Learning the Art of Electronics 2ed/1W Worked Examples: DC circuits

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This is a chapter of Learning the Art of Electronics 2ed.

1W Worked Examples: DC circuits

1W.1 Design a voltmeter, current meter

• you have a 100 µA meter movement which has an internal resistance of 2.25 kΩ
  • what shunt resistor would you need to make a 1 A ammeter?
  • what series resistor would you need to make a 10 V voltmeter?
• in addition to giving you some experience with ammeters and voltmeters, this exercise also helps build your intuition for when to use approximations

See: AoE 1.2.5, Multimeters, ex 1.8

1A ammeter

See: Learning the Art of Electronics: 1W.1 Designing an Ammeter.

"100 µA meter movement" means that the meter will read full scale (1 A) when 100 µA flows through it. To achieve this, we need to shunt the remaining current (1 A - 100 µA = 0.9999 A) around the meter movement using a shunt resistor.

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Figure 1W.1 Shunt resistance allows sensitive meter movement to measure total current of 1 A.

In addition to understanding the currents involved we know the resistance of the meter too, being 2.25 kΩ. We can use this information to find the voltage across the meter movement when it is at full scale:

$V=I\cdot R=100\mu A\cdot 2.25k\mathrm{\Omega }=0.225V$

Now that we know current and voltage we can choose R_shunt. The voltage is 0.225 V, and the current through the shunt resistor is 0.9999 A, being 1 A total minus the 100 µA through the meter movement, but we can approximate this as 1 A since the difference is negligible. So instead of:

$R=\frac{V}{I}=\frac{0.225V}{0.9999A}=0.2250025002\mathrm{\Omega }$

we can just say:

$R=\frac{V}{I}=\frac{0.225V}{1A}=0.225\mathrm{\Omega }$

10V voltmeter

• in this exercise we build a 10 V voltmeter from a 100 µA meter movement with an internal resistance of 2.25 kΩ
• we need to add a series resistor to the meter movement to make it read 10 V at full scale

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Figure 1W.2 Voltmeter requires added series resistance.

• V_total = 10 V
• R_m = 2.25 kΩ
• I_m = I_s = 100 µA
• V_m = I_m * R_m = 100 µA * 2.25 kΩ = 0.225 V

• for the voltmeter, we want 100 µA to flow through the meter movement when the voltage across it is 10 V
• we can use Ohm's law to find the total resistance needed in either of two ways:
  1. R_s must drop 9.775 V out of 10 V when 100 µA flows through it, so R_s = V_s / I_s = 9.775 V / 100 µA = 97.75 kΩ
  2. R_total must drop 10 V when 100 µA flows through it, so R_total = V_total / I_total = 10 V / 100 µA = 100 kΩ, and since R_total = R_s + R_m, we can find R_s = R_total - R_m = 100 kΩ - 2.25 kΩ = 97.75 kΩ
• either way we find that R_s = 97.75 kΩ, which we can approximate as 100 kΩ from our 10% resistor series
  • or better we could pick a 1% tolerance resistor of 97.6 kΩ from the E96 series, which is a more accurate choice than the 100 kΩ from the 10% series

1W.2 Resistor power dissipation

References:

• AoE 1.2.2.C Power in resistors, ex 1.5 and 1.6

As a reminder here are the 10% resistor values from the E12 series:

• 10 Ω | 12 Ω | 15 Ω | 18 Ω | 22 Ω | 27 Ω | 33 Ω | 39 Ω | 47 Ω | 56 Ω | 68 Ω | 82 Ω | 100 Ω...

Problem: What is the smallest 1/4 W resistor one should put across a 5 V supply?

Solution:'

$P=\frac{V^2}{R}$

$R=\frac{V^2}{P}$

$R=\frac{5^2}{1/4}=\frac{25}{0.25}=100\mathrm{\Omega }$

So the smallest 1/4 W resistor one should put across a 5 V supply is 100 Ω.

Problem: What is the lowest value surface-mount 1/8 W resistor one should put between -15 V and +15 V?

Solution:

$P=\frac{V^2}{R}$

$R=\frac{V^2}{P}$

$R=\frac{(15-(-15){)}^2}{1/8}=\frac{900}{0.125}=7200\mathrm{\Omega }$

So the lowest value surface-mount 1/8 W ES12 resistor one should put between -15 V and +15 V is 8.2 kΩ.

Problem: What is the maximum voltage one ought to put across a 10 W, 10 Ω power resistor?

Solution:

$P=\frac{V^2}{R}$

$V=\sqrt{P\times R}$

$V=\sqrt{10\text{ W}\times 10\text{ Ω}}=\sqrt{100}=10\text{ V}$

So the maximum voltage one ought to put across a 10 W, 10 Ω power resistor is 10 V.

Problem: Electric power is transmitted at very high voltages to reduce power loss in the transmission lines. By what factor are long-distance power line losses reduced if the power company manages to raise its transmission voltage from 100,000 volts to 1 million volts? (We assume the power company delivers the same amount of power in either case, and that the resistance of the transmission line is unchanged.)

Solution:

Power losses are proportional to $V\times I$ in the line, or, equivalently, to $I^2\times R$, where R is the resistance of a unit length of the line.

Stepping up the voltage by a factor of 10 steps down the current by the same factor (while transmitting a given level of power). Since power loss is proportional to the square of the current, the power dissipated will fall by $(1/10{)}^2=1/100$. A big reward. The change sounds worthwhile, but the high voltages are troublesome to insulate, so 1 MV is about the practical limit.

Problem: Why does high-voltage transmission militate strongly against Thomas Edison's program of DC power transmission?

Solution:

AC transmission is favored, for most purposes, because stepping AC voltages up and down is easy, requiring only transformers. Stepping DC voltages up and down is more difficult, requiring more complex and expensive equipment. This was a major factor in the "War of Currents" between Edison (DC) and Tesla/Westinghouse (AC) in the late 19th century, which ultimately led to the widespread adoption of AC for power transmission. High-voltage DC transmission is possible and is used in some cases (e.g., for long-distance undersea cables), but it is generally more expensive and less efficient than AC transmission for most applications.

1W.3 Working around imperfections of instruments

• you can't use a voltmeter and ammeter on a circuit together at the same time without having one affect the other

Problem: if the analog meter movement is as described in §1W.1: what percentage error in the voltage reading results, if the voltage probe is connected as in circuit A of Figure 1W.3 below, when the load is:

• 200 Ω
• 20 kΩ
• 2 MΩ

Solution:

The question is easier than it may appear. The load resistance does not matter, the VOM drops the same voltage across itself regardless of the load, so the percentage error is the same for all three loads.

Problem: same question as above, but concerning the current measurement error, if the voltmeter is connected as in circuit C of Figure 1W.3 below, when the load is:

• 200 Ω
• 20 kΩ
• 2 MΩ

Solution:

If we move the DVM to the top of the resistor, then the voltage reading becomes correct: we are measuring what we mean to measure. But now the current meter is reading a little high: it measures not only the resistor current but also the DVM current, which flows parallel to the load, R.

The size of this error depends on the load resistance. The higher the load resistance, the smaller the current through it, and the larger the percentage error in the current reading.

• if R is 200 Ω then the current error is 200 parts in 20,000,000, or 1 part in 100,000, which is 0.001%
• if R is 20 kΩ then the current error is 20,000 parts in 20,000,000, or 1 part in 1,000, which is 0.1%
• if R is 2 MΩ then the current error is 2,000,000 parts in 20,000,000, or 1 part in 10, which is 10%

Conclusion concerning imperfections of instruments

There is no general answer to the question "which is the better way to place the DVM in this circuit?" It depends on the load resistance, the applied voltage, and the ammeter range setting.

The results depend on the ammeter range setting because of a concept we call Electronic Justice. The idea is that The Greedy Will Be Punished. If we want high resolution on the VOM you will pay a price: the meter will alter results more than if you looked for less resolution.

With a VOM or DVM as ammeter: the larger the reading, the larger the voltage error introduced; VOM as voltmeter: the larger the deflection at a given V_in the lower the input impedance.

If you want the current meter to swing nearly all the way across the dial, giving the best resolution, whereby small changes in current cause relatively large needle movement, then you will get nearly the full voltage drop across the ammeter. The same goes for the DVM as ammeter, if you understand that 'full scale' for the DVM means filling its digital range, which may be something like "3 1/2 digits" or something like that. So if you set your DVM current range so that your reading is 0.093 you have poor resolution, but if you can choose a setting that makes the same current look like 0.930 then you have improved the resolution tenfold. But you have also increased the voltage drop across the ammeter by a factor of ten; for the DVM, like the analog VOM, drops 1/4 volt full-scale, and proportionally less for smaller "deflection" (in the VOM) or smaller fractions of the full-scale range (for the DVM).

Prac for this section

In this prac we build the four circuits shown in Figure 1W.3 below, and measure the voltage and current in each case, using both a VOM and a DVM, to see how the readings differ depending on how we connect the meters. Conclusion? John is very confused. It's not clear to him how to interpret these results. We may need to revisit this experiment.

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Figure 1W.3 Using VOM and DVM by themselves or together.

Circuit	Load	VOM reading	DVM reading	VOM reading 2	DVM reading 2
A	200 Ω	100 mA	19.46 V	100 mA	19.60 V
A	20 kΩ	1 mA	19.99 V	1 mA	20.00 V
A	2 MΩ	10 µA	20.00 V	10 µA	20.00 V
B	200 Ω		19.47 V		19.62 V
B	20 kΩ		19.99 V		20.00 V
B	2 MΩ		20.00 V		20.00 V
C	200 Ω	100 mA	20.09 V
C	20 kΩ	1 mA	20.09 V
C	2 MΩ	10 µA	20.09 V
D	200 Ω	100 mA
D	20 kΩ	1 mA
D	2 MΩ	10 µA

1W.4 Thévenin models

References:

1W.5 "Looking through" a circuit fragment, and R_in, R_out

References:

1W.6 Effects of loading

References: