Figure DP 3.2 is the equivalent resistance of a pressure transducer. This resistance is specified to be 200 Ω ± 5 percent. That is, 190 Ω ≤ RL ≤ 210 Ω. The voltage source is a 12 V ± 1 percent source capable of supplying 5 W. Design this circuit, using 5 percent, 1/8-watt resistors for R1 and R2, so that the voltage across RL is

            vo = 4V ± 10%

(A 5 percent, 1/8-watt 100-Ω resistor has a resistance between 95 and 105 Ω and can safely dissipate 1/8-W continuously.)

 

Figure DP 3.2

 

To simplify problem let , which means an open circuit.  Then our circuit becomes

 

From Voltage division, we know

 

Solving for R2

 

 

 

Check power, the voltage across R2 is 8 V s0

 

 

If instead of one 400 Ω we go with 2 200 Ω resistors in series, then each would have 4 V dropped across them.  Their power would be

 

 

So we satisfy the 1/8 W limit on individual resistors.

 

Now check that the power supply can deliver enough power for whole circuit.

 

 

So power is ok, now we need to consider if the resistors are off, can we produce the required voltage within tolerances.

 

V0 will be lowest when R2 is largest, while RL and Vsupply are at their lowest

 

 

 

V0 will be highest when R2 is lowest, while RL and Vsupply are at their largest

 

 

 

 

 

 

Provides

 

 

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This page last updated on January 24, 2019