Download Electrical Measure and Test P3 Task 2 As always I want to give you

Electrical Measure and Test P3 Task 2
As always I want to give you a bit of background information so that you appreciate some of the
fundamentals involved. This exercise involves using test equipment to find out the internal
resistance of the laboratory power supply.
The lab power supply is the Thurlby PL320 with a maximum output voltage of 30 volts and
output current of 2A.
Consider the following diagram:
o The voltage source the is voltage supplied by the power supply
o The internal resistance is the resistance of the power supply itself
o The electrical load is the application that we are driving using the power supply e.g. a
computer
o The ammeter which is actually part of the power supply itself measures the current
drawn by the circuit. Ammeters are always connected in series
o The voltmeter measures the voltage available to the electrical load (computer)
voltmeters are always connected in parallel
Now here’s the thing! Think about when you turn you computer on. At first when you are
running no applications the current draw will be relatively little. However imagine you power
up the latest 3D game. You will probably hear the cooling fans kicking into overdrive as the CPU
begins to draw close to maximum current from the supply. Therefore electronics applications
cannot really be modeled as a constant resistance because the current being drawn is changing
all the time.
The power supply has to be able to supply all of this current over a large range without the
output voltage dropping too much. This specification is called Load Regulation.
The load regulation specification for the Thurlby PL310 is give as < 0.01 of the maximum output
for a 90% load change. From the data sheet, remember this because we will be visiting it a little
bit later on.
Now back to our original circuit, let’s apply a little bit of Ohm’s law to prove the importance of
load regulation and the internal resistance of the power supply:
Imagine the following:
o Supply voltage = 5 volts
o Internal resistance = 5Ω
o External Resistance = 50Ω (best case scenario)
o External resistance = 5Ω (worst case scenario)
Firstly we find the total resistance of the circuit. For a series circuit we simply add both of the
resistances together:
RT = 5 + 50 = 55Ω
The current (I) drawn by this circuit is given by:
I = V/RT
= 5/55
I = 0.09 Amps
To find the voltage dropped across each of the resistances we need to rearrange Ohm’s law to
make voltage the subject:
I = V/R
Rearranged
V=IxR
The voltage across the internal resistance is given by
V=IxR
= 0.09 x 5
= .45 volts
The voltage across the electrical load is given by
0.09 x 50
= 4.5 volts
Therefore although the power supply is making 5 volts available, only 4.5 volts reaches the
electrical load (computer). The situation gets far worse if we use the worst case scenario
external load. Try the calculation for yourself and then try to imagine how your computer
would react to a supply voltage jumping all over the place like that. I can assure you it would
not like it one bit.
This is why the internal resistance of a power supply is such an important specification. Given
the calculations above and what they show. What do you think the internal resistance of an
ideal power supply should be?
Setting up the Experiment
Now I’m going to show you how to set up the experiment and why it won’t work as we might
have hoped.
Initially we would set up the circuit as described at the beginning of the document.
However we need to find an electrical load 30/2 = 15Ω in order to test the power supply in the
worst case scenario when it is supplying its full rated current of 2A.
In the worst case scenario (current at 2A, voltage at 30 volts) the power dissipated in the
electrical load can be worked out using any of the following:
o Power = Volts x Current = V x I = 30 x 2 = 60 Watts
o Power = I2R = 2 x 2 x 15 = 60 Watts
o Power = V2/R = (30 x 30)/R = 900/15 = 60 Watts
If you try using any of the lab resistors on their own you will fry them in an instant. Even if you
try a large parallel combination, for example say 20 resistors in parallel you will fry them. As
shown in the next photograph.
Please see next page
Moral of the story
Things don’t always work out as planned in the field of engineering. Keep an open mind and
try to apply the things you have learned.
Resistors cooking
This is because the standard lab resistors have a maximum power dissipation of .25 watts.
Therefore to test the lab supply you would need:
60/.25 = 240
240 standard resistors in parallel
Therefore testing the power supply internal resistance in this way is not really feasible because
to do it properly you need to use the full rated current of the supply.
Therefore I propose
You set up the same experiment but use a 1.5 volt battery as the power source
Check the batteries maximum rating in mA hours. If there is no rating printed you can use this
source of information to decide on a suitable current draw for the battery
http://www.allaboutcircuits.com/vol_1/chpt_11/3.html I would suggest perhaps 100 mA
Select an electrical load that will draw the chosen current from the battery (bearing in mind the
power dissipation limitations incurred by the laboratory resistors)
When a battery is not connected to a circuit we say it is either off load or in a no load condition
Use a voltmeter to measure the no load terminal voltage of the battery as shown here.
Connect the circuit as shown. Measure the current drawn and the voltage dropped across the
resistor. Use Ohm’s law to calculate the approximate internal resistance of the battery.
Finally (You don’t need to read this it’s for interest only)
There is another way to calculate the approximate internal resistance of the lab power supply.
It goes back to the load regulation figure. For the Thurlby the stated load regulation is 0.01% of
the full rated output for a 90% load change.
The absolute minimum load for this power supply is 15Ω which would draw the maximum
current from the device.
If the power supply was driving a 28.5Ω load, then a switch to 15 Ω would represent a change in
load of 90%.
The absolute maximum expected change in output voltage is given as 0.01 x 30 = 0.3 volts
In the first instance:
30/28.5 = 1.052 Amps
In the second instance, assuming a terminal voltage change of 0.3 volts caused by a 90% change
in load:
29.7/15 = 1.98 Amps
Since Ohm’s law for resistance = V/I we can use the changes in voltage and current to find the
approximate internal resistance of the power supply
R = ∆V/∆I (Where ∆ means the change in a given value)
0.3/(1.98 – 1.052)
0.3/0.928 = 0.323 Ω
So even in a worst case scenario we can see that the internal resistance of this power supply is
pretty good, not very far away from the ideal.
Hope you found it interesting and informative and that it helps with P3 task 2