Download ELECTRONICS 4 – Fundamentals of Electronics I

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El 04- Electronics 4
Laboratory Exercise 6 – Ohm’s Law, Part 3
OBJECT: To examine how Ohm’ Law illustrates a voltage "drop" across a resistor
based upon the current through it. This is outside of the applied voltage of the circuit.
MATERIALS: Protoboard; Power supply; Multimeter; Resistors – 1K, 1.5K, 2.2K,
3.3K, 4.7K, 5.6K, 6.8K, 10K, 12K, 15K, 20K.
DISCUSSION: One of the important aspects of Ohm’s Law describes what is casually
referred to as an IR Drop. This is a voltage that appears across a resistor as a result of
current through it, according to the Ohm’s Law formula V = IR. That is, the voltage that
appears across a resistor is based upon the value of the resistor in ohms and the current
that is flowing through it.
While previous exercises have usually provided a voltage directly from a power supply,
in this case we are interested in the voltage that would appear across a resistor
somewhere within a complex circuit. Our example assumes that the resistor in whose
voltage we are interested is just one of many in a complicated circuit that might include
semiconductors and other devices. Most of the time, technicians will make voltage
measurements on individual parts with the understanding that the voltage indicated by
their meter is the result of the device’s placement in the circuit and not directly on the
power supply attached to the circuit elsewhere.
The term IR Drop is an old one that will be clearer to you in the discussion on series
circuits coming up shortly. If current is flowing through a resistor, the device will "drop"
the voltage according to Ohm’s Law so that all the voltages in the entire circuit add up to
0.
In this exercise, we duplicate the idea of a complex circuit by applying a constant voltage
to a constant resistor. We then place the resistor under test, that is, the resistor in which
we are interested, in series with the power supply and fixed resistor. We can then make
measurements on that one device that is connected to a more complicated setup. We are
dealing with the one resistor specifically, not the entire circuit.
PROCEDURE:
1. In this exercise, we are going to make our measurements first and then do the
calculations. This is because our calculations are based on current measurements.
To make use of one multimeter for both current and voltage measurements, we
build the circuit first with the meter set up to measure current. We swap resistors
to get a variety of readings. We then remove the meter as a current meter and use
it as a voltmeter, and, again, we swap the resistors. These procedures will fill out
the second and third columns of the chart. Here is the chart to use:
R2
Measured Current
V2 Measured
V2 Calculated
1K
1.5K
2.2K
3.3K
4.7K
5.6K
6.8K
10K
12K
15K
20K
2. Obtain the supplies and equipment and set these up on the workbench.
3. Connect the circuit below. Set the voltage applied to 20 volts from the power
supply. DO NOT change the power supply settings or the 10K resistor R1 during
the procedures.
4. What range will you use for the milliampmeter? HINT: What is the total current
of the fixed resistor R1 and the largest size for the variable resistor R2 for the
applied voltage?
Range? ______________________
5. Apply power to the circuit. In turn, place a new value of device into the R2
position as directed on the chart. Make a current reading and place this value into
the appropriate location the chart to match the resistor. This would be in the
Measured Current column.
6. When you have made all the current readings, remove power from the circuit –
DO THIS BY removing the lead from the supply – DON’T tweak the power
supply’s knob.
7. Now, remove the current meter from the circuit and replace the break with a
jumper or by moving the leads of the resistors accordingly.
8. Next, set up the meter to read voltage. What range will you use?
_____________________
Attach the meter ACROSS the resistor under test.
9. Again, starting at the top of the chart with the 1K device, complete the circuit
shown here:
10. Apply power to this circuit. For each resistor value, make a voltage reading and
fill in the third column on the chart.
11. When you have completed your readings, remove power from the circuit.
12. Next, calculate the voltage you should have seen across the R2 devices for each
resistor value using the formula V = IR. The value of R2 is given on the devices
and in the table, and the value of current was measured in step 5. Fill in the fourth
column of the chart with these calculated voltages.
13. When completed, restore the supplies and equipment to their regular locations.
Then, answer the following questions in writing for submission to the instructor
for credit.
REVIEW QUESTIONS:
1. If the resistance of a device doubles and the current through it remains the same,
what would we see in the voltage across the device?
2. Is the voltage across R2 equal to the supply voltage? Why?
3. Is the IR Drop you measured directly or inversely proportionate to the resistance
value of the device?
4. How did the calculated values for voltage compare to the measured ones?
5. Define IR DROP.