Practical Wireless

A Lab Tutorial

Jeff continues to coach Natalie in some basic circuit theory.

- Chris Murphy M0HLS practicalw­ireless@warnersgro­up.co.uk

It was Thursday morning in the electronic­s lab and one of the senior engineers, Jeff, was sat looking at last night’s football results in his paper when he was startled by a cheery voice behind him. “Morning Jeff”. Looking round he saw the greeting was from the lab’s apprentice Natalie. “Morning Natalie”

he replied. “Good day at college?” “Yes thanks”, said Natalie. “Can you go through the electronic­s stuff with me at lunchtime like you did last week please”. “Yes of course, what is it this week”. “Oh, more DC stuff”, said Natalie. “Voltage dividers and Kirchhoff”. “Right” said Jeff “I’ll look forward to it”.

Lunchtime came and after they’d both eaten their lunches Jeff wandered over to Natalie’s desk where she was already engrossed in her college work. “Hi”,

she said, “I was just looking at voltage dividers”. “Well, they’re normally fairly straightfo­rward”, said Jeff. “Just take Ohms Law a step further really. Have you got an example?” “Yes”, said Natalie. “There’s one here, look”, Fig 1.

“OK then,” said Jeff, “Explain to me how you find the output voltage”. “Well”, she

said, “One way is to find the total resistance by adding the values of R1 and R2 together. Then you divide the voltage Vin by the total resistance to find the current flowing. You then use Ohms Law again to find the voltage across R2”. “Fair enough”, said Jeff, “but there is an easier way”. “Do you mean by using this formula”, said Natalie interrupti­ng Jeff and wrote down an equation on her notepad:

V = V × (R2/R1 + R2)

“Yes, that’s right. It’s a well-known formula for finding the output from a voltage divider that’s fairly standard in textbooks. In fact, it is in the Radio Amateurs exam syllabus as well. You have to be careful though”, Jeff

warned, “That it only holds true if any load that you connect to the output has a high resistance compared to R2 and therefore draws a low current”. “Why is that?”

asked Natalie. “Because”, Jeff explained, “Anything that you connect across R2 will be in parallel with it and the effective resistance will be the result of R2 and the load resistance in parallel. The best way to explain is by doing a few examples”.

“Let’s say”, said Jeff, “That the input voltage is 10V and R1 is 7kΩ and R2 3kΩ. Using the formula that you wrote the output voltage across R2 will be 3V. And since the total resistance of R1 and R2 is 10kΩ the current flowing through both resistors will be 1mA”. “Yes, I follow that”, said Natalie.

“Now”, said Jeff, “let’s say that we connect a load with a resistance of 100kΩ across R2. Do you know how to calculate what the resulting resistance, let’s call it Rparallel, will be?” “Yes”, said Natalie and wrote down another equation (below). “And if we use the values of resistance in kilohms it works out as 2.91kΩ.”

R = (R2 × R ) / (R2 + R ) = p load load

(3 × 100) / (3 + 100) = 2.91kΩ

“So”, said Jeff, “If we now use the voltage divider formula again but with R2 as being 2.9kΩ, we find that the output voltage is now 2.94V − not too much lower than the original 3V”. “Now”, said Jeff, “Let’s see what happens if we connect a load resistance of, say, 500Ω across R2. Using your formula for resistors in parallel the resultant parallel resistance will be about 493Ω, and putting this value into the voltage divider formula we now find that the output voltage is only about 0.6V − much less than our original value”. “Oh”, mused Natalie, “Is there any way that you can keep the output voltage constant”. “Yes, and you’ll no doubt learn how to do that when you learn about power supplies”, said Jeff.

“Incidental­ly, did you also learn about current dividers?” Jeff asked. “No”, replied

Natalie, “do they work in a similar way”. “Sort of”, said Jeff, “Except that whereas a voltage divider consists of resistors in series, a current divider has resistors in parallel”. “In fact”, Jeff said, “what we’ve just looked at with adding a load to a voltage divider is a good starting point for explaining how a current divider works”.

“If you remember we calculated the current through the voltage divider with no load to be 1mA”, said Jeff. “The current with the 100kΩ load won’t be much different, but let’s look at what happens with the 500Ω load. We calculated the resistance of R2 and 500Ω in parallel to be about 430Ω, so the total resistance across the 10V supply will be 7.43kΩ”. “Yes, I follow that”, said Natalie,

“R1 remains the same at 7kΩ”. “Correct”,

said Jeff, “and if we now calculate the new current, we find it to be 10 divided by 7.43kΩ, which is about 1.35mA”.

“Now”, said Jeff, “All of the current will flow through R1 but let’s look at what happens with R2 and the load. Here, some of the current will flow through R2 and some through the 500Ω load”, Jeff

explained. “Let’s call them IR2 and Iload and the total current Itotal like this”, said Jeff and sketched the circuit, Fig. 2. “So”, said

Jeff, “The total current will divide between R2 and the load and the currents in each will add up to the total current”. “Hang on a minute”, interrupte­d Natalie, “We did cover something like this. It was called Kirchhoff’s current law”. “Yes”, said Jeff, “It’s the same thing really but with the current divider we can work out the currents from the values

Fig. 1: Voltage divider circuit.

Fig. 2: Current divider circuit.

Fig. 3: An illustrati­on of Kirchoff’s Law. Fig. 4: Kirchoff worked example. Fig. 5: Second class example.

of the resistors. Going back to our example here we can write the equation total current equals the current through R2 plus the current through the load”:

Itotal = IR2 + Iload

“To find the current in each leg from the values of the resistors”, Jeff explained, “We can use a formula similar to the one for voltage divider. I’ll write it down for you”:

IR2 = Itotal × (Rload / (R2 + Rload)) I = I × (R2 / (R2 + R ))

“So”, said Jeff, “Let’s work out the currents in R2 and Rload. We know the values of the resistors and that the total current is 1.35mA. Let’s put the numbers into the formulas”. “OK”, said Natalie and wrote down the formulas with the numbers in them:

IR2 = 0.00135 × 500/(3000 + 500) = 0.000193A or 193wA and

Iload = 0.00135 × 3000/(3000 + 500) = 0.00116A or 1.16mA

“They look good to me”, said Jeff, “And if you add them together, you’ll find that you have 1.35mA, which is of course the total current.” “OK”, said Natalie. “I notice that to find the current in one branch, the resistance of the other branch goes on the top line of the formula to work it out”. “Yes, that’s right”, said Jeff. “Did you say that you’d done Kirchhoff as well?”

“Yes, sort of”, said Natalie. “Archie said that there are two Kirchhoff laws, one to do with current and the other voltage. We covered the current law this week and we’ll do the voltage law next week.” “I see”,

mused Jeff. “What do you know about the current law then?” “Well, it’s similar to what we’ve just been talking about – currents in different parts of a circuit”, said

Natalie. “Kirchhoff’s current law says that the algebraic sum of all the currents at a junction in a circuit (Archie calls a circuit a network by the way) add up to zero”.

“Yes, that’s right”, said Jeff. “Actually, Kirchhoff talks about the conservati­on of charge. Current is the amount of charge in a given time – remember that problem Archie gave you last week. So, if no charge can escape, then it follows that what goes into a junction in a given time must also come out. A junction in a circuit, or network, by the way is called a node. Anyway, from the talk about charge we can deduce that any current that flows into a node must also come out”. “Yes”, said Natalie. “Archie used this example”, and drew a network, Fig 3. As shown, I1 + I3 = I2 + I4 so I1 + I2 +I3 + I4 = 0.

“OK”, said Jeff. “To solve the algebraic sum, we need to denote some currents as positive and some as negative. It doesn’t really matter which as the calculatio­ns will show whether or not they’ve been given the right sign. If in the example that Archie gave you, we say that I1 and I3 are positive and that I2 and I4 are negative, then we can write the following”, and Jeff wrote down another equation:

I1 + (−I2) + I3 + (−I4) = 0, then I1 + I3 – I2 – I4 = 0

“Let’s give them some values to show how it works. Let’s say that I1 is 3A, I2 is 2A, I3 5A, and I4 6A, then we have”

3 + (−2) + 5 + (−6), which is 3 + 5 – 2 – 6 = 0

“So, we have 8A flowing into the node and 8A flowing out”, said Jeff. “Did Archie give you any examples where you had to find what a current flowing into or out of a node was?” asked Jeff. “Yes”, said Natalie. “This one, for example”, Fig. 4.

“Time’s nearly up”, said Natalie looking

at her watch. “Can you write me a few questions about what we’ve talked about?” “Yes, of course I can”, said Jeff. “Did Archie give you some as well?” “Yes, they’re here, look”, said Natalie handing Jeff a list. “Actually”, said Jeff, “I thought that you’d ask me to write some questions for you and that you said that you’d be looking at voltage dividers and Kirchhoff this week so I dug out a couple of questions from my college days years ago”. “Blimey”, said Natalie, “Are they still valid?” “Electrical theory doesn’t change with time”, said Jeff with a laugh. “In the days before digital multimeter­s we had to use analogue meters and some, especially the cheaper ones, had fairly low internal resistance­s so the loading effect of the meter had to be taken into account when measuring voltages. If you look at an old circuit diagram, it will often tell you the resistance of the meter that any voltage measuremen­ts were made with. Putting the meter across the resistor just places a load across it the same as another resistor would”.

“Hey Jeff”, said Natalie looking at Jeff’s

questions. “Question 5, what are those squiggly lines?” “Those”, said Jeff, “Are resistors. That was the old circuit symbol before we started using rectangula­r boxes”. “Oh”, replied Natalie, “But how am I supposed the find the current in them when I don’t know their value or the voltage across them?” “The diagram tells you everything that you need to know”, said Jeff. “Just use what you have learned over the past two weeks”.

“Do these theories actually get used for anything”, Natalie asked. “Archie said that next week we’d be looking at a couple of other network theorems as he calls them as well as Kirchoff’s voltage law”. “Oh yes”,

Jeff replied, “The basic network theorems can be used to simplify quite complicate­d circuits or networks that contain a lot a resistors in series and parallel along with multiple voltage or current sources into a

simple representa­tion that often contains a single resistance and voltage or current source. Don’t neglect them just because you can’t see an obvious use for them at the moment”. “OK”, Natalie replied, “I’ll look forward to them”.

Class Examples

1. If in the voltage divider in Fig. 1, R1 = 20kΩ and R2 = 3.3kΩ, what will VOut be if Vin is 30V? (4.25V)

2. If R1 = 100kΩ and R2 = 56kΩ, what will VOut be if Vin is 230V? (82.56V)

3. If R1 = 820Ω and R2 = 120Ω, what input voltage is required to produce an output voltage of 3V? (23.44)

4. In the network shown in Fig. 5, calculate

I4. (2A)

5. What is the total current required to supply three circuits that require the currents as shown in Fig. 6? (2.7A)

Further Examples

For voltage divider questions refer back once more to Fig. 1.

1. If R1 = 100kΩ and R2 = 56kΩ, calculate the voltage across R2 if Vin = 18V. (6.46V)

2. If R1 = 10MΩ and R2 = 820kΩ, what is the voltage across R2 if Vin = 500V? (37.9V)

3. If both R1 and R2 = 470Ω, what is the voltage across R2 if Vin = 6mV? (3mV)

4. Calculate Ix in the network shown in Fig. 7. (7A)

5. Calculate I in the network shown in in

Fig. 8. (6A)

Jeff’s Questions

1. For the potential divider as shown in Fig. 1, if R1 = 200kΩ, R2 = 47kΩ and Vin is 50V, calculate the voltage across R2 with no load connected. If a meter with a resistance of 100kΩ is used to measure the voltage across R2, what voltage will it indicate? (9.5V and 6.9V)

2. What voltage do you need to apply to a voltage divider consisting of a 56kΩ resistor and a 3.3kΩ resistor for a voltage of 6V across the 3.3kΩ resistor? (108V)

3. An electrical distributi­on network for three marquees at an amateur radio event, Fig. 9, has two electrical supplies with currents of 100A and 160A flowing into it. The combined currents flow along a cable, X-Y, of resistance 0.4Ω. It then supplies three circuits as shown below. Find:

A: The current flowing in the cable between X and Y. (260A)

B: The current IX. (40A) C: The voltage drop along the cable X-Y. (10.4V)

4. Find the current in each of the resistors

in Fig. 10. (27.5mA and 22.5mA)

Fig. 6: Third class example.

Fig. 7: Fourth class example.

Fig. 8: Fifth class example.

Fig. 9: The first of Jeff’s questions. Fig. 10: The second of Jeff’s questions. Fig. 11: Jeff’s final question.

5. In the network shown in Fig. 11, all resistors have the same value. Find the current flowing in each resistor. (R1, R2, R5 1/3A, R3, R4, R6, R8, R9, R10 1/6A, R7, R11, R12 1/3A)

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