A Lab Tutorial
Chris Murphy M0HLS introduces us to Jeff as he walks a colleague through some basic electrical theory.
Chris Murphy M0HLS introduces us to Jeff as he walks a colleague through some basic electrical theory.
Jeff walked back into the electronics lab having replenished his huge mug with fresh coffee for the second time that lunchtime. Walking back to the far end of the lab where he and several other engineers had their desks, he walked past the lab’s newest recruit Natalie, an apprentice who’d joined them three weeks ago. As Jeff walked past her he noticed that Natalie was engrossed in what looked to be some paperwork containing electronic calculations.
“You’re looking very studious today young Natalie” Jeff remarked as he passed her. “No Facebook to look at?” “Oh” said Natalie looking up from her notes. “I was just looking at some notes that we were given at college yesterday. We have four subjects this year, Maths, Engineering Science, Electronic Principles and General Studies – how to write reports and all that sort of thing. I plan to go through the notes again at lunchtimes – one subject each day apart from Wednesdays, which is college day”
“Oh yes, it was your first day at college yesterday wasn’t it. How did it go?” Jeff enquired.
“Not too bad” replied Natalie, “A bit boring really though since it was mainly about Ohm’s Law, which I’d already done at school”. “Oh”
said Jeff “Then since you’ve done Ohm’s Law at school and now at college, you’ll have no trouble in telling me what Ohm’s Law states”. “Sure I can” said Natalie, “Ohm’s Law says that the current through a conductor is directly proportional to the voltage applied across that conductor, Like this”. Natalie pulled her notepad towards her and drew a graph, Fig 1.
“Well, yes it does imply that” said Jeff.
“Anything else though?” “Not really” said
Natalie with a frown. “I suppose that you can juggle the words about to say that the resistance of a conductor is equal to the voltage divided by the current, or to put it in mathematical terms R equals V over I. In fact, there’s a magic triangle that you can use to find either voltage, current or resistance if you know the other two” she said and drew a triangle as
shown in Fig. 2. “If you cover up the unknown quantity, you’re left with the answer of what to do with the other two. For example, if you want to find the voltage you cover up the letter V and you’re left with I and R. So, V equals I times R.”
Ohms’ Law
“Mmmm” mused Jeff, “some years ago, back in the late 1970s or early 1980s there was an article in a magazine called Practical Wireless about a question that had been presented in the City and Guilds Radio Amateurs exam that asked what Ohm’s Law said. Four choices of answer were given and at least three of them did sort of represent Ohm’s Law but none of them was technically correct. One said the same as what you just said about the current being proportional to voltage and the mathematical equation of R equals V over I was there. But, as I say, none of the answers actually stated Ohm’s Law” said Jeff. “How come”
Natalie asked, “What’s wrong with those answers? I suppose the one using the words is the proper answer”. “Well yes” said Jeff, “It’s nearly right and what it says does hold true
– in many cases anyway. But Ohm’s law also talks about temperature and says that for a metallic conductor the current flowing through it is proportional to the voltage applied to it provided that the temperature of the conductor doesn’t change”.
“The temperature” said Natalie with a
frown, “What’s that got to do with it?” “Metals”
said Jeff “generally have what is called a positive coefficient of temperature. What that means is that as the temperature of a metal increases so will its electrical resistance. Other materials such as carbon and silicon have negative coefficients of temperature, so their resistance decreases with a rise in temperature” “How do you know what this coefficient is?” Natalie asked. “There are tables that list different materials such as copper, aluminium, silver and other metals and also materials such as carbon, silicon, germanium, and glass” said Jeff. “They very often quote the coefficient at a temperature of 20° Centigrade because that is more or less room temperature. Others quote it at other temperatures such as zero degrees Centigrade so be careful” Jeff warned. “The tables also usually state another electrical property called resistivity”. “Yes, I’ve heard of that. It’s also to do with resistance isn’t it, what’s the difference between resistance and resistivity?” Natalie asked.
“Resistivity is a physical property of a material that defines the opposition to a flow of electrons for a sample of that material of a certain length and cross-sectional area and at a standard temperature. The resistivity of a material is usually quoted as being the resistance of the material as measured across two opposite faces of a cube. It is quoted in Ohms per metre, or more commonly for conductors milli-Ohms per metre (mΩ/m). For example, the resistivity of copper at 20º Centigrade is 1.7 x 10-8 Ω/m. The resistivity of silver is 1.6 x 10-8 Ω/m, and glass has a resistivity of around 1012 Ω/m”.
“Oh” said Natalie, “So silver is a better conductor than copper whereas glass is obviously an insulator”. “Correct” said Jeff. “The values of resistivity can be found from tables in data books or from the internet and are given at a standard temperature, often 20º centigrade. It is given the Greek symbol Rho (ρ) Anyway, going back to your question about the differ
ence between resistivity and resistance, I’ve explained that resistivity is a physical property of a material but the resistance of something like an electrical conductor is also dependent upon other things such as its length and how thick it is.”
“You mean like how they give the size of three core cable as being one millimetre squared or two point five millimetres squared”
said Natalie. “Yes” said Jeff. “Basically, the greater the cross-sectional area, or thickness of the cable the less resistance a length of that cable will have. You can work out the resistance of a length of cable using a formula which says that the resistance is equal to the resistivity times the length all divided by the cross-sectional area”. Jeff wrote the formula down:
“Let’s do an example” Jeff said. “Let’s say that we’ve got a length of copper wire one hundred metres long that has a cross sectional area of one millimetre squared. What will its resistance be? Using what we know about the resistivity of copper we can use the formula to work it out as follows. Note how we have to write the cross-sectional area in terms of square metres, not millimetres”. Jeff wrote down the equation:
“So, the resistance of one hundred metres of one-millimetre square copper cable is one point seven Ohms. Such calculations aren’t used an awful lot in electronics” said Jeff, “But they are in things like electrical installation calculations where the resistance of cables is important so that the voltage drop along the cable can be calculated”.
“Incidentally” said Jeff, “There is another property of conductors and insulators that is a sort of opposite or inverse of resistance called conductance, which has the symbol G and is measured in Siemens. And here is the formula for it”. Jeff added yet another formula to his notepad:
“So” said Natalie “a resistance of ten Ohms will have a conductance of nought point one Siemens”. “That’s right” said Jeff “and if you think back to how you calculate resistors in series and parallel, the formulas are reversed. Conductances in parallel simply add up but for conductances in series you need to add the reciprocals as for resistors in parallel” Jeff wrote down two more formulas:
“Also” said Jeff, “When using Ohms Law for finding conductances the formula is turned upside down so instead of using voltage divided by current as we do for resistance, for conductance we use current divided by voltage”.
“But, let’s go back to temperature. As with most things in electronics there’s a formula that is used to describe how the resistance of a material varies with temperature. It looks quite daunting at first but it only requires very basic maths” said Jeff as he scribbled down the following formula on Natalie’s notepad:
“Rx is the change in resistance” explained
Jeff, “and Rs is a standard resistance or resistivity of the material at a reference temperature of say zero degrees Centigrade or 20º Centigrade. The symbol Alpha represents the coefficient value and t is the change in temperature. Both the standard resistance and the coefficient are obtained from the tables that I told you about”. “Best way to explain is by doing an example” said Jeff walking over to his desk where he opened a drawer and took out a book.
“According to this table here copper has a temperature coefficient of 0.004041 per degree Centigrade (4.041 x 10-3 per °C) at 20º Centigrade” said Jeff. “Now let’s assume that we’ve got a length of copper wire whose standard resistance at 20º Centigrade is 5Ω but we need to know what its resistance will be at 100º Centigrade”.
“So, we know the temperature coefficient, the standard resistance, and the change in temperature, which is 80 degrees” said Jeff.
“Let’s put these values into the formula” said
Jeff and wrote the equation:
Natalie picked up her calculator and tapped in the figures. “That works out” she said “to be six point six one six Ohms” and wrote the figures next to Jeff’s equation.
“Of course” said Jeff, “Not all things obey Ohm’s law and these are sometimes called non-Ohmic devices. Sometimes it is the effect of temperature that causes them to be nonOhmic” “Like what” said Natalie, “Give me an example”. “OK” said Jeff, “Diodes, a filament lamp, thermistors”. “Thermistors” said Natalie with a thoughtful look on her face. “I’ve heard of those but I’m not sure what they are or what they do”.
“Thermistors” said Jeff, “Are electronic devices similar to resistors whose resistance changes by a known amount over a certain temperature range. You can get thermistors where the resistance increases as temperature increases or where the temperature decreases as temperature increases. They are said to have positive temperature coefficients and negative temperature coefficients respectively”. “What are they used for” Natalie asked.
“Well, an obvious application” said Jeff “Is for temperature measurement. Another not so obvious application, but quite an important one is temperature compensation”. “What does that mean?” asked Natalie. “Well, let’s say that we have something like an oscillator”
said Jeff “whose frequency is dependent upon the values of some resistors. If the surrounding temperature changes, so will the value of the resistors and even though it might only be a small change it will cause the oscillator to drift or change frequency. If, however a thermistor that has the opposite temperature coefficient to the resistors is incorporated into the circuit, it can be used to compensate for the change in resistance and keep the oscillator frequency stable” “This is the circuit symbol for a thermistor” said Jeff and drew the symbol on his note pad, Fig. 3:
Charge and Power
“Going back to Ohm’s Law” said Natalie with a bemused look on her face, “We were given a question to work through in a form that I’ve never seen before at school. Here it is. Look.”
‘Over a time period of ten seconds a charge of 30 Coulombs in transferred into a circuit consisting of a 5Ω resistor. What is the voltage that is applied to the resistor?’
“Yes” said Jeff looking at the question
“what’s the problem with that?” “Well” said
Natalie, “the question doesn’t say what the current is. Just talks about charge and time” “I see” mused Jeff. “what you need to do here is to go back to what might be called basics. You may have come across this at school”. Jeff picked up his pen and wrote on the notepad:
Charge (Q) = Current (I) x Time (t)
“Charge, Q in Coulombs equals current, I in Amps times time, t in seconds” said Jeff.
“So, since you know the charge and the time you can find the current” explained Jeff. “Manipulating the formula I’ve written will give Current equals Charge divided by the time. With a charge of thirty Coulombs and a time of ten seconds a current of three Amps must have been flowing. Now you can use Ohm’s Law to find the voltage”.
“Oh, right, I see now” exclaimed Natalie, “So the voltage is three Amps times five Ohms.
That gives fifteen Volts” “Correct” said Jeff. “Wasn’t that difficult was it? Anyway, you said that you covered power as well”.
“Yes” said Natalie “Nothing new there though. Just the same formulas that I’d learned at school, which you can use depending upon what you know about the problem – voltage, current, resistance etc” “Oh, you mean these” said Jeff and jotted down three formulas:
P = I xV P = I2 x R P= V2 / R
“Those cover most situations” said Jeff.
“In fact, people taking the radio amateur examinations learn these three equations at various stages of their learning” “Our lecturer said though that these formulas only work for direct current and that we would learn how to calculate power in alternating current circuits later. What is the difference” Natalie asked.
“Yes” said Jeff “Your lecturer is right, or nearly right. These formulas assume that the voltage and current are in phase, which for simple direct current and resistive alternating current circuits is true. But when you start introducing inductors and capacitors the voltage and current are not in phase so it isn’t just a simple matter of multiplying amps and volts to find the power. It’s a little bit more complicated. You will no doubt learn how to find power in alternating current circuits in a later lesson.”
“Oh, OK” said Natalie. “It seemed that everyone knew how to find the power by using voltage and current but a few had never seen the other formulas, which use resistance, so he gave us a couple of worked examples of them” Natalie explained. “They’re here, look”:
If a current of 2mA flows through a 1kΩ resistor, what power will it produce?
P = I2R, hence 0.0022 X 1000 = 0.004W or 4mW
If a current of 0.5A produces a power of 2W in a resistor, what is the value of the resistor?
P = I2R so R = P / I2 so R = 2 / (0.5 x 0.5) = 8Ω
A 240Ω resistor has a voltage of 12V across it. What power will it produce?
P = V2 / R so P = 122 / 240 = 0.6W
What voltage is required to produce a power of 18W in a 56Ω resistor?
P = V2 / R so V = √P x R so V = √18 x 56 = 31.75V
SomeWorked Examples
“For the rest of the lesson, Archie, he’s our lecturer by the way, just went through some more worked examples of Ohm’s Law and power with us” said Natalie. “He says that will be the format of his classes. Worked example in class then two sets of further examples to do in our own time. One set will have worked answers and the other set just the answers. That’s what I was looking at when you came over” “Sorry to have interrupted” said Jeff.
“No prob” smiled Natalie. “Say Jeff, could you write me a few examples of what we’ve talked about today? You know, resistivity and temperature coefficients” “Sure I can” said Jeff “I’ll add them to the end of the ones that you got in class, but make sure that you do the college ones first. Anything that I tell you should be regarded as outside of the curriculum” “Of course” said Natalie. “I suppose we’d better get back to work. Here’s the questions that Archie gave us. Just add yours to the end”
Class worked examples
1. A 30Ω resistor has a voltage of 12V across it. What is the current in the resistor?
I = V/R so I = 12/30 = 0.40A
2. What is the voltage across a 200Ω resistor that has a current of 3mA flowing through it?
V = I x R so V = 0.003 x 200 V = 0.60V
3. A current of 1.3A flows through a resistor when a voltage of 24V is connected across it. What is the value of the resistor?
R = V/I so R = 24/1.3 = 18.46Ω
4. An electrical machine draws a current of 15A from a 230V supply. What is the power consumed by the machine?
P = V x I so P = 230 x 15 = 3450W or 3.45kW
5. A current of 10mA flows through a 47Ω resistor. What power is generated in the resistor?
P = I2 x R so P = 0.0102 x 47 = 0.0047W or 4.7mW
6. A 150Ω resistor has a voltage of 16V connected across it. What power will be generated by the resistor?
P = V2/R so P = 162/150 = 1.71W
Further examples
1. What current will flow through a resistor of 22 Ohms if it has a voltage of 18V across it? (0.818A)
2. What current will cause a voltage of 100V to be developed across a resistance of 35 Ohms? (2.857A)
3. If 0.004A flows through a 2.4kΩ resistor, what is the voltage across it? (9.6V)
4. What is the voltage across a 0.3Ω resistor if a current of 0.25A flows through it? (0.075V)
5. What value of resistor will have a voltage of 50V across it when a current of 0.1A flows? (500Ω)
6. 7 mA flows through a resistor which has a voltage of 4V across it. What is the value of the resistor? (571Ω)
7. An electric heater draws a current of 6A from a 200V supply. What power is produced by the heater? (1200W)
8. What power is produced by a 15Ω resistor when a current of 20mA flows through it? (0.006W or 6mW)
9. A 200Ω resistor has a voltage of 48V across it. What power is produced by the resistor? (11.52W)
10. What current flows through a 1.5kΩ resistor if it produces a power of 0.5W? (18.25mA)
11. What voltage will result in a power of 20W being produced by a 120Ω resistor? (48.99V)
12. A resistor with 150V across it produces a power of 1.25W. What is the value of the resistor? (18kΩ)
Jeff’s extra questions
1. What is the resistance of a cable made from aluminium that is 50 metres long and has a cross sectional area of 4mm2 if the resistivity of aluminium is 2.82 x 108Ω/m (0.353Ω)
2. Calculate the resistance of a piece of Tungsten wire 2 metres long with a cross sectional area of 0.5mm2 if the resistivity of Tungsten is 5.60 x 10-8Ω/m (0.224Ω)
3. If the resistivity of gold is 2.44 x 10-8 Ω/m, what is the resistance of a piece of gold wire 15m long with a cross sectional area of 0.1mm2 (3.66Ω)
4. A conductor has a resistance of 1.3Ω at 20°C. What will its resistance be at 95°C if the temperature coefficient at 20°C is 0.0038/°C? (1.67Ω)
5. A relay coil has a resistance of 120Ω at 20°C. If it is known that the temperature coefficient of the relay coil is 0.0042, what will its resistance be at a temperature of 56°C? (138.14Ω).